Answer:
5120 m/s
Explanation:
The acceleration due to gravity is:
g = MG / r²
where M is the mass of the earth, G is the universal constant of gravitation, and r is the distance from the earth's center to the object's center.
Here, r = h + R, where h is the height of the chest above the surface and R is the radius of the earth.
g = MG / (h + R)²
Acceleration is the derivative of velocity:
dv/dt = MG / (h + R)²
Using chain rule, we can say:
(dv/dh) (dh/dt) = MG / (h + R)²
(dv/dh) v = MG / (h + R)²
Separate the variables:
v dv = MG / (h + R)² dh
Integrating:
∫₀ᵛ v dv = MG ∫₀ʰ dh / (h + R)²
½ v² |₀ᵛ = -MG / (h + R) |₀ʰ
½ (v² − 0²) = -MG / (h + R) − -MG / (0 + R)
½ v² = -MG / (h + R) + MG / R
½ v² = MGh / (R(h + R))
v² = 2MGh / (R(h + R))
Given:
M = 5.98×10²⁴ kg
R = 6.37×10⁶ m
h = 1.69×10⁶ m
G = 6.67×10⁻¹¹ m³/kg/s²
Plugging in:
v² = 2 (5.98×10²⁴) (6.67×10⁻¹¹) (1.69×10⁶) / ((6.37×10⁶) (1.69×10⁶ + 6.37×10⁶))
v² = 2 (5.98) (6.67) (1.69) / ((6.37) (1.69 + 6.37)) × 10⁷
v ≈ 5120 m/s
Notice that if we had approximated g as a constant 9.8 m/s², we would have gotten an answer of:
v² = v₀² + 2a(x - x₀)
v² = (0 m/s)² + 2 (9.8 m/s²) (1.69×10⁶ m - 0 m)
v ≈ 5760 m/s
So we know that our calculated velocity of 5120 m/s is a reasonable answer.
The Newton's second law and the law of universal gravitation allows to find the result for the speed of the chest when reaching the ocean is:
The velocity is v = 5120 m / s
The law of universal gravitation is stable that the gravitational force between bodies is attractive and is proportional to the mass of the bodies and inversely proportional to the square of the distance.
[tex]F= - G \frac{Mm}{r^2}[/tex]
Where M and m are the mass of the two bodies and r is the distance.
Indicate the height from where the chest falls h= 1690 km = 1,690 10⁶ m, they also give the radius and the mass of the earth.
Newton's second law establishes a relationship between force, mass and the acceleration of bodies.
F = m a
[tex]- G \frac{Mm}{r^2} = m a \\a= - G \frac{M}{r^2}[/tex]
The distance of the body from the center of the planet is
r = R + h
The acceleration is defined as the variation of velocity with time.
[tex]a= \frac{dv}{dt} \\ \frac{dv}{dt} = - G \frac{M}{(h+R)^2}[/tex]
Let's use the chain rule
[tex]\frac{dv}{dh}\ \frac{dh}{dt} = - GM \frac{1}{(h+R)^2 }[/tex]
The velocity is the derivative of the position with respect to the time.
[tex]v=\frac{dh}{dt} \\ v \ \frac{dv}{dh} = - GM \ \frac{1}{(h+R)^2}[/tex]
To solve we use the method of separation of variables and we integrate.
[tex]\int\limits^v_0 {v} \, dv = -GM \int\limits^0_h {\frac{1}{(h+R)^2} } \, dh \\\frac{1}{2} ( v^2 - 0) = -GM (-1) [ \frac{1}{(0+R)} - \frac{1}{(h+R)}] \\\frac{1}{2} v^2 = GM \ \frac{h}{(h+R)R}[/tex]
[tex]v^2 = 2GM \ \frac{h}{(h+R) R }[/tex]
Let's calculate
v² = [tex]2 \ 6.67 \ 10^{-11} \ 5.98 \ 10^{24}} \ \frac{1.690 }{(1.609 + 6.370) 6.370} \ 10^{-6}[/tex]
v = 5120 m/s
In conclusion we use Newton's second law and the universal gravitation's law we can find the result for the speed of the chest when reaching the ocean is;
The velocity is v = 5120 m / s
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An attacker at the base of a castle wall 3.62 m high throws a rock straight up with speed 8.60 m/s at a height of 1.59 m above the ground. (a) Will the rock reach the top of the wall? Yes No (b) If so, what is the rock's speed at the top? If not, what initial speed must the rock have to reach the top? m/s (c) Find the change in the speed of a rock thrown straight down from the top of the wall at an initial speed of 8.60 m/s and moving between the same two points. m/s (d) Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upwards between the same elevations? Explain physically why or why not.
Providing a technical understanding of the results from the aspects of physics. Through the use of relevant formulae and principles, it's been deduced under the right conditions and assuming no externalities, the rock could reach the top and the change in speed for the upward and downwards thrown rocks would theoretically be the same.
Explanation:The subject of the question is related to the concepts of kinematics and energy in Physics.
(a) To determine if the rock can reach the top of the castle wall, we need to calculate the maximum possible height. By taking into account the initial speed and height of the rock, we can find out the greatest height it can reach. Using the formula for time-independent vertical motion: Δy = v₀t + 0.5gt² , where Δy is the maximum height, v₀ is the initial velocity (8.60m/s), 'g' is the acceleration due to gravity (≈9.80m/s²), and 't' is time.
(b) In order to calculate the speed of the rock at the top of the wall, we first need to determine if the rock reaches the top using the formula above. If the rock reaches the top, we can use energy considerations (similar to the 15.0 m/s rock thrown from the 20.0 m bridge) to determine the speed.
(c) When calculating the change in speed for a rock thrown straight down from the top of the wall, when the initial speed is 8.60 m/s, we can use similar calculations with respective sign conventions and constraints.
(d) The change in speed for both the upward and the downward-moving rock would theoretically be the same, assuming no other external forces (like air resistance or wind). This is primarily because the law of conservation of energy states that energy cannot be created or destroyed – only transformed.
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A spring has a natural length of 14 cm. If a 23-N force is required to keep it stretched to a length of 20 cm, how much work W is required to stretch it from 14 cm to 17 cm? (Round your answer to two decimal places.)
Answer:
1.78 J
Explanation:
Find the spring coefficient using Hooke's law:
F = k Δx
23 N = k (0.20 m − 0.14 m)
k = 383.33 N/m
The work is the change in energy:
W = PE₂ − PE₁
W = ½ kx₂² − ½ kx₁²
W = ½ k (x₂² − x₁²)
W = ½ (383.33 N/m) ((0.17 m)² − (0.14 m)²)
W = 1.78 J
A bicyclist notes that the pedal sprocket has a radius of rp = 9.5 cm while the wheel sprocket has a radius of rw = 6.5 cm. The two sprockets are connected by a chain which rotates without slipping. The bicycle wheel has a radius R = 65 cm. When pedaling the cyclist notes that the pedal rotates at one revolution every t = 1.1 s. When pedaling, the wheel sprocket and the wheel move at the same angular speed.
The angular speed of the wheel when pedaling is approximately [tex]\(3.73 \, \text{s}^{-1}\)[/tex].
Given the information about the radii and the rotations, we can analyze the relationships between the angular speeds of the sprockets and the wheel.
The angular speed of the pedal sprocket can be found using the information that it rotates at one revolution every t = 1.1 seconds.
The formula for angular speed omega is:
[tex]\[ \omega = \frac{{2\pi \text{ (revolutions)}}}{{\text{time}}} \][/tex]
For the pedal sprocket:
[tex]\[ \omega_{\text{pedal}} = \frac{{2\pi}}{{1.1 \, \text{s}}} \][/tex]
Now, knowing the relationship between the angular speeds of the wheel sprocket and the pedal sprocket:
[tex]\[\omega_{\text{wheel sprocket}} = \omega_{\text{pedal}}\][/tex]
The linear speed of a point on the wheel (where the chain makes contact) is given by the product of the angular speed and the radius:
[tex]\[ v = \omega \times r \][/tex]
Since the chain rotates without slipping, the linear speed of the chain where it contacts the wheel is equal to the linear speed of the wheel.
For the wheel sprocket:
[tex]\[ \omega_{\text{wheel sprocket}} \times r_{\text{wheel sprocket}} = \omega_{\text{wheel}} \times r_{\text{wheel}} \][/tex]
Given:
- [tex]\( r_{\text{pedal}} = 9.5 \, \text{cm} \)[/tex]
- [tex]\( r_{\text{wheel sprocket}} = 6.5 \, \text{cm} \)[/tex]
- [tex]\( R_{\text{wheel}} = 65 \, \text{cm} \)[/tex]
We have:
[tex]\[ \omega_{\text{pedal}} = \frac{{2\pi}}{{1.1 \, \text{s}}} \][/tex]
[tex]\[ \omega_{\text{wheel sprocket}} = \omega_{\text{pedal}} = \frac{{2\pi}}{{1.1 \, \text{s}}} \][/tex]
[tex]\[ \omega_{\text{wheel sprocket}} \times r_{\text{wheel sprocket}} = \omega_{\text{wheel}} \times R_{\text{wheel}} \][/tex]
Let's solve for the angular speed of the wheel.
[tex]\[ \omega_{\text{wheel sprocket}} \times r_{\text{wheel sprocket}} = \omega_{\text{wheel}} \times R_{\text{wheel}} \][/tex]
Given:
- [tex]\( r_{\text{wheel sprocket}} = 6.5 \, \text{cm} \)[/tex]
- [tex]\( R_{\text{wheel}} = 65 \, \text{cm} \)[/tex]
We found earlier that [tex]\( \omega_{\text{wheel sprocket}} = \frac{2\pi}{1.1 \, \text{s}} \)[/tex]. Substituting the known values into the equation:
[tex]\[ \frac{2\pi}{1.1 \, \text{s}} \times 6.5 \, \text{cm} = \omega_{\text{wheel}} \times 65 \, \text{cm} \][/tex]
Now, solve for [tex]\( \omega_{\text{wheel}} \)[/tex], the angular speed of the wheel:
[tex]\[ \omega_{\text{wheel}} = \frac{\frac{2\pi}{1.1 \, \text{s}} \times 6.5 \, \text{cm}}{65 \, \text{cm}} \][/tex]
[tex]\[ \omega_{\text{wheel}} = \frac{2\pi \times 6.5}{1.1 \times 65} \, \text{s}^{-1} \][/tex]
[tex]\[ \omega_{\text{wheel}} = \frac{13 \pi}{11} \, \text{s}^{-1} \][/tex]
[tex]\[ \omega_{\text{wheel}} \approx 3.73 \, \text{s}^{-1} \][/tex]
Therefore, the angular speed of the wheel when pedaling is approximately [tex]\(3.73 \, \text{s}^{-1}\)[/tex].
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Approximately 8.35 rad/s is the angular speed of the wheel sprocket in radians per second.
We'll first find the angular speed of the pedal sprocket, then use that to find the angular speed of the wheel sprocket and the bicycle wheel.
Given:
- Radius of the pedal sprocket, [tex]\( r_p = 9.5 \, \text{cm} \)[/tex]
- Radius of the wheel sprocket, [tex]\( r_w = 6.5 \, \text{cm} \)[/tex]
- Radius of the bicycle wheel, R = 65 cm
- Time taken for one revolution of the pedal, t = 1.1 s
1. Calculate the angular speed of the pedal sprocket [tex]\( \omega_p \)[/tex]:
[tex]\[ \omega_p = \frac{2\pi}{1.1 \, \text{s}} \][/tex]
[tex]\[ \omega_p \approx 5.71 \, \text{rad/s} \][/tex]
2. Calculate the linear speed of a point on the pedal sprocket:
[tex]\[ v_p = r_p \cdot \omega_p \][/tex]
[tex]\[ v_p = 9.5 \, \text{cm} \cdot 5.71 \, \text{rad/s} \][/tex]
[tex]\[ v_p \approx 54.295 \, \text{cm/s} \][/tex]
3. Find the angular speed of the wheel sprocket and the bicycle wheel:
[tex]\[ \omega_w = \frac{v_w}{r_w} = \frac{v_p}{r_w} \][/tex]
[tex]\[ \omega_w = \frac{54.295 \, \text{cm/s}}{6.5 \, \text{cm}} \][/tex]
[tex]\[ \omega_w \approx 8.35 \, \text{rad/s} \][/tex]
The angular speed of the wheel sprocket and the bicycle wheel is approximately 8.35 rad/s.
Complete Question:
A bicyclist notes that the pedal sprocket has a radius of rp = 9.5 cm while the wheel sprocket has a radius of rw = 6.5 cm. The two sprockets are connected by a chain which rotates without slipping. The bicycle wheel has a radius R = 65 cm. When pedaling the cyclist notes that the pedal rotates at one revolution every t = 1.1 s. When pedaling, the wheel sprocket and the wheel move at the same angular speed.
Calculate the angular speed of the wheel sprocket in radians per second.
Calculate the moment of inertia for a solid cylinder with a mass of 100g and a radius of 4.0 cm
Solution here,
mass of cylinder(m)=100 g
radius of cylinder(r)= 4 cm
momoent of inertia(I)=?
we have,
I=mr^2=100×4^2=1600 g/cm^2
in an eslastic ,the momentum is_______and the mechanical energy is
Answer:
In an elastic collision, the momentum is conserved and the mechanical energy is conserved too.
Explanation:
There are two types of collisions:
- Elastic collision: in an elastic collision, the total momentum before and after the collision is conserved; also, the total mechanical energy before and after the collision is conserved.
- Inelastic collision: in an inelastic collision, the total momentum before and after the colllision is conserved, while the total mechanical energy is not conserved (in fact, part of the energy is converted into other forms of energy such that thermal energy, due to the presence of frictional forces)
Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 25.9 THz . Calculate the wavelength of the infrared radiation. Round your answer to 3 significant digits.
Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light, the wavelength is 2.325 x 10¹ um.
The wavelength can be calculated by the information given:
- Frequency f = 25.9 THz = [tex]\(25.9 \times 10^{12}\)[/tex] Hz
- Speed of light c = [tex]\(3.00 \times 10^8\)[/tex] m/s
We'll use the formula [tex]\(\lambda = \frac{c}{f}\)[/tex] to calculate the wavelength ([tex]\(\lambda\)[/tex]).
Substitute the values:
[tex]\[\lambda = \dfrac{3.00 \times 10^8 \, \text{m/s}}{25.9 \times 10^{12} \, \text{Hz}}.\][/tex]
Calculate the value of [tex]\(\lambda\)[/tex]:
[tex]\[\lambda = 1.157 \times 10^{-5} \, \text{m}.\][/tex]
Now, convert the wavelength to micrometers:
[tex]\[\lambda = 1.157 \times 10^{-5} \, \text{m} \times \dfrac{10^6 \, \mu m}{1 \, \text{m}}.\][/tex]
Calculate the value in micrometers:
[tex]\[\lambda \approx 2.325 \, \mu m.\][/tex]
Thus, the wavelength is 2.325 x 10¹ um or 2.325μm.
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The wavelength of the infrared radiation emitted from young stars with a frequency of 25.9 THz can be calculated using the formula relating the speed of light, wavelength, and frequency (c = λv). By rearranging the formula to λ = c/v and substituting the values, we find that the wavelength is approximately 11.6 micrometers.
Explanation:To calculate the wavelength of the infrared radiation, we must use the formula for the relation between the speed of light, wavelength, and frequency. This formula is c = λv, where c is the speed of light, λ (lambda) is the wavelength, and v is the frequency.
The speed of light is a constant and is approximately 3.00 x 108 m/s. The frequency provided is 25.9 THz or 25.9 x 1012 Hz.
To find the wavelength, we rearrange the formula to λ = c/v. We substitute the given values to get λ = (3.00 x 108 m/s) / (25.9 x 1012 Hz). After performing the calculation, the wavelength comes out to approximately 1.16 x 10-5 meters, or 11.6 micrometers when rounded to three significant figures.
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Rosa goes into the hospital with her son.
Her son does not recall what happened so Rosa describes the following scene to the doctor.
“He was sitting on the couch watching a movie when his muscles got really tense. He started to drool and grunt and then started convulsing. He then urinated on himself. After a few seconds, he stopped convulsing and complained of being very tired.”
What most likely caused Rosa’s son to have that experience?
a stroke
a metallic taste
feeling anxious
a seizure
The answer would be a seizure because the child showed signs of seizure such as shaking drooling and urinating on himself
Final answer:
Rosa's son likely experienced a seizure, as indicated by his muscle tension, involuntary movements, and postictal fatigue. These symptoms align with common seizure characteristics rather than a stroke, anxiety, or a metallic taste.
Explanation:
Based on the symptoms described by Rosa of her son, it is most likely that her son experienced a seizure. The sudden onset of muscle tenseness, involuntary movements, drooling, grunting, convulsions, and subsequent loss of bladder control, followed by extreme fatigue, are all characteristic signs of a seizure. A seizure is a burst of uncontrolled electrical activity between brain cells that causes temporary abnormalities in muscle tone or movements, sensations, behavior, or consciousness.
These symptoms do not align with those of a stroke, which are typically more localized to one side of the body and can include speech difficulties, facial drooping, and weakness, nor do they match the symptoms of feeling anxious or experiencing a metallic taste, which are not typically associated with loss of consciousness or convulsive behavior.
The Sun has a lifetime of approximately 10 billion years. If you could determine the rate of nuclear fusion for a star with twice the mass of the Sun, which of the following would best describe how its fusion rate would compare to the Sun? Choose the best possible response to complete the sentence given below. A star with twice the mass of the Sun would have a rate of nuclear fusion that is ________ the rate of fusion in the Sun. a) less than b) a little more than c) twice d) more than twice Explain your reasoning for the choice you made.
Final answer:
A star with twice the mass of the Sun would have a nuclear fusion rate that is significantly more than twice the rate of the Sun. This occurs because the fusion rate is highly sensitive to core temperature and density, which increases more than linearly with mass. Additionally, the lifetime of such a star is inversely proportional to the square of its mass, resulting in a much shorter lifespan.
Explanation:
A star with twice the mass of the Sun would have a rate of nuclear fusion that is more than twice the rate of fusion in the Sun. This is because the rate of nuclear fusion increases with both the mass and the core temperature of the star. Since a star's core temperature and its density increase with greater mass, the rate of nuclear fusion increases exponentially, not linearly. The relationship between the temperature and the nuclear reaction rate is to the power of four. Therefore, a star with twice the mass of the Sun would have a considerably higher core temperature, leading to a significantly increased reaction rate, well over twice that of the Sun's.
To put this into perspective using the proportional relationship of mass (M), luminosity (L), and lifetime (T) of a star; for a star with twice the mass of the Sun, we use the equation T = 10¹⁰ y where M is the mass relative to the Sun, and L is the luminosity relative to the Sun. Considering that the luminosity increases dramatically with mass, the rate of fuel consumption would be much higher. Hence, while a massive star has more fuel, its higher fusion rate due to increased core temperature leads to a much shorter lifetime than less massive stars like our Sun.
Furthermore, we understand that the lifetime of a star is inversely proportional to its mass squared, as demonstrated by the worked example where the lifetime of a star with twice the Sun's mass is a quarter of the Sun's lifetime.
can someone help me?!!!!!
Answer:
Parabola
Explanation:
The motion of a projectile consists of two independent motions:
- A uniform motion along the horizontal (x) direction, with constant velocity [tex]v_x[/tex]. In fact, there are no forces acting along this direction (if we neglect air resistance), so the acceleration is zero and the velocity is constant
- An uniformly accelerated motion along the vertical (y) direction, with constant acceleration [tex]g=9.8 m/s^2[/tex] downward (acceleration due to gravity). This acceleration is due to the force of gravity that pulls the projectile downward.
The composition of these two motions gives a parabolic trajectory. In fact, the equations of the motion along the two directions are:
[tex]x(t) = v_x t[/tex] (1)
[tex]y(t) = h + v_y t - \frac{1}{2}gt^2[/tex] (2)
Solving for t in eq.(1),
[tex]t=\frac{x}{v_x}[/tex]
and substituting into (2)
[tex]y(t) = h + \frac{v_y}{v_x}x - \frac{1}{2}g(\frac{x}{v_x})^2[/tex]
which is the equation of a parabola.
A stone is thrown vertically downwards from the top of a skyscraper with an initial speed of 5.10m/s. The stone hits the street below 7.60 seconds after being thrown.How tall is the skyscraper?
Answer:
322 m
Explanation:
y = y₀ + v₀ t + ½ gt²
If we take down to be positive, then:
y = 0 + (5.10) (7.60) + ½ (9.8) (7.60)²
y = 322 m
The skyscraper is 322 m tall.
A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing experiments, one of their tasks is to demonstrate the concept of the escape speed by throwing rocks straight up at various initial speeds. With what minimum initial speed ????esc will the rocks need to be thrown in order for them never to "fall" back to the asteroid? Assume that the asteroid is approximately spherical, with an average density ????=2.02×106 g/m3 and volume ????=3.09×1012 m3 . Recall that the universal gravitational constant is ????=6.67×10−11 N·m2/kg2 .
Answer:
9.60 m/s
Explanation:
The escape speed of an object from the surface of a planet/asteroid is given by:
[tex]v=\sqrt{\frac{2GM}{R}}[/tex]
where
G is the gravitational constant
M is the mass of the planet/asteroid
R is the radius of the planet/asteroid
In this problem we have
[tex]\rho = 2.02\cdot 10^6 g/m^3[/tex] is the density of the asteroid
[tex]V=3.09\cdot 10^{12}m^3[/tex] is the volume
So the mass of the asteroid is
[tex]M=\rho V=(2.02\cdot 10^6 g/m^3)(3.09\cdot 10^{12} m^3)=6.24\cdot 10^{18} g=6.24\cdot 10^{15} kg[/tex]
The asteroid is approximately spherical, so its volume can be written as
[tex]V=\frac{4}{3}\pi R^3[/tex]
where R is the radius. Solving for R,
[tex]R=\sqrt[3]{\frac{3V}{4\pi}}=\sqrt[3]{\frac{3(3.09\cdot 10^{12} m^3)}{4\pi}}=9036 m[/tex]
Substituting M and R inside the formula of the escape speed, we find:
[tex]v=\sqrt{\frac{2(6.67\cdot 10^{-11})(6.24\cdot 10^{15})}{(9036)}}=9.60 m/s[/tex]
An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 9.0 μC flows to the positive plate. The 4.0 V battery is then disconnected and replaced with a 5.0 V battery, with the positive and negative terminals connected in the same manner as before. Part A How much additional charge flows to the positive plate? Express your answer in microcoulombs.
Answer:
[tex]2.25\mu C[/tex]
Explanation:
At the beginning, we have:
V = 4.0 V potential difference across the capacitor
[tex]Q=9.0 \mu C=9.0\cdot 10^{-6}C[/tex] charge stored on the capacitor
Therefore, we can calculate the capacitance of the capacitor:
[tex]C=\frac{Q}{V}=\frac{9.0 \cdot 10^{-6} C}{4.0 V}=2.25\cdot 10^{-6} F[/tex]
Later, the battery is replaced with another battery whose voltage is
V = 5.0 V
Since the capacitance of the capacitor does not change, we can calculate the new charge stored:
[tex]Q=CV=(2.25\cdot 10^{-6} F)(5.0 V)=11.25 \cdot 10^{-6} C=11.25 \mu C[/tex]
Since the capacitor has been connected exactly as before, we have that the charge on the positive plate has increased from [tex]9.0 \mu C[/tex] to [tex]11.25 \mu C[/tex]. Therefore, the additional charge that moved to the positive plate is
[tex]\Delta Q = 11.25 \mu C-9.0 \mu C=2.25 \mu C[/tex]
In a uncharged capacitor when connected to 5.0 V battery, the additional charge flows to the positive plate, is [tex]2.25\rm \mu C[/tex] .
What is capacitance of capacitor?The capacitance of capacitor is the ratio of the electric charge stored inside the capacitance to the potential difference. The capacitance of capacitor can be given as,
[tex]C=\dfrac{Q}{V}[/tex]
Here, (Q) is the electric charge and (V) is the potential difference.
Given information-
The potential difference of the first battery is 4.0 V.
The charge of the first battery is 9.0 μC.
The potential difference of the second battery is 5.0 V.
The capacitance of the first battery is,
[tex]C=\dfrac{9.0\times10^{-6}}{4}\\C=2.25\times10^{-6} \rm F[/tex]
Let the charge of the second battery is (q). Thus The capacitance of the first battery is,
[tex]C=\dfrac{q}{5}\\[/tex]
As the capacitance of the capacitor remain same. Thus put the value of C in the above equation as,
[tex]2.25\times10^{-6}=\dfrac{q}{5}\\q=11.25\rm \mu C[/tex]
The additional charge flows to the positive plate is the difference of the charge flows to the positive plate and second battery to the first battery. Thus,
[tex]\Delta q=11.25-9\\\Delta q=2.25\rm \mu C[/tex]
Thus the additional charge flows to the positive plate is [tex]2.25\rm \mu C[/tex] .
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6. A large balloon containing 2.00×103 m3 of helium gas at 1.00 atm and a temperature 15.0◦ C rises rapidly from ground level to an altitude where the atmospheric pressure is only 0.900 atm. Assume that the helium behaves as an ideal gas and that the balloon rises so rapidly that no heat is exchanged with the surrounding atmosphere. a. Calculate the volume of the helium gas at the higher altitude. b. Calculate the temperature of the gas at the higher altitude. c. What is the change in the internal energy of the helium as the balloon rises to the higher altitude?
Answer:
2130 m³, 3.11 °C, 1.25×10⁷ J
Explanation:
No heat is exchanged, so this is an adiabatic process. For an ideal gas, this means:
PV^((f+2)/f) = constant,
where P is pressure, V is volume, and f is degrees of freedom.
For a monatomic gas, f=3. For a diatomic gas, f=5. Since helium is monatomic, f=3.
Therefore:
PV^(5/3) = PV^(5/3)
(1.00 atm) (2.00×10³ m³)^(5/3) = (0.900 atm) V^(5/3)
V = 2130 m³
For an ideal gas in an adiabatic process, we can also say:
VT^(f/2) = constant
Therefore:
(2.00×10³ m³) (15.0 + 273.15 K)^(3/2) = (2130 m³) T^(3/2)
T = 276.26 K
T = 3.11 °C
Finally, the change in internal energy is:
ΔU = (f/2) nRΔT
We need to find the number of moles, n, using ideal gas law:
PV = nRT
n = PV/(RT)
n = (1.00 atm) (2.00×10³ m³) / ((8.21×10⁻⁵ atm m³ / mol / K) (15.0 + 273.15 K))
n = 84,500 mol
So the change in internal energy is:
ΔU = (3/2) (84,500 mol) (8.314 J/mol/K) (15.0 - 3.11) K
ΔU = 1.25×10⁷ J
Tina is driving her sports car down US1, traveling 27m/s. She sees her friend Rita up ahead, waiting to cross the street at the next intersection. Tina quickly slams down on the horn. If the horn has a frequency of 400Hz, which of the following is the only possible frequency heard by Rita, as Tina drives toward Rita?400Hz, 434Hz or 371Hz
Answer:
434 Hz
Explanation:
According to the Doppler effect, when a source of a wave is moving towards an observer at rest, then the observer will observe an apparent frequency which is higher than the original frequency of the source.
In this situation, Tina is driving towards Rita. Tina is the source of the sound wave (the horn), while RIta is the observer. Since the original frequency of the sound is 400 Hz, Rita will hear a sound with a frequency higher than this value.
The only choice which is higher than 400 Hz is 434 Hz, so this is the frequency that Rita will hear.
Answer:b
Explanation:test
If you use an extension cord, current travels from the 120 V outlet, along one wire inside the cord, through the appliance you've plugged into the cord, and back to the outlet through a second wire in the cord. The total resistance of the two wires in a light-duty extension cord is 0.44Ω; the current through such a cord should be limited to 13 A. A 13 A shop vacuum is powered using this cord. What is the voltage drop across the vacuum?
The vacuum experiences a voltage drop of 114.28 V after accounting for the 5.72 V drop due to the resistance of the extension cord.
To calculate the voltage drop across the vacuum, we can use Ohm's law, which states that Voltage (V) = Current (I) times Resistance (R). Given that the total resistance of the wires in the extension cord is 0.44Ω and the current is limited to 13 A, we first find the voltage drop across the extension cord itself:
V = I times R = 13 A times 0.44 Ω = 5.72 V.
Since the outlet provides 120 V and the voltage drop across the cord is 5.72 V, the voltage drop across the vacuum is:
120 V - 5.72 V = 114.28 V.
This voltage drop across the vacuum is how much voltage is available for the vacuum to operate with.
It is important to note that the higher the resistance of the cord or the greater the current, the larger the voltage drop will be, leading to a decrease in the appliance's performance.
How much heat is needed to raise the temperature of 8g of water by 15oC?
Explanation:
The heat (thermal energy) needed in to raise the temperature in a process can be found using the following equation:
[tex]Q=m.C.\Delta T[/tex] (1)
Where:
[tex]Q[/tex] is the heat
[tex]m=8 g[/tex] is the mass of the element (water in this case)
[tex]C[/tex] is the specific heat capacity of the material. In the case of water is [tex]C=4.184\frac{J}{g\°C}[/tex]
[tex]\Delta T=15\°C[/tex] is the variation in temperature (which is increased in this case)
Knowing this, let's rewrite (1) with these values:
[tex]Q=(8 g)(4.184\frac{J}{g\°C})(15\°C)[/tex] (2)
Finally:
[tex]Q=502.08 J[/tex]
The highest frequency that a healthy ear can typically hear is 2.0 × 104 Hz. Assume that a sound wave with this frequency travels at 346 m/s and passes through a doorway that has a width of 0.94 m. (a) Determine the angle that locates the first minimum to either side of the central maximum in the diffraction pattern for the sound. (b) Suppose that yellow light (wavelength = 569 nm, in vacuum) passes through a doorway and that the first dark fringe in its diffraction pattern is located at the angle determined in part (a). How wide would this hypothetical doorway have to be?
The first part involves calculating the angle for the first minimum in the diffraction pattern of a sound wave and we get approximately 1.05°. For the second part, we calculate that the width of a hypothetical doorway for yellow light would need to be around 30.9 micrometers to match this diffraction angle.
To answer the student’s question, we need to determine the angle for the first minimum in the diffraction pattern for sound waves passing through a doorway. The formula for the angle of the first minimum is given by:
sin(θ) = λ / a
where λ is the wavelength and 'a' is the width of the doorway.
Part (a)
Calculate the wavelength (λ) of the sound using the formula: v = fλ:Part (b)
When yellow light (wavelength = 569 nm or 569 × 10-9 m) passes through a hypothetical doorway, and the first dark fringe is at the same angle:
Using the angle θ from part (a) which is approximately 1.05°, we calculate the width of the doorway 'a' for yellow light:sin(θ) = λ / aThus, the width of the hypothetical doorway for yellow light would need to be approximately 30.9 micrometers.
During a test, a NATO surveillance radar system, operating at 37 GHz at 182 kW of power, attempts to detect an incoming stealth aircraft at 104 km. Assume that the radar beam is emitted uniformly over a hemisphere. (a) What is the intensity of the beam when the beam reaches the aircraft's location? The aircraft reflects radar waves as though it has a cross-sectional area of only 0.22 m2. (b) What is the power of the aircraft's reflection? Assume that the beam is reflected uniformly over a hemisphere. Back at the radar site, what are (c) the intensity, (d) the maximum value of the electric field vector, and (e) the rms value of the magnetic field of the reflected radar beam?
(a) [tex]2.68\cdot 10^{-6} W/m^2[/tex]
The intensity of an electromagnetic wave is given by
[tex]I=\frac{P}{A}[/tex]
where
P is the power
A is the area of the surface considered
For the waves in the problem,
[tex]P=182 kW = 1.82\cdot 10^5 W[/tex] is the power
The area is a hemisphere of radius
[tex]r=104 km=1.04\cdot 10^5 m[/tex]
so
[tex]A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2[/tex]
So, the intensity is
[tex]I=\frac{1.82\cdot 10^5 W}{6.8\cdot 10^{10}m^2}=2.68\cdot 10^{-6} W/m^2[/tex]
(b) [tex]5.9\cdot 10^{-7} W[/tex]
In this case, the area of the reflection is
[tex]A=0.22 m^2[/tex]
So, if we use the intensity of the wave that we found previously, we can calculate the power of the aircraft's reflection using the same formula:
[tex]P=IA=(2.68\cdot 10^{-6} W/m^2)(0.22 m^2)=5.9\cdot 10^{-7} W[/tex]
(c) [tex]8.7\cdot 10^{-18} W/m^2[/tex]
We said that the power of the waves reflected by the aircraft is
[tex]P=5.9\cdot 10^{-7} W[/tex]
If we assume that the reflected waves also propagate over a hemisphere of radius
[tex]r=104 km=1.04\cdot 10^5 m[/tex]
which has an area of
[tex]A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2[/tex]
Then the intensity of the reflected waves at the radar site will be
[tex]I=\frac{P}{A}=\frac{5.9\cdot 10^{-7} W}{6.8\cdot 10^{10} m^2}=8.7\cdot 10^{-18} W/m^2[/tex]
(d) [tex]8.1\cdot 10^{-8} V/m[/tex]
The intensity of a wave is related to the maximum value of the electric field by
[tex]I=\frac{1}{2}c\epsilon_0 E_0^2[/tex]
where
c is the speed of light
[tex]\epsilon_0[/tex] is the vacuum permittivity
[tex]E_0[/tex] is the maximum value of the electric field vector
Solving the equation for [tex]E_0[/tex],
[tex]E_0=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(8.7\cdot 10^{-18} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=8.1\cdot 10^{-8} V/m[/tex]
(e) [tex]1.9\cdot 10^{-16} T[/tex]
The maximum value of the magnetic field vector is given by
[tex]B_0 = \frac{E_0}{c}[/tex]
Substituting the values,
[tex]B_0 = \frac{(8.1\cdot 10^{-8} V/m)}{3\cdot 10^8 m/s}=2.7\cdot 10^{-16} T[/tex]
And the rms value of the magnetic field is given by
[tex]B_{rms} = \frac{B_0}{\sqrt{2}}=\frac{2.7\cdot 10^{-16} T}{\sqrt{2}}=1.9\cdot 10^{-16} T[/tex]
Nerve cells in your body can be electrically stimulated; a large enough change in a membrane potential triggers a nerve impulse. Certain plants work the same way. A touch to mimosa pudica, the "sensitive plant," causes the leaflets to fold inward and droop. We can trigger this response electrically as well. In one experiment, investigators placed electrodes on the thick tissue at the base of a leaf. The electrodes were 3.5 mm apart. When the electrodes were connected to a 47 μF capacitor charged to 1.5 V, this stimulated a response from the plant.Eventually, all the charge on the capacitor was transferred to the plant. How much charge was transferred?q=What was the approximate electric field between the electrodes?E=
1. [tex]70.5\cdot 10^{-6} F[/tex]
The relationship between capacitance, charge and voltage across a capacitor is:
[tex]Q=CV[/tex]
where
Q is the charge
C is the capacitance
V is the voltage
In this problem,
[tex]C=47\mu F=47\cdot 10^{-6}F[/tex] is the capacitance
V = 1.5 V
Substituting, we find the charge on the capacitor, that is eventually transferred to the plant:
[tex]Q=(47\cdot 10^{-6}F)(1.5 V)=70.5\cdot 10^{-6} F[/tex]
2. 428.6 V/m
The electric field between the electrodes is given by
[tex]E=\frac{V}{d}[/tex]
where
V = 1.5 V is the potential difference across the electrodes
d = 3.5 mm = 0.0035 m is the distance between the electrodes
Substituting into the equation, we find
[tex]E=\frac{1.5 V}{0.0035 m}=428.6 V/m[/tex]
The charge transferred to the plant is 70.5 μC, and the electric field between the electrodes is approximately 428.57 V/m.
To determine how much charge was transferred to the plant, we can use the formula for the charge stored in a capacitor: q = C * V, where C is the capacitance and V is the voltage.
Given a capacitance C of 47 μF (which is 47 x 10⁻⁶ F) and a voltage V of 1.5 V:
q = 47 * 10⁻⁶ * 1.5
Thus, the charge q is:
q = 70.5 * 10⁻⁶ C or 70.5 μC
To find the electric field between the electrodes, use the formula: E = V / d, where V is the voltage and d is the distance between the electrodes.
Given V is 1.5 V and d is 3.5 mm (which is 3.5 x 10⁻³ m):
E = 1.5 / 3.5 * 10⁻³
Thus, the electric field E is: E = 428.57 V/m
A horse of weight w presses down on the ground with a magnitude F and becomes airborne as a result. The magnitude of the force the ground exerted on the horse must have been _____.
equal to w and less than F
equal to w and equal to F
more than w and equal to F
more than w and more than F
If anybody can let me know this answer asap, it would be appreciated.
Answer: Third option
More than w and equal to F
Explanation:
If the horse presses the floor with a force F, then it follows that by Newton's third law, the floor makes a force -F on the horse.
Then if this force F is greater than the weight w of the horse, then the horse accelerates upwards and rises from the ground.
We know that after pressing the floor with a force F, the horse gets up off the floor. Then it is true that:
The force exerted by the floor on the horse is equal to F and greater than w
The answer is the third option more than w and equal to F
To become airborne, the force exerted by the ground on the horse must be greater than both the horse's weight and the force applied by the horse.
When a horse becomes airborne after pressing down on the ground, the magnitude of the force that the ground exerted on the horse must have been more than w (weight of the horse) and more than F (force applied by the horse on the ground). This is because for the horse to become airborne, the ground must exert a force greater than the horse's weight, which is the normal reaction force, to overcome gravitational pull. Additionally, the applied force F would be a combination of the horse's weight (mg) and the additional force caused by the horse's muscles to initiate the jump.
The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 X 10^3 N with an effective perpendicular lever arm of 3.10 cm, producing an angular acceleration of the forearm of 125 rad/s2. What is the moment of inertia of the boxer's forearm? kg · m2
Answer:
[tex]0.496 kg m^2[/tex]
Explanation:
The torque exerted is given by
[tex]\tau = Fd[/tex]
where
[tex]F=2.00 \cdot 10^3 N[/tex] is the force applied
d = 3.10 cm = 0.031 m is the length of the lever arm
Substituting,
[tex]\tau=(2.00\cdot 10^3 N)(0.031 m)=62 Nm[/tex]
The equivalent of Newton's second law for rotational motion is:
[tex]\tau = I \alpha[/tex]
where
[tex]\tau = 62 Nm[/tex] is the net torque
I is the moment of inertia
[tex]\alpha = 125 rad/s^2[/tex] is the angular acceleration
Solving the equation for I, we find
[tex]I=\frac{\tau}{\alpha}=\frac{62 Nm}{125 rad/s^2}=0.496 kg m^2[/tex]
Suppose your vehicle is moving 12m/s as it crosses the 7m line. It took 6 seconds to get to the 7m line. What was its acceleration?
Answer:
3.61 m/s²
Explanation:
Given:
v = 12 m/s
x = 7 m
x₀ = 0 m
t = 6 s
Find:
a
x = x₀ + vt - ½ at²
7 m = 0 m + (12 m/s) (6 s) - ½ a (6 s)²
a = 3.61 m/s²
Can someone help me??!!!!!
Answer:
III quadrant
Explanation:
First of all, let's write the components of each vector along the two directions. We have:
[tex]B_x = B cos 33^\circ = (5.6) cos 33^\circ=-4.7\\B_y = B sin 33^\circ = (5.6) sin 33^\circ=-3.0[/tex]
where we put a negative sign on both components, since they are in the negative x- and y- direction.
Similarly,
[tex]C_x = C sin 22^\circ = (4.8) sin 22^\circ=1.8\\C_y = C cos 22^\circ = (4.8) cos 22^\circ=-4.5[/tex]
where we put a negative sign on the y-direction, since it is along the negative direction.
Now we sum the components along each axis:
[tex]B_x + C_x = -4.7+1.8=-2.9\\B_y + C_y = -3.0 + (-4.5)=-7.5[/tex]
We see that both components of (B+C) are negative, so this vector must be in the 3rd quadrant.
Monochromatic light is incident on a metal surface, and electrons are ejected. If the intensity of the light increases, what will happen to the ejection rate of the electrons? Monochromatic light is incident on a metal surface, and electrons are ejected. If the intensity of the light increases, what will happen to the ejection rate of the electrons? The ejection rate will remain constant. Changes in the ejection rate cannot be determined without additional information. The ejection rate will decrease. The ejection rate will increase.
The photoelectric effect is a phenomenon that consists of the emission of electrons by certain metals when a beam of light impacts on its surface.
For this phenomenon to occur, certain conditions must be met, such as when the photon collides with the electron, in order to "pull it" from the metal, the photon must have a minimum energy equal to the ionization energy of the atom, so that the electron can leave the influence of the nucleus.
This is achieved with the adequate intensity of the incident radiation, which is related to the number of photons that impact the metal.
This means:
The greater the intensity, the greater the number of photons, hence the greater number of electrons emitted.
Increasing the intensity of monochromatic light incident on a metal surface causes an increase in the rate of electron ejection due to the larger number of photons arriving on the surface per unit time. However, the kinetic energy of the ejected electrons doesn't change with changing light intensity.
Explanation:The phenomenon described here is called the photoelectric effect, which occurs when monochromatic light is incident on a metal surface causing the ejection of electrons, also known as photoelectrons.
When the intensity of the light increases, the rate of electron ejection also increases. This is because the increased intensity of light means a larger number of photons are arriving on the surface per unit time, thus more electrons can gain energy from the photons and be ejected.
The kinetic energy of the ejected electrons, however, does not change with changing light intensity. The energy of an ejected electron equals to the energy of a single photon (which depends only on the light's frequency, not its intensity) minus the binding energy of the electron. This means, even though there are more electrons ejected per unit time due to increased intensity, they have the same kinetic energy as before.
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A vertical spring (spring constant =160 N/m) is mounted on the floor. A 0.340-kg block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. The block is not attached to the spring. (a) Obtain the frequency (in Hz) of the motion. (b) Determine the amplitude at which the block will lose contact with the spring.
(a) 3.5 Hz
The angular frequency in a spring-mass system is given by
[tex]\omega=\sqrt{\frac{k}{m}}[/tex]
where
k is the spring constant
m is the mass
Here in this problem we have
k = 160 N/m
m = 0.340 kg
So the angular frequency is
[tex]\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s[/tex]
And the frequency of the motion instead is given by:
[tex]f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz[/tex]
(b) 0.021 m
The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at
x = A
where A is the amplitude of the motion.
The maximum displacement is given by Hook's law:
[tex]F=kA[/tex]
where
F is the force applied initially to the spring, so it is equal to the weight of the block:
[tex]F=mg=(0.340 kg)(9.81 m/s^2)=3.34 N[/tex]
k = 160 N/m is the spring constant
Solving for A, we find
[tex]A=\frac{F}{k}=\frac{3.34 N}{160 N/m}=0.021 m[/tex]
a. The frequency (in Hz) of the motion is equal to 3.45 Hertz.
b. The amplitude at which the block will lose contact with the spring is 0.021 meters.
Given the following data:
Spring constant =160 N/mMass of block = 0.340 Kga. To find the frequency (in Hz) of the motion:
Since the spring initiates simple harmonic motion, we would determine its angular frequency.
Mathematically, angular frequency of a spring is given by the formula:
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
Where:
k is the spring constant.m is the mass of the spring.Substituting the given parameters into the formula, we have;
[tex]\omega = \sqrt{\frac{160}{0.340} }\\\\\omega = \sqrt{470.59}\\\\\omega = 21.69 \;rad/s[/tex]
Now, we can find the frequency of the motion by using the formula:
[tex]F = \frac{\omega}{2\pi} \\\\F = \frac{21.69}{2 \times 3.142} \\\\F = \frac{21.69}{6.284}[/tex]
Frequency, F = 3.45 Hz
b. To determine the amplitude at which the block will lose contact with the spring:
[tex]Force = kA\\\\mg = kA\\\\A = \frac{mg}{k}[/tex]
Where:
A is the amplitude.k is the spring constant.m is the mass of the spring.Substituting the parameters into the formula, we have;
[tex]A = \frac{0.340 \times 9.8}{160} \\\\A = \frac{3.332}{160}[/tex]
Amplitude, A = 0.021 meters
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Answer:
110 m
Explanation:
First of all, let's find the initial horizontal and vertical velocity of the projectile:
[tex]v_{x0}=v cos 30^{\circ}=(25 m/s)(cos 30^{\circ})=21.7 m/s[/tex]
[tex]v_{y0}=v sin 30^{\circ}=(25 m/s)(sin 30^{\circ})=12.5 m/s[/tex]
Now in order to find the time it takes for the projectile to reach the ground, we use the equation for the vertical position:
[tex]y(t)=h+v_{0y}t-\frac{1}{2}gt^2[/tex]
where
h = 65 m is the initial height
t is the time
g = 9.8 m/s^2 is the acceleration due to gravity
The time t at which the projectile reaches the ground is the time t at which y(t)=0, so we have:
[tex]0=65+12.5 t - 4.9t^2[/tex]
which has 2 solutions:
t = -2.58 s
t = 5.13 s
We discard the 1st solution since its negative: so the projectile reaches the ground after t=5.13 s.
Now we know that the projectile travels horizontally with constant speed
[tex]v_x = 21.7 m/s[/tex]
So, the horizontal distance covered (x) is
[tex]x=v_x t = (21.7 m/s)(5.13 s)=111.3 m[/tex]
So the closest option is
110 m
Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudinal load of 46500 N (10400 lbf). Assume Vf = 0.3, Vm = 0.7, Ef = 131 GPa and Em = 2.4 GPa. (a) Calculate the fiber-matrix load ratio. (b) Calculate the actual load carried by fiber phase. (c) Calculate the actual load carried by matrix phase. (d) Compute the magnitude of the stress on the fiber phase. (e) Compute the magnitude of the stress on the matrix phase. (f) What strain is expected by the composite?
(a) 23.4
The fiber-to-matrix load ratio is given by
[tex]\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}[/tex]
where
[tex]E_f = 131 GPa[/tex] is the fiber elasticity module
[tex]E_m = 2.4 GPa[/tex] is the matrix elasticity module
[tex]V_f=0.3[/tex] is the fraction of volume of the fiber
[tex]V_m=0.7[/tex] is the fraction of volume of the matrix
Substituting,
[tex]\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4[/tex] (1)
(b) 44,594 N
The longitudinal load is
F = 46500 N
And it is sum of the loads carried by the fiber phase and the matrix phase:
[tex]F=F_f + F_m[/tex] (2)
We can rewrite (1) as
[tex]F_m = \frac{F_f}{23.4}[/tex]
And inserting this into (2):
[tex]F=F_f + \frac{F_f}{23.4}[/tex]
Solving the equation, we find the actual load carried by the fiber phase:
[tex]F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N[/tex]
(c) 1,906 N
Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:
[tex]F=F_f + F_m[/tex] (2)
Using
F = 46500 N
[tex]F_f = 44594 N[/tex]
We can immediately find the actual load carried by the matrix phase:
[tex]F_m = F-F_f = 46,500 N - 44,594 N=1,906 N[/tex]
(d) 437 MPa
The cross-sectional area of the fiber phase is
[tex]A_f = A V_f[/tex]
where
[tex]A=340 mm^2=340\cdot 10^{-6}m^2[/tex] is the total cross-sectional area
Substituting [tex]V_f=0.3[/tex], we have
[tex]A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2[/tex]
And the magnitude of the stress on the fiber phase is
[tex]\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa[/tex]
(e) 8.0 MPa
The cross-sectional area of the matrix phase is
[tex]A_m = A V_m[/tex]
where
[tex]A=340 mm^2=340\cdot 10^{-6}m^2[/tex] is the total cross-sectional area
Substituting [tex]V_m=0.7[/tex], we have
[tex]A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2[/tex]
And the magnitude of the stress on the matrix phase is
[tex]\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa[/tex]
(f) [tex]3.34\cdot 10^{-3}[/tex]
The longitudinal modulus of elasticity is
[tex]E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa[/tex]
While the total stress experienced by the composite is
[tex]\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa[/tex]
So, the strain experienced by the composite is
[tex]\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}[/tex]
The fiber-matrix load ratio, actual load carried by the fiber and matrix phase, stress on the fiber and matrix phase, and the strain in the composite can be calculated using the given volume fractions and elastic moduli.
Explanation:SolutionGiven the load P = 46500N, volume fraction of fiber (Vf) = 0.3, volume fraction of matrix (Vm) = 0.7, elastic modulus of fiber (Ef) = 131 GPa = 131 * 10⁹ Pa, and elastic modulus of matrix (Em) = 2.4 Gpa = 2.4 * 10⁹ Pa, the fiber-matrix load ratio is given by the ratio of the product of the volume fraction and elastic modulus of fiber to the product of the volume fraction and elastic modulus of matrix, i.e., (Vf * Ef)/(Vm * Em). (b) The actual load carried by the fiber phase can be calculated by multiplying load P with Vf and the ratio of Ef to the sum of Vf * Ef and Vm * Em. (c) The actual load carried by the matrix phase can be calculated by subtracting the load carried by fiber from total load P. (d) The stress on the fiber phase can be calculated as the load carried by the fiber phase divided by the cross-sectional area. (e) The stress on the matrix phase can be calculated as the load carried by the matrix phase divided by the cross-sectional area. (f) The strain in the composite can be calculated by dividing the total applied stress by the equivalent stiffness of the composite material (i.e., (Vf * Ef + Vm * Em)).
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can someone help me?!!!!!
Answer:
33 m/s
Explanation:
By analyzing the vertical motion, we can find what is the time of flight of the projectile. The vertical position is
[tex]y(t) = h + v_{0y}t - \frac{1}{2}gt^2[/tex]
where
h = 20 m is the initial height
[tex]v_{0y} = 26 m/s[/tex] is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
By putting y(t)=0, we find the time t at which the projectile hits the ground:
[tex]0=20 + 26 t - 4.9t^2[/tex]
which has 2 solutions:
t = -0.7 s
t = 6.0 s
We discard the negative solution since it has no physical meaning. So, we know that the projectile hits the ground 6.0 s later after the launch.
The vertical velocity is given by
[tex]v_y (t)= v_{0y} -gt[/tex]
So we can find the vertical velocity when the projectile reaches point Q, by substituting t=6.0 s into this equation:
[tex]v_y = 26 m/s - (9.8 m/s^2)(6.0 s)=-32.8 m/s \sim -33 m/s[/tex]
and the negative sign means the direction is downward.
In which situation is the acceleration of the car negative?
A) The velocity of a car was 75 km/h over 4 hours.
B) The velocity of a car reduced from 50 km/h to 35 km/h over one minute.
C) The velocity of a car increased from 40 km/h to 75 km/h over 15 minutes.
Eliminate
D) The velocity of a car was 45 km/h at 2:00 pm and at 4:00 pm the velocity of the car was 85 km/h.
B) The velocity of the car reduced from 50k/m over one min.
Answer:
Answer b
Explanation:
An element has a half-life of 2.5 days. If it begins with 100 atoms, how many will be left after 5 days?
Answer:
25
Explanation:
Half life equation is:
A = A₀ (½)^(t / T)
where A is the final amount, A₀ is the initial amount, t is time, and T is the half life.
Given that A₀ = 100, t = 5, and T = 2.5:
A = 100 (½)^(5 / 2.5)
A = 100 (½)^2
A = 100 (¼)
A = 25
There will be 25 atoms left.
There’s 25 atoms left