Answer:
[tex]\Delta t =1.31\ s[/tex]
Explanation:
given,
coefficient of kinetic friction, μ = 0.25
Speed of sled at point A = 8.6 m/s
Speed of sled at point B = 5.4 m/s
time taken to travel from point A to B.
we know,
J = F Δ t
J is the impulse
where F is the frictional force.
t is the time.
we also know that impulse is equal to change in momentum.
[tex]J = m(v_f - v_i)[/tex]
frictional force
F = μ N
where as N is the normal force
now,
[tex]F\Delta t = m(v_f -v_i)[/tex]
[tex]\mu m g \times \Delta t = m(v_f-v_i)[/tex]
[tex]\mu g \times \Delta t = v_f-v_i[/tex]
[tex]\Delta t =\dfrac{v_f-v_i}{\mu g}[/tex]
[tex]\Delta t =\dfrac{8.6-5.4}{0.25\times 9.8}[/tex]
[tex]\Delta t =1.31\ s[/tex]
time taken to move from A to B is equal to 1.31 s
Answer:
Time taken by the sled is 1.31 s
Solution:
As per the question:
Coefficient of kinetic friction, [tex]\mu_{k} = 0.25[/tex]
Velocity at point A, [tex]v_{A} = 8.6\ m/s[/tex]
Velocity at point A, [tex]v_{B} = 5.4\ m/s[/tex]
Now,
To calculate the time taken by the sled to travel from A to B:
According to the impulse-momentum theorem, impulse and the change in the momentum of an object are equal:
Impulse, I = Change in momentum of the sled, [tex]\Delta p[/tex] (1)
[tex]I = Ft[/tex] (2)
where,
F = Force
t = time
p = momentum of the sled
Force on the sled is given by:
[tex]F = \mu_{k}N[/tex]
where
N = normal reaction force = mg
where
m = mass of the sled
g = acceleration due to gravity
Thus
[tex]F = \mu_{k}mg[/tex] (3)
Using eqn (1), (2) and (3):
[tex]\mu_{k}mgt = m\Delta v[/tex]
[tex]\mu_{k}gt = v_{A} - v_{B}[/tex]
[tex]t = \frac{v_{A} - v_{B}}{\mu_{k}g}[/tex]
[tex]t = \frac{8.6 - 5.4}{0.25\times 9.8}[/tex]
t = 1.31 s
a projectile is shot at an inclination of 45 frin tge horizontial with a speed of 250 m/s. how far will it travek ub the horizontal direction
Answer:
6250 m
Explanation:
When an object is projected into the air, the distance along the horizontal direction is called the Range.
The Range of a projectile is expressed as;
[tex]R = \frac{u^{2}sin2\alpha }{g}[/tex]
Where,
R is the range of the projectile
α is the angle of inclination with the horizontal
g is the acceleration due to gravity = 9.8 m/s ≈ 10 m/s
Given; α =45° , u = 250 m/s
[tex]R = \frac{250^{2}sin2(45)}{10}[/tex]
[tex]R = \frac{62500sin90}{10}[/tex]
[tex]R = \frac{62500}{10}[/tex]
R = 6250 m
The range is 6250 m
In a projectile motion, the given object travel 6377.55 m in the horizontal direction.
In a projectile motion, the distance of the object along the horizontal direction is called the Range.
The Range of a projectile
[tex]\bold {R = \dfrac {u^2 sin2\alpha }{g}}[/tex]
Where,
R - range of the projectile
u - initial speed = 250 m/s
α - angle of inclination with the horizontal = 45°
g - gravitational acceleration = 9.8 m/s
Put the values in the formula,
[tex]\bold {R = \dfrac {(250)^2 sin2(45) }{9.8}}\\\\\bold {R = \dfrac {62500\ sin 90 }{9.8}}\\\\\bold {R =6377.55}[/tex] Since, sin 90 = 1
Therefore, in a projectile motion, the given object travel 6377.55 m in the horizontal direction.
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Newton's law of motions worksheet answer solutions What is the mass of an object that needs a force of 6 600 N to increase its speed from rest to 107 m/s in 2.3 seconds?
Answer:
141.87 kg.
Explanation:
Deduction From Newton's second law of motion.
F = ma....................... Equation 1
Where F = Force acting on the object, m = mass of the object, a = acceleration of the object.
Making m the subject of the equation,
m = F/a .................. Equation 2
But
a = (v-u)/t............... Equation 3
Where v = final velocity, u = initial velocity, t = time.
Given: v = 107 m/s, u = 0 m/s ( fro rest), t = 2.3 s.
Substituting into equation 3
a = (107-0)/2.3
a = 107/2.3
a = 46.52 m/s².
Also Given, F = 6600 N
Substitute into equation 2
m = 6600/46.52
m = 141.87 kg.
Hence the mass of the object = 141.87 kg.
If the gap between C and the rigid wall at D is initially 0.15 mm, determine the magnitudes of the support reactions at A and D when the force P
Answer / Explanation
The question in the narrative is incomplete.
Kindly find the complete question below:
If the gap between C and the rigid wall at D is initially 0.15 mm, determine the support reactions at A and D when the force is applied. The assembly is made of A36 steel
Procedure
Recalling the the equation of equilibrium and referencing the free body diagram of the assembly,
Therefore, ∑fₓ = 0 ,
where, 20 ( 10³) - Fₐ - Fₙ = 0 --------------equation (1)
Now, recalling the compatibility equation, while utilizing the superposition method,
Therefore, δₓ - δfₓ
= 0.15 = 200(10³)(600) ÷ π/4 (0.05²)(200)(10⁹) - [ Fₐ (600) / π/4 (0.05²)(200)(10⁹) + Fₐ (600) π / 4 (0.05²)(200)(10⁹) ]
Solving this further,
We get: Fₐ = 20365.05 N
Which is equivalent to = 20.4 kN.
Now, substituting the answer (Fₐ) into equation (1)
Fₙ = 179634.95 N
= 180 kN
A block with velocity v>0 slides along the floor (with no friction). It hits an ideal spring at time t=0 (configuration #1). The spring starts to compress until the block comes to a (momentary) stop (configuration #2). (Figure 1) (Later, the spring will of course expand, pushing the block back). Here we show you some plots relating to the motion of the block and spring. You will need to identify what these plots represent. In each plot, the point we label as "1" refers to configuration #1 (when the block first comes in contact with the spring). The point we label "2" refers to configuration #2 (which is the moment the block comes to rest, with the spring fully compressed). Here, "force" refers to the x-component of the force of the spring on the block and "position" (and "velocity") refer to the x-components of the position (and velocity) of the block. In all cases, consider the origin to be (0,0); that is, the x-axis represents y=0 and the y-axis represents x=0.
Part A
Look first a t graph A. (Figure 2)
Which of the choices given could this graph represent?
1. position (x) vs. time
2. velocity (v) vs. time
3. force (F) vs. time
4. force (F) vs. position
Part B
Now look at graph B. (Figure 3)
Which of the choices given could this graph represent?
1. position (x) vs. time
2. velocity (v) vs. time
3. force (F) vs. time
4. force (F) vs. position
Part C
Next look at graph C. (Figure 4)
Which of the choices given could this graph represent?
1. position (x) vs. time
2. velocity (v) vs. time
3. force (F) vs. time
4. force (F) vs. position
Answer:
(A) position vs time
(B) Force vs position
(C) velocity vs time
Explanation:
Part A
This graph shows that the position of the block increases with time along the x-axis exponentially (that is it increases in unequal amounts in equal time intervals). This is because the velocity of the block is changing with time and as a result the position changes in unequal amounts per time
PartB
The force on the spring increases in a negative direction going from zero to a negative value. This is because the spring is being compressed from configuration 1 to 2. The force of compression on a spring is usually taken to have a negative sign and expansion to have a positive sign. So in this case force becomes increasingly negative with time.
Part C
The velocity of the block decreases from a positive nonzero value (v>0) to zero because the spring resists the motion of the block. As a result the block comes to a stop momentarily. The velocity decreases exponentially because the acceleration of the block is also changing with time since the force of the block is decreasing with time.
Thank you for reading.
A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Sun. If that star has four times the mass of our Sun, how does the orbital period of the planet compare to Earth's orbital period? A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Sun. If that star has four times the mass of our Sun, how does the orbital period of the planet compare to Earth's orbital period? The planet's orbital period will be four times Earth's orbital period. The planet's orbital period will be one-half Earth's orbital period. The planet's orbital period will be one-fourth Earth's orbital period. The planet's orbital period will be equal to Earth's orbital period. The planet's orbital period will be twice Earth's orbital period.
Answer:
The planet´s orbital period will be one-half Earth´s orbital period.
Explanation:
The planet in orbit, is subject to the attractive force from the sun, which is given by the Newton´s Universal Law of Gravitation.
At the same time, this force, is the same centripetal force, that keeps the planet in orbit (assuming to be circular), so we can put the following equation:
Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r
As we know to find out the orbital period, as it is the time needed to give a complete revolution around the sun, we can say this:
ω = 2*π / T (rad/sec), so replacing this in the expression above, we get:
Fg = Fc ⇒ G*mp*ms / r² = mp*(2*π/T)²*r
Solving for T²:
T² = (2*π)²*r³ / G*ms (1)
For the planet orbiting the sun in Andromeda, we have:
Ta² = (2*π)*r³ / G*4*ms (2)
As the radius of the orbit (distance to the sun) is the same for both planets, we can simplify it in the expression, so, if we divide both sides in (1) and (2), simplifying common terms, we finally get:
(Te / Ta)² = 4 ⇒ Te / Ta = 2 ⇒ Ta = Te/2
So, The planet's orbital period will be one-half Earth's orbital period.
The planet´s orbital period should be considered as the one-half Earth´s orbital period.
What is planet in orbit:
It is subjected to the attractive force from the sun that we called as the
Newton´s Universal Law of Gravitation.
Also, the following equation should be used
Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r
Now
ω = 2*π / T (rad/sec),
So,
Fg = Fc ⇒ G*mp*ms / r² = mp*(2*π/T)²*r
Now
T² = (2*π)²*r³ / G*ms (1)
And,
Ta² = (2*π)*r³ / G*4*ms (2)
So,
(Te / Ta)² = 4 ⇒ Te / Ta = 2 ⇒ Ta = Te/2
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Two point sources produce waves of the same wavelength that are in phase. At a point midway between the sources, you would expect to observea) constructive or destructive interference depending on the wavelength.b) destructive interferencec) no interferenced) constructive interferencee) alternating constructive and destructive interference interference
Answer:
Constructive Interference.
Explanation:
Constructive Interference.
Definition:
Two waves meet in such a way their highs(Crests) combine to form a new waves whose magnitude is the sum of magnitude of combining waves.
Since two waves have same wavelength and are in phase so when they combine they well form a way which has the magnitude equal to the sum of the magnitude of both waves.
Reasons why it is Constructive Interference:Waves
Have Same wavelengthAre in phase (Can have phase difference of 2πHave crests aligned with each other (Appear at same point)Masses are stacked on top of the block until the top of the block is level with the waterline. This requires 20 g of mass. What is the mass of the wooden block
Answer:
Mass of the wooden Block is 20g.
Explanation:
The buoyant force equation will be used here
Buoyant Force= ρ*g*1/2V Here density used is of water
m*g= ρ*g*1/2V
Simplifying the above equation
2m= ρ*V Eq-1
Also we know from the question that
ρ*V = m + 0.020 Eq-2 ( Density = (Mass+20g)/Volume )
Equating Eq-1 & Eq-2 we get
2m = m+0.020
m = 0.020kg
m = 20g
A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a friction force be required? If so, how much and in what direction?
Final answer:
Determining if a friction force is required for a car rounding a banked curve depends on the car's speed relative to the curve's ideal speed. At 95 km/h, a frictional force may be needed if this speed is not the ideal speed for the 68 m radius curve banked at 16 degrees.
Explanation:
When a 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees, we need to determine if a friction force is required when the car is traveling at 95 km/h. If the car is traveling at the correct banked curve speed, it could complete the turn without any frictional force. However, if the car travels at a speed higher or lower than this optimal speed, a frictional force will be necessary either to prevent the car from slipping outward or to prevent it from falling inward towards the center of the curve.
To find out whether a friction force is needed, we first need to calculate the ideal speed for this banked turn. This involves calculating the speed at which the components of the normal force provide enough centripetal force for the turn. The ideal speed is reached when no friction force is needed to keep the car on the path, meaning the force of gravity, the normal force, and the centripetal force are in perfect balance.
However, if the car is indeed traveling at 95 km/h, faster or slower than this ideal speed, then either a static frictional force acting upwards along the bank or a static frictional force opposite to the car's direction would be required to maintain its circular path without slipping.
A 372-g mass is attached to a spring and undergoes simple harmonic motion. Its maximum acceleration is 17.6 m/s2 , and its maximum speed is 1.75 m/s. a)Determine the angular frequency. b)Determine the amplitude. c)Determine the spring constant.
Answer with Explanation:
We are given that
Mass , m=372 g=[tex]\frac{372}{1000}=0.372 Kg[/tex]
1 kg=1000g
Maximum acceleration, a=[tex]17.6 m/s^2[/tex]
Maximum speed ,v=1.75 m/s
a.We know that
Maximum acceleration, a=[tex]A\omega^2[/tex]
Maximum speed, v=[tex]\omega A[/tex]
[tex]17.6=A\omega^2[/tex]
[tex]1.75=A\omega[/tex]
[tex]\frac{17.6}{1.75}=\frac{A\omega^2}{A\omega}=\omega[/tex]
Angular frequency,[tex]\omega=10.06 rad/s[/tex]
b.Substitute the value of angular frequency
[tex]1.75=A(10.06)[/tex]
[tex]A=\frac{1.75}{10.06}=0.17 m[/tex]
Hence, the amplitude=0.17 m
c.Spring constant,k=[tex]m\omega^2[/tex]
Using the formula
[tex]k=0.372\times (10.06)^2[/tex]
Hence, the spring constant,k=37.6 N/m
Final answer:
The angular frequency is approximately 7.686 rad/s, the amplitude is approximately 0.227 m, and the spring constant is approximately 6.5472 N/m.
Explanation:
To determine the angular frequency, we can use the formula:
ω = √(k/m)
where ω is the angular frequency in radians per second, k is the spring constant in Newtons per meter, and m is the mass in kilograms.
Given that the maximum acceleration is 17.6 m/s^2 and the mass is 372 g (or 0.372 kg), we can calculate the spring constant:
k = m * a
k = 0.372 kg * 17.6 m/s^2 = 6.5472 N/m
Now we can find the angular frequency:
ω = √(6.5472 N/m / 0.372 kg) ≈ 7.686 rad/s
To determine the amplitude, we can use the formula:
A = vmax / ω
where A is the amplitude, vmax is the maximum speed, and ω is the angular frequency.
Given that the maximum speed is 1.75 m/s and the angular frequency is 7.686 rad/s, we can calculate the amplitude:
A = 1.75 m/s / 7.686 rad/s ≈ 0.227 m
Therefore, the angular frequency is approximately 7.686 rad/s, the amplitude is approximately 0.227 m, and the spring constant is approximately 6.5472 N/m.
To get up on the roof, a person (mass 92.0 kg) places a 5.60 m aluminum ladder (mass 14.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes (in N) of the forces on the ladder at the top and bottom?
Answer:
Down
F1ₓ = 219.6N
[tex]F1_{y}[/tex] = 1038.8 N
Top
F2ₓ = 219.6 N
[tex]F2_{y}[/tex] = 0
Explanation:
For this exercise we must make a free body diagram of the ladder, see attached, then use the balance equations on each axis
Transnational Balance
X axis
F1ₓ -F2ₓ = 0
F1ₓ = F2ₓ
Y Axis
[tex]F1_{y}[/tex] - [tex]F2_{y}[/tex] - W - W_man = 0 (1)
Rotational balance
The reference system is placed at the bottom of the stairs and we can turn the anti-clockwise direction of rotation as positive
F2ₓ y - [tex]F2_{y}[/tex] x - W x - W_man x_man = 0
Let us write the data they give, the masses of the ladder (m = 14.0 kg), the mass of man (m_man = 92 kg), the center of mass of the ladder that is 2m from the bottom (the height) and the position of the man which is 3 m high
Let's look with trigonometry for distances
The angle of the stairs is
cos θ = x / L
θ = cos⁻¹ x / L
θ = cos⁻¹ 2 / 5.6
θ = 69⁰
Height y
tan 69 = y / x
y = x tan 69
y = 2 tan 69
y = 5.21 m
Distance x
tan 69 = 2 / x
x = 2 / tan 69
x = 0.7677 m
The distance x_man
x_man = 3 / tan 69
x_man = 1,152 m
They indicate that between the scalars and the support there is no friction so the vertical force at the top is zero
[tex]F2_{y}[/tex] = 0
Let's replace in the translational equilibrium equation
F2ₓ y - [tex]F2_{y}[/tex] x - W x - W_man x_man = 0
F2ₓ 5.21 -0 - 14.0 9.8 0.7677 - 92.0 9.8 1,152 = 0
F2ₓ = 1143.97 / 5.21
F2ₓ = 219.6 N
We use equation 1
[tex]F1_{y}[/tex] + 0 - W - W_man = 0
[tex]F1_{y}[/tex] = W + W_man
[tex]F1_{y}[/tex] = (m + m_man) g
[tex]F1_{y}[/tex] = (14 +92) 9.8
[tex]F1_{y}[/tex] = 1038.8 N
We can write the force on each part of the ladder
Down
F1ₓ = 219.6N
[tex]F1_{y}[/tex] = 1038.8 N
Top
F2ₓ = 219.6 N
[tex]F2_{y}[/tex] = 0
If we know the moon's position in the sky and its phase, we can estimate the ____________. In general, knowing any two of the following three things allows us to estimate the third:
Answer:
we can estimate the _time__.
The three things are;
1. Moon's position in the sky
2. The moon phase
3. The time
Explanation:
In general, knowing any two of the following three things; (the moon's position in the sky, the moon phase, and the time) will allows us to estimate the third. Yes, this statement is true because, position and phase of moon is used to determine the hour of the day and night most especially in the morning. Example of this is the determination of prayer time by Muslims community as it was the major time determinant in the past before the advent of clock or watch.
Also, if will know the time and moon position, we can determine the phase of the moon likewise using time and moon phase to know the moon position.
The different phases of the moon cause changes in the size of the moon. If we know the moon's position in the sky and its phase. we can easily estimate the time.
What is the moon phase?The moon changes shape every day. This is due to the fact that the celestial body has no light of its own and can only reflect sunlight.
Only the side of the moon facing the sun can reflect this light and seem bright. The opposite side appears black. this is a full moon.
We can only see the black section when it lies between the sun and the earth when a new moon occurs. We witness intermediate phases like a half-moon and crescent in between these two extremes.
In general, we may estimate the third by knowing any two of the following three things
(1) the moon's position in the sky
(2) the moon phase
(3) the time
Yes, this statement is correct since the moon's location and phase are utilized to define the time of day and night, particularly in the morning.
Also, if we know the time and moon position, we can figure out the moon phase by utilizing the time and moon phase to figure out the moon position.
Hence If we know the moon's position in the sky and its phase. we can easily estimate the time.
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What is the electric field 3.3 m from the center of the terminal of a Van de Graaff with a 7.20 mC charge, noting that the field is equivalent to that of a point charge at the center of the terminal?
Answer:
Electric Field is [tex]5.943801*10^6 N/C[/tex]
Explanation:
Electric Field:
It originates from positive charge and ends at negative charge.
General Formula for electric Field:
[tex]E=\frac{kq}{r^2}[/tex]
where:
k is the Coulomb Constant
q is the charge
r is the distance
Given:
q=7.20 mC
r=3.3 meters
k=[tex]8.99*10^9 N.m^2/C^2[/tex]
Find:
Electric Field=?
Solution:
[tex]E=\frac{kq}{r^2}[/tex]
[tex]E=\frac{(8.99*10^9)(7.20*10^{-3})}{3.3^2}\\E=5943801.653 N/C\\E=5.943801653*10^6 N/C[/tex]
Electric Field is [tex]5.943801*10^6 N/C[/tex]
When energy is converted from one form to another in a chemical or physical change, which of the following also changes by a measureable amount? Select the correct answer below:
a.The total mass in the system
b.The force of gravity
c.The total energy
d.None of the above
Answer:
None of the above
Explanation:
When energy is converted from one form to another in a chemical or physical change, none will change. This is due to the law of conservation of energy. It states that the total energy of the system remains constant. It only changes energy from one form of energy to another. So, the correct option is (d) "none of the above".
Answer: None of the above
Explanation:
The total mass of a system does not change during normal (non-nuclear) chemical reactions or during other processes where energy changes form. The force of gravity is constant and will not change when energy is converted. While energy can be converted between forms or exchanged between species, no additional energy can be created or removed.
You are 9.0 m from the door of your bus, behind the bus, when it pulls away with an acceleration of 1.0 m/s2. You instantly start running toward the still-open door at 5.7 m/s.
How long does it take for you to reach the open door and jump in?
What is the maximum time you can wait before starting to run and still catch the bus?
Final answer:
To determine the time it takes for you to reach the open door and jump in, we can use the equations of motion. We know that you are initially 9.0 m away from the door, and the bus is accelerating at 1.0 m/s². The final velocity of the bus is not given, so we can't find the exact time it takes for you to reach the door. However, we can find the maximum time you can wait before starting to run and still catch the bus.
Explanation:
To determine the time it takes for you to reach the open door and jump in, we can use the equations of motion. We know that you are initially 9.0 m away from the door, and the bus is accelerating at 1.0 m/s². The final velocity of the bus is not given, so we can't find the exact time it takes for you to reach the door. However, we can find the maximum time you can wait before starting to run and still catch the bus.
To find the maximum time you can wait, we need to calculate when the distance between you and the door is equal to 0. Since you are moving towards the door at a constant speed of 5.7 m/s and the bus is accelerating away from you, the distance between you and the door will decrease over time. Let's call the time you wait before starting to run as 't'.
The distance traveled by the bus can be calculated using the equation:
S = ut + (1/2)at^2
Where S is the distance traveled, u is the initial velocity which is 0 m/s, a is the acceleration which is 1.0 m/s², and t is the time.
The distance traveled by you can be calculated using the equation:
S = vt
Where S is the distance traveled, v is your constant velocity which is 5.7 m/s, and t is the time.
After the time 't', both the bus and you will be at the same position which is the door. So the total distance traveled by you and the bus will be equal. We can set up the equation:
ut + (1/2)at^2 = vt
Simplifying this equation, we get:
(1/2)at^2 - vt = 0
Since we know that you can wait a maximum of 't' seconds, we need to solve this quadratic equation for 't'.
Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac))/2a
Where a = (1/2)a, b = (-v), and c = 0
Substituting the values, we get:
t = (-(-v) ± sqrt((-v)^2 - 4(1/2)a(0)))/2(1/2)a
t = v + sqrt(v^2)/a
Now we can substitute the values of 'v' and 'a' to find the maximum time you can wait before starting to run:
t = 5.7 + sqrt((5.7)^2)/1.0
t = 5.7 + sqrt(32.49)/1.0 ≈ 5.7 + 5.7 = 11.4 seconds
Therefore, the maximum time you can wait before starting to run and still catch the bus is approximately 11.4 seconds.
Solar energy strikes earth’s atmosphere at 343 W m-2 . About 30% is reflected and the rest is absorbed
Answer:
The black body temperature of Earth T_e= 180.4 K
Explanation:
Assuming we have to find Black body temperature of the earth.
[tex]T_e =(\frac{s_o(1-\alpha)}{4\sigma})^{0.25}[/tex]
S0= solar energy striking the earth= 343 Wm^{-2}
\alpaha = 30% = 0.3
\sigma = stephan boltsman constant.= 5.67×10^{-8} Wm^{-2}K^4
[tex]T_e =(\frac{343(1-0.3)}{4\times5.67×10^{-8}})^{0.25}[/tex]
T_e= 180.4 K
One pipe has successive harmonics of 165 Hz, 275 Hz, and 385 Hz, and another pipe has successive harmonics of 165 Hz, 220 Hz, and 275 Hz. (a) Which pipe is closed at one end and which is open at both ends
Answer:
165 Hz, 220 Hz, and 275 Hz belongs to pipe open at both ends
165 Hz, 275 Hz, and 385 Hz belongs to pipe closed at one end
Explanation:
Open ended pips have harmonic frequencies that are multiple of the fundamental frequency
Find the fundamental frequency for each of the samples:
165Hz, 275Hz, 385Hz
(275-165)=110
(385-275)= 110
165 Hz, 220 Hz, and 275 Hz
(220-165)=55
(275-220)=55
F= 55
Note that 165 =3f
220=4f
275=5f
SO these frequencies are multiples of the fundamental frequency
Assume that you have a rectangular tank with its top at ground level. The length and width of the top are 14 feet and 7 feet, respectively, and the bottom of the tank is 15 feet beneath ground level. How much work W does it take to empty the tank by pumping the liquid back to ground level once the tank is full?
To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of [tex] 62 lb / ft ^ 3 [/tex] (Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as
[tex]W = \gamma A * \int_0^15 dy[/tex]
Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing
[tex]W = (62)(14*7)\int^{15}_0 (15-y)dy[/tex]
[tex]W = (14*7*62)\big [15y-\frac{y^2}{2}\big ]^{15}_0[/tex]
[tex]W = (14*7*62)[15(15)-\frac{(15)^2}{2}][/tex]
[tex]W = 683550ft-lbs[/tex]
Therefore the total work in the system is [tex]683550ft-lbs[/tex]
A small segment of wire contains 10 nC of charge. The segment is shrunk to one-third of its original length. A proton is very far from the wire. What is the ratio Ff/Fi of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk?
The ratio of the electric force on the proton after the wire segment is shrunk to three times its original length to the force before the segment was shrunk is 3.
The electric force between a point charge and a segment of wire with a distributed charge is given by Coulomb's law.
The formula for the electric force on a point charge q due to a segment of wire with charge Q distributed along its length L is:
[tex]F=\frac{k.q.Q}{L}[/tex]
where:
F is the electric force on the point charge,
k is Coulomb's constant ( 8.988 × 1 0⁹ Nm²/ C²),
q is the charge of the point charge,
Q is the charge distributed along the wire segment, and
L is the length of the wire segment.
When the wire segment is shrunk to one-third of its original length, the new length becomes 1/3 L.
The charge distribution remains the same, only the length changes.
So, the new electric force [tex]F_f[/tex] on the proton after the segment is shrunk becomes:
[tex]F_f=\frac{k.q.Q}{\frac{1}{3}L}[/tex]
The original electric force [tex]F_i[/tex] on the proton before the segment was shrunk is:
[tex]F_i = \frac{k.q.Q}{L}[/tex]
let's find the ratio [tex]\frac{F_f}{F_i}[/tex] :
[tex]\frac{F_f}{F_i}=\frac{\frac{k.q.Q}{\frac{1}{3}L}}{\frac{k.q.Q}{L}}[/tex]
[tex]\frac{F_f}{F_i}=3[/tex]
Hence, the ratio of the electric force on the proton after the wire segment is shrunk to the force before the segment was shrunk is 3.
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The ratio of the electric force on the proton after the wire segment is shrunk is equal to the ratio of their charges.
Explanation:The ratio of the electric force on the proton after the wire segment is shrunk to the force before the segment was shrunk can be found using Coulomb's law. Coulomb's law states that the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
In this case, the charges involved are the charge of the wire segment and the charge of the proton. Since the wire segment contains 10 nC of charge, we can consider it as one of the charged objects. The proton is very far from the wire, so we can assume that the distance between them remains the same before and after the wire segment is shrunk. Therefore, the ratio of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk is equal to the ratio of their charges.
Let's assume that the initial force on the proton is Fi and the final force on the proton is Ff. Using the given information, we have:
Fi = k(q1 * q2) / r^2
where k is the electrostatic constant, q1 and q2 are the charges of the wire segment and the proton respectively, and r is the distance between them.
After the wire segment is shrunk to one-third of its original length, the charge of the wire segment remains the same and the distance between the wire segment and the proton also remains the same. Therefore, the ratio Ff/Fi can be calculated as:
Ff/Fi = (q1 * q2) / (q1 * q2) = 1
A neutron star is the remnant left after certain supernovae (explosions of giant stars). Typically, neutron stars are about 17 km in diameter and have around the same mass as our sun. What is a typical neutron star density in g/cm3
To solve this concept we will apply the mathematical equations concerning the calculation of Volume in a sphere and the relation of density as a function of mass and volume, that is
The volume of the neutron star is
[tex]V = \frac{4\pi }{3}R^3[/tex]
[tex]V = \frac{4 \pi}{3} (\frac{17*10^{5}cm}{2})^3[/tex]
[tex]V = 25.72^{17}cm^3[/tex]
Now the density of the neutron star is
[tex]\rho = \frac{M}{V}[/tex]
[tex]\rho = \frac{1.989*10^{30}kg(\frac{10^3g}{1kg})}{25.72*10^{17}cm^3}[/tex]
[tex]\rho = 7.733*10^{14}g/cm^3[/tex]
Therefore the density of the neutron star is [tex]\rho = 7.733*10^{14}g/cm^3[/tex]
The pressure rise p associated with wind hitting a window of a building can be
estimated using the formula p = rho(V2
/2), where rho is density of air and V is the speed of the
wind. Apply the grid method to calculate pressure rise for rho = 1.2 kg/m3
and V = 100 km/h.
(20%)
a. Express your answer in pascals.
b. Express your answer in meters of water column (m-H2O).
Answer:
a)P=462.70 Pa
b)h = 0.047 m of water
Explanation:
Given that
Pressure ,[tex]P=\dfrac{1}{2}\rho V^2[/tex]
[tex]\rho = 1.2\ kg/m^3[/tex]
V= 100 km/h
[tex]V=100\times \dfrac{1000}{3600}\ m/s[/tex]
V=27.77 m/s
The pressure P
[tex]P=\dfrac{1}{2}\rho V^2[/tex]
[tex]P=\dfrac{1}{2}\times 1.2\times 27.77^2\ Pa[/tex]
P=462.70 Pa
We know that density of the water [tex]\rho=1000\ kg/m^3[/tex]
Lets height of the water column = h m
We know that
[tex]_P=\rho _w g h[/tex]
462.70 = 1000 x 9.81 h
[tex]h=\dfrac{462.7}{1000\times 9.81}\ m[/tex]
h = 0.047 m of water
a)P=462.70 Pa
b)h = 0.047 m of water
a)The presuure rise will be P=462.70 Pa
b) The height of the water column h = 0.047 m of water
What will be the pressure rise and the height of the water column of the fluid?It is given that
Pressure,
[tex]p= \dfrac{1}{2} \rho v^2[/tex]
Here [tex]\rho =1.2 \ \dfrac{kg}{m^3}[/tex]
[tex]V=100\ \frac{km}{h} =\dfrac{100\times 1000}{3600} =27.77 \ \dfrac{m}{s}[/tex]
Now to calculate the pressure P
[tex]P=\dfrac{1}{2} \rho v^2[/tex]
[tex]P= \dfrac{1}{2}\times 1.2\times (27.77)^2[/tex]
[tex]P=462.70 \ \frac{N}{m^2}[/tex]
As we know that the density of water
[tex]\rho = 1000\ \frac{kg}{m^3}[/tex]
Lets height of the water column = [tex]h_m[/tex]
As We know that
[tex]P= \rho_w gh[/tex]
[tex]462.70=1000\times 9.81\times h_m[/tex]
[tex]h_m=0.047\ m \ of \ water[/tex]
Thus
a)The presuure rise will be P=462.70 Pa
b) The height of the water column h = 0.047 m of water
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If the distance between two charges is doubled, by what factor is the magnitude of the electric force changed? F_e final/F_e, initial =____
To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,
[tex]F_{initial} = \frac{kq_1q_2}{r^2}[/tex]
Here,
k = Coulomb's constant
[tex]q_{1,2}[/tex] = Charge at each object
r = Distance between them
As the distance is doubled so,
[tex]F_{final} = \frac{kq_1q_2}{( 2r )^2}[/tex]
[tex]F_{final} = \frac{ kq_1q_2}{ 4r^2}[/tex]
[tex]F_{final} = \frac{1}{4} \frac{ kq_1q_2}{r^2}[/tex]
[tex]F_{final} = \frac{1}{4} F_{initial}[/tex]
[tex]\frac{F_{final}}{ F_{initial}} = \frac{1}{4}[/tex]
Therefore the factor is 1/4
Which of the following is a small electronic component made up of transistors (tiny switches) and other miniaturized parts?
a.
Peripheral
b.
Integrated circuit (IC)
c.
Tablet
d.
Mouse
e.
Vacuum tube
Answer:
Integrated circuit (IC)
Explanation:
An integrated circuit ( IC ) is a semiconductor which contains multiple electronic components
interconnected to form a complete electronic function. Integrated circuits are the most essential part of all electronic products.
Modern integrated circuits contain as much as billions of circuit components such as transistors , diodes , resistors , and capacitors
onto a single monolithic die .
The correct answer is option b. An Integrated Circuit (IC), or microchip, is a small electronic component consisting of transistors and miniaturized parts etched onto a piece of silicon, pivotal in the development and miniaturization of electronic devices
The small electronic component made up of transistors (tiny switches) and other miniaturized parts is called an Integrated Circuit (IC). An Integrated Circuit, sometimes referred to as a microchip, is an electronic circuit of transistors etched onto a small piece of silicon. This technology allows for complex circuitry to be compacted into a tiny chip, which is crucial for the functionality of modern electronic devices like computers and cell phones. The invention of the IC was pivotal in launching the modern computer revolution because it significantly reduced the size and complexity of electronic devices, replacing bulky vacuum tubes and complicated wiring with a compact, efficient solution.
Integrated Circuits are designed to handle both analog and digital signals, but in the realm of digital electronics, they are essential for managing binary code, the series of ones and zeroes that computers use to process data. This is achieved through the hundreds, thousands, or even millions of transistors that can act as on-off switches within a single IC. The miniaturization and efficiency of ICs have been fundamental in advancing the technological capabilities of electronic devices, making them smaller, faster, and more accessible to the general public.
A rocket is launched upward with a constant acceleration of 165 m/s^2. After 8.00 seconds of ascension a passenger on the rocket drops a rock out of it. How long does it take for this rock to hit the ground?
To solve this problem we will apply the concepts related to the linear kinematic movement. We will start by finding the speed of the body from time and the acceleration given.
Through the position equations we will calculate the distance traveled.
Finally, using this same position relationship and considering the previously found speed, we can determine the time to reach your goal.
For time (t) and acceleration (a) we have to,
[tex]t = 8s, a = 165m/s^2[/tex]
The velocity would be,
[tex]u = a*t \\u = 165*8\\u = 1320m/s[/tex]
Now the position is,
[tex]h= \frac{1}{2} at^2[/tex]
[tex]h = \frac{1}{2} 165*8^2[/tex]
[tex]h = 5280m[/tex]
Now with the initial speed and position found we will have the time is,
[tex]h=ut +\frac{1}{2} at^2[/tex]
[tex]-5280=1320t - \frac{1}{2} 9.8t^2[/tex]
[tex]4.9t^2-1320t-5280=0[/tex]
Solving the polynomian we have,
[tex]t = 273.33s = 4.56minutes[/tex]
Therefore the rocket will take to hit the ground around to 4.56min
First, we need to determine the velocity of the rocket at the time the rocket is dropped after 8 seconds of powered ascent.
Given the constant acceleration of 165 m/s^2, the velocity (v) at 8 seconds can be found using the formula
v = at
where 'a' is the acceleration and 't' is the time.
Therefore, v = 165 m/s^2 x 8 s = 1320 m/s.
Now that the rocket is dropped, it will initially have the velocity of the rocket at that instant, which is 1320 m/s upward. To find out how long it takes for the rock to reach the highest point, we use the formula
v = u + at
where 'u' is initial velocity, 'v' is final velocity (0 m/s at the highest point), 'a' is the acceleration due to gravity (which is negative since it is in the opposite direction to the initial motion), and 't' is the time.
Solving for time, we get
t = -u/g. With g ≈ 9.81 m/s^2, the time to reach the highest point is
t ≈ -1320 m/s / -9.81 m/s^2 ≈ 134.6 s.
After reaching the highest point, the rocket will start falling back to the ground. Since the rocket starts from rest at the highest point, we can use the formula
s = 0.5gt^2
where 's' is the distance and 't' is time, to calculate the time it takes to fall to the ground.
But since we already have the time to reach the highest point, we can simply double that time to find the total time taken for the round trip, because the time to go up is the same as the time to come down in free fall. So the total time taken for the rocket to hit the ground is
134.6 s up + 134.6 s down = 269.2 s.
If the magnitude of the resultant force acting on the eyebolt is 570 N and its direction measured clockwise from the positive x axis is θ = 33 ∘, determine the magnitude of
Answer:
F1 = 1210.65 N
Q = 65.7081 degrees
Explanation:
Sum of Forces in x - direction:
F1 * cos (Q) + F2*sin(30) - F3*(3/5) = Fres*cos (theta)
F1 * cos (Q) + (500)*(0.5) - 450*(3/5) = 570*cos(33)
F1*cos (Q) = 498.0422237 N .... Eq1
Sum of Forces in y - direction:
F1 * sin (Q) - F2*sin (60) - F3 * (4/5) = Fres*sin (theta)
F1 * sin (Q) - 500*sin(60) - 450 * (4/5) = 570*sin (33)
F1 * sin (Q) = 1103.45692 N .... Eq 2
Divide Eq 2 by Eq 1
tan (Q) = 2.21558916
Q = arctan (2.21558916) = 65.7081 degrees
F1 = 1210.65 N
A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 11000 N/C. The mass of the water drop is 3.37 × 10-9 kg. How many excess electrons or protons reside on the drop?
To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.
[tex]E = \frac{mg}{q}[/tex]
Here,
m = mass
g = Acceleration due to gravity
Rearranging to find the charge,
[tex]q = \frac{mg}{E}[/tex]
Replacing,
[tex]q = \frac{(3.37*10^{-9})(9.8)}{11000}[/tex]
[tex]q = 3.002*10^{-12}C[/tex]
Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is
[tex]q = ne[/tex]
Here,
n = Number of electrons
e = Charge of each electron
[tex]n = \frac{q}{e}[/tex]
Replacing,
[tex]n = \frac{3.002*10^{-12}}{1.6*10^{-19}}[/tex]
[tex]n = 2.44*10^7[/tex]
Therefore the number of electrons that reside on the drop is [tex]2.44*10^7[/tex]
To determine the number of excess electrons or protons residing on the water drop, we can use the principle of electrostatics. When an electric field is applied to a charged object, the electrostatic force acting on it can be calculated using the equation:
[tex]F = qE[/tex]
Where:
F is the electrostatic force
q is the charge on the object
E is the electric field strength
In this case, the electrostatic force acting on the water drop is balanced by the gravitational force, so we have:
F = mg
Where:
m is the mass of the water drop
g is the acceleration due to gravity
We can equate these two forces and solve for the charge q:
qE = mg
From this equation, we can isolate the charge q:
q = mg / E
Now we can calculate the charge on the water drop:
m = 3.37 × 10^(-9) kg
g = 9.8 m/s^2
E = 11000 N/C
Substituting the values into the equation:
q = (3.37 × 10^(-9) kg * 9.8 m/s^2) / 11000 N/C
Calculating this expression:
q = 3.037 × 10^(-15) C
The elementary charge of an electron or proton is approximately 1.602 × 10^(-19) C. To find the number of excess electrons or protons, we can divide the calculated charge by the elementary charge:
Number of excess electrons or protons = q / elementary charge
Number of excess electrons or protons = (3.037 × 10^(-15) C) / (1.602 × 10^(-19) C)
Calculating this expression:
Number of excess electrons or protons ≈ 1.895 × 10^(4)
Therefore, the water drop has approximately 18,950 excess electrons or protons.
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If the neutral atom of an element has only 5 valence electrons it must be in which group? 1. VIIA 2. VA 3. IVB 4. VIA 5. IVA 6. IIIA
Answer:
2. VA
Explanation:
The valency electron or outer electron of a neutral atom of an element determines the group at which an element belong. The electronic configuration of an atom that have valency of 5 can be represented as 2 5 , 2 8 5 etc.
The electronic configuration 2 5 represent Nitrogen atom while 2 8 5 represent phosphorus atom. The valency 5 depicts the element belongs to group 5A(VA).
This means atoms of valency electron of 3 belong to group IIIA, valency electrons of 4 belongs to group IVA, valency electron of 6 belong to group VIA and valency electrons of 7 belong to group VIIA.
When the spring, with the attached 275.0 g mass, is displaced from its new equilibrium position, it undergoes SHM. Calculate the period of oscillation, T , neglecting the mass of the spring itself.
Answer:
The period of oscillation is 1.33 sec.
Explanation:
Given that,
Mass = 275.0 g
Suppose value of spring constant is 6.2 N/m.
We need to calculate the angular frequency
Using formula of angular frequency
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
Where, m = mass
k = spring constant
Put the value into the formula
[tex]\omega=\sqrt{\dfrac{6.2}{275.0\times10^{-3}}}[/tex]
[tex]\omega=4.74\ rad/s[/tex]
We need to calculate the period of oscillation,
Using formula of time period
[tex]T=\dfrac{2\pi}{\omega}[/tex]
Put the value into the formula
[tex]T=\dfrac{2\pi}{4.74}[/tex]
[tex]T=1.33\ sec[/tex]
Hence, The period of oscillation is 1.33 sec.
Based on the scaling laws, how many times greater is the strength-to-weight ratio of a nanotube (D = 10 nm) than the leg of a flea (D = 100 μm)? Than the leg of an elephant (D = 2m)?
To solve this problem we will apply the laws of proportion between the resistance / weight ratio that may occur depending on the cases mentioned. The strength-to-weight ratio for each object is,
For the nano tube,
[tex]\frac{S}{W} = \frac{1}{10*10^{-9}} =1*10^{8}[/tex]
For the leg of a flea,
[tex]\frac{S}{W}= \frac{1}{100*10^{-6}}=10000[/tex]
For the elephant,
[tex]\frac{S}{W} = \frac{1}{2}= 0.5[/tex]
From this we can conclude that the resistance / weight ratio of the nano tube is 10 thousand times better than the flea's leg and [tex]10 ^ 8[/tex] times better than the elephant's leg.
You measure the intensity of a sound wave to be 9.80 W/m2 . The power output of the signal is 75 W and the signal is emitted in all directions.
Part A
How far away from the source are you?
r = ?
Answer:
r = 0.78 m
Explanation:
If the sound source is emitting the signal evenly in all directions (as from a point source) this means that at any time, the source power is distributed over the surface of a sphere of radius r.
At a distance r of the source, the intensity of the sound is defined as the power per unit area:
I = P/A
As the area is the area of a sphere, we can say the following:
I = P / 4*π*r²
Replacing I and P by the values given, we can solve for r (which is the distance from the listener to the source) as follows:
r² = P / I*4*π ⇒ r = [tex]\sqrt{P/(4*\pi*I)}[/tex] = 0.78 m
The distance between you and the source is 0.78 m.
The given parameters:
Intensity of the sound, I = 9.8 W/m²Output power of the signal, P = 75 WThe area of the source is calculated as follows;
[tex]A = \frac{P}{I} \\\\ A = \frac{75}{9.8} \\\\ A = 7.65 \ m^2[/tex]
The distance between you and the source is calculated as follows;
[tex]A = 4\pi r^2\\\\ r^2 = \frac{A}{4\pi} \\\\ r = \sqrt{\frac{A}{4\pi}} \\\\ r = \sqrt{\frac{7.65}{4\pi}} \\\\ r = 0.78 \ m[/tex]
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Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 44.0mph and half the distance at 65.0 . On her return trip, she drives half the time at 44.0mph and half the time at 65.0mph .
1)What is Julie's average speed on the way to Grandmother's house?
2)What is her average speed on the return trip?
*Express your answer with the appropriate units.
Answer:
52.47706 mph
54.5 mph
Explanation:
The average speed is given by
[tex]V_{av}=\dfrac{Distance}{Time}[/tex]
[tex]V_{av}=\dfrac{100}{\dfrac{50}{44}+\dfrac{50}{65}}\\\Rightarrow V_{av}=52.47706\ mph[/tex]
Julie's average speed on the way to Grandmother's house is 52.47706 mph
[tex]V_{av}=\dfrac{44+65}{2}\\\Rightarrow V_{av}=54.5\ mph[/tex]
Average speed on the return trip is 54.5 mph