Answer:
If the cell is placed in a surrounding solution which is hypotonic in nature.
Then the water from outside of the cell to the inside of the cell. The water will keep on moving from the outside of the cell to the inside of the cell.
The flow of water will take place until the outside environment of the cell and the inside of the cell becomes equal.
The flow of water will take place from the outside of the cell to the inside of the cell.
Answer:
The direction of motion of water molecules will be into the cell.
Explanation:
A hypotonic solution is one which has the less concentration of solute in the solvent as compared to the solution on the other side of the semi-permeable membrane. This creates an osmotic pressure gradient across the semi-permeable membrane which is responsible for the flow of water molecules across the membrane until the concentration becomes equal for both the solutions.(A semi-permeable membrane is a sheet or a plane barrier which does not allows the molecules over certain size to pass through it. Here the membrane does not allows the molecules larger in size than that of water molecules to pass through it.)An object is in simple harmonic motion. The rate at which the object oscillates may be described using the period T, the frequency f, and the angular frequency . If the angular frequency decreases, what is the effect on the period and the frequency?
Explanation:
The relation between angular frequency and frequency is directly proportional:
[tex]\omega=2\pi f[/tex]
So, if the angular frequency decreases, the frequency also decreases.
The relation between angular frequency and period is inversely proportional:
[tex]\omega=\frac{2\pi}{T}[/tex]
So, if the angular frequency decreases, the period increases.
When the angular frequency of an object in simple harmonic motion decreases, the period increases and the frequency decreases.
Explanation:When the angular frequency of an object in simple harmonic motion decreases, the effect on the period and frequency is as follows:
Arnold Strongman and Suzie Small each pull very hard on opposite ends of a massless rope in a tug-of-war. The greater force on the rope is exerted by
A. Arnold, of course
B. Suzie, surprisingly
C. both the same, interestingly enough
Answer:
The greater force on the rope is exerted by:
C. both the same, interestingly enough
Explanation:
The option c is correct as the rope is mass-less. Both Arnold and Suzie are pulling hard on opposite ends. But the tension of the rope is same at every point so that means both are exerting the same force on both ends of rope.The analogy for the given situation is that as when a vehicle moves on a road, the force exerted by the wheel of vehicle on the road is equal to the force exerted by the road on the wheel of vehicle.The options A and B can be possible if anyone of them exert more power on the ground through feet in this way, one can exert more force and can win.An experiment on the earth's magnetic field is being carried out 1.00 m from an electric cable. What is the maximum allowable current int eh cable if the experiment is to be accurate?
The maximum allowable current in the cable for the experiment to be accurate, considering Earth's magnetic field is 0.5 Amps. This is calculated using the magnetic field strength formula and rearranging it to find the current.
Explanation:To answer the question, we would need to use the formula to determine the magnetic field strength generated by a current carrying wire which is B = μ0 * I / 2πr, where B is the magnetic field strength generated by the wire, I is the current, r is the distance from the wire and `μ0` is the permeability of free space which is a constant equal to 4π * 10^-7 T m/A.
Let's rearrange this equation to make I the subject:
I = B * 2πr / μ0
We are told in the question that the maximum magnetic field (B) that does not adversely affect the experiment is 1.0 * 10^-4 T, and the distance (r) is 1.00 m. So substituting these numbers in:
I = 1.0 * 10^-4 T * 2π * 1.00 m / 4π * 10^-7 T m/A
This simplifies to 0.5 A. Therefore, the maximum allowable current in the wire is 0.5 Amps.
Learn more about Magnetic Fields and Current here:https://brainly.com/question/10677063
#SPJ12
An electromagnetic wave has an electric field given by
E? (y,t)=[3.90×105V/mcos(ky−(1.200×1013rad/s)t)]k^
In which direction is the wave traveling?What is the wavelength of the wave?
Answer:
λ = 157 μm
Explanation:
given,
[tex]\vec{E}(y,t)=[ 3.9\times 10^5\ V/m cos(ky-(1.20\times 10^{13}\ rad/s )t)]\vec{k}[/tex]
now,
ω = 1.2 x 10¹³ rad/s
[tex]f =\dfrac{\omega}{2\pi}[/tex]
[tex]f =\dfrac{1.2\times 10^{13}}{2\pi}[/tex]
f = 1.91 x 10¹² Hz
determine the wavelength of the E M wave
[tex]\lambda = \dfrac{c}{f}[/tex]
[tex]\lambda = \dfrac{3\times 10^8}{1.91\times 10^{12}}[/tex]
λ = 1.57 x 10⁻⁴ m
λ = 157 x 10⁻⁶ m
λ = 157 μm
hence, the wavelength of the wave is equal to λ = 157 μm
Given that,
[tex]\vec{E}(y,t) = [3.9\times 10^5 \ V/mcos] (ky-(1.20\times 10^{13} \ rad/s)t)] \vec{k}[/tex]here,
[tex]\omega = 1.2\times 10^{13} \ rad/s[/tex]As we know,
→ [tex]f = \frac{\omega}{2 \pi}[/tex]
By substituting the values, we get
[tex]= \frac{1.2\times 10 ^{13}}{2 \pi}[/tex]
[tex]= 1.91\times 10^{12} \ Hz[/tex]
hence,
The wavelength will be:
→ [tex]\lambda = \frac{c}{f}[/tex]
[tex]= \frac{3\times 10^8}{1.91\times 10^{12}}[/tex]
[tex]=157\times 10^{-6} \ m[/tex]
[tex]= 157 \ \mu m[/tex]
Thus the above response is correct.
Learn more about electric field here:
https://brainly.com/question/13872620
boy pulls a 5.0-kg sled with a rope that makes a 60.0° angle with respect to the horizontal surface of a frozen pond. The boy pulls on the rope with a force of 10.0 N; and the sled moves with constant velocity. What is the coefficient of friction between the and the ice?
(a) 0.09
(b) 0.12
(c) 0.18
(d) 0.06
(e) 0.24
Answer:
0.1
Explanation:
mass, m = 5 kg
θ = 60°
Force, F = 10 N
velocity is constant , it means the net force is zero.
So, the component of force along the surface is equal to the friction force
FCosθ = friction force
10 x cos 60 = μ x m x g
where, μ is the coefficient of friction
5 = μ x 5 x 9.8
μ = 0.1
Thus, the coefficient of friction is 0.1
How much work will a machine with a power rating of 1.1 × 103 watts do in 2.0 hours?
A. 1.5 × 106 joules
B. 2.1 × 106 joules
C. 4.0 × 106 joules
D. 8.1 × 106 joules
Answer:
D.
Explanation:
8.1X10 to the power of 6 Joules.
Answer:
7.92×10^6Joules
Explanation:
Power is defined as the rate of change in work done. Mathematically, Power = Workdone/Time
Given power = 1.1×10³watts
Time = 2.0hours = 2×60×60
= 7,200seconds
Workdone = power × time
Work done = 1.1×10³×7200
Workdone = 7,920,000Joules
= 7.92×10^6Joules
Assume that the earth is a uniform sphere and that its path around the sun is circular.
(a) Calculate the kinetic energy that the earth has because of its rotation about its own axis. For comparison, the total energy used in the United States in one year is about 9.33 multiplied by 109 J.
(b) Calculate the kinetic energy that the earth has because of its motion around the sun.
Explanationhe rotational kinetic energy is
[tex]K_{r} =\frac{1}{2} Iω^{2}[/tex]
The moment of inertia I for a sphere is ( 2 / 5 ) m r ^2
. Substituting this in the equation yields
Kr=1/2( ( 2 / 5 ) m r ^2 )([tex](\frac{v}{r})^{2}[/tex]
1/5mv^2
1/5*5.97 × 10 ^24 *(2[tex]\pi[/tex]*6.38*10^6/86400)^2
2.57 × 10 ^29 J
b. kinetic energy of the sun
K.E=1/2*mv^2
the distance from the earth to the sun is given as
.
Answer:
a. 7.43 × 10³⁴ J b. 3.51 × 10³⁸ J
Explanation:
a. The gravitational force of attraction of a body on the surface of the earth equals the centripetal force on it due to the earth.
So, GMm/R² = mRω²
ω = √(GM/R³) where ω = angular speed of the earth. M = mass of earth = 5.972 × 10²⁴ kg, R = radius of earth = 6.4 × 10⁶ m and G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²
The rotational kinetic energy of earth K.E = 1/2Iω² where I = rotational inertia = 2/5MR²
K.E = 1/2Iω²
= 1/2 × 2/5MR² × GM/R³
= GM²/5R
= 6.67 × 10⁻¹¹ Nm²/kg² × (5.972 × 10²⁴ kg)² /(6.4 × 10⁶ m × 5)
= 7.43 × 10³⁴ J
b. Similarly, the rotational kinetic energy of the earth around the sun is
K.E = GM²/5R where M = mass of sun = 1.989 × 10³⁰ kg and R = distance of earth from sun = 1.5047 × 10¹¹ m
K.E = GM²/5R
= 6.67 × 10⁻¹¹ Nm²/kg² × (1.989 × 10³⁰ kg)² / (1.5047 × 10¹¹ m × 5)
= 3.5073 × 10³⁸ J ≅ 3.51 × 10³⁸ J
Which wavelengths of light drive the highest rates of photosynthesis?
Answer:
400-450 nm and 670-680 nm
Explanation:
Which wavelengths of light drive the highest rates of photosynthesis?
400-450 nm and 670-680 nm
Light in the violet-blue and red portions of the spectrum is most effective in driving photosynthesis
photosynthesis is the process through which green plant manufacture there food through sunlight, they synthesize nutrient in the this process through chlorophyll thereby releasing oxygen as a bye-product. a spectrum contains 7 colours combined . the violet-blue and red portions of the spectrum is most effective
When a sound wave travels directly toward a hard wall, the incoming and reflected waves can combine to produce a standing wave. There is an antinode right at the wall, just as at the end of a closed tube, so the sound near the wall is loud. You are standing beside a brick wall listening to a 80Hz tone from a distant loudspeaker.
How far from the wall must you move to find the first quiet spot? Assume a sound speed of 340 m/s.
Answer:
L= 1.06 m
Explanation:
frequency of a wave in a pipe closed at one end is given by
[tex]f= \frac{v}{4L}[/tex]
L= length of the pipe
v= velocity of wave
therefore, the length of the pipe
[tex]L= \frac{v}{4f}[/tex]
[tex]L= \frac{340}{4\times80}[/tex]
L= 1.06 m
L= 1.06 m far from the wall must you move to find the first quiet spot.
The distance from the wall(i.e the pipe length) you need to move to find the first quiet spot is 1.06 m.
What is the frequency of a sound wave in a closed pipe?The smallest standing wave pattern, i.e the first fundamental frequency or harmonic for a closed pipe usually consists of one node and an antinode. When the wavelength of a closed pipe of length L is 4 L, the basic standing wave is formed.
The frequency of a sound wave for a closed pipe at one end may thus be represented using the following formula:
[tex]\mathbf{f = \dfrac{v}{4L}}[/tex]
where;
L = pipe length = ???v = velocity of the wave = 340 m/sf = frequency = 80 HzThus, the length of the pipe can be computed as:
[tex]\mathbf{L = \dfrac{340 \ m/s}{4\times 80 \ Hz}}[/tex]
L = 1.06 m
Learn more about the frequency of sound waves here:
https://brainly.com/question/15666150
A solid cone is 10 cm high. where is its center of mass?
Answer:
The center of mass of a cone is located along a line. This line is perpendicular to the base and reaches the apex. The center of mass is a distance 3/4 of the height of the cone with respect to the apex.
Explanation:
The center of mass of a solid cone is located one-third of the way up from the base.
Explanation:The center of mass of a solid cone is located at one-third of its height from the base. In this case, the cone is 10 cm high, so the center of mass is located 10 cm * (1/3) = 3.33 cm from the base.
Learn more about Center of mass here:https://brainly.com/question/33607198
#SPJ2
Unless otherwise posted, the maximum speed limit in any urban or residential district is ____ mph.
A. 45
B. 40
C. 30
Answer:
C 30 mph
Explanation:
The speed limits in various area's can be concluded as
1. 25- 30 mph in urban residential areas and school districts.
2.55 mph on rural highways, and
3. 70 mph on rural Interstate highways.
so, the correct answer is c
Slow moving vehicles must display ___________ emblem at the rear to warn of their low speed.A. a red square B. a yellow triangular C. an orange triangular
Answer:
option C
Explanation:
The correct answer is option C
An orange triangular sign on the rear of the vehicle will show that the vehicle will move at a slow speed.
Warning Sign is an important method of conveying the message. Warning sign helps for the smooth movement of the traffic.
A car with an orange sign on the rear will inform the fellow driver that the vehicle is slow and they can overtake it.
A biker to move from speed of 0 m/s to final speed of 25 m/s in 10 s. What is the acceleration of the bike?
Answer:
a = 2.5 m/s²
Explanation:
given,
initial speed = 0 m/s
final speed = 25 m/s
time = 10 s
acceleration of bike is equal to change in velocity per unit time
[tex]a = \dfrac{\Delta v}{t}[/tex]
[tex]a = \dfrac{v_f-v_i}{t}[/tex]
[tex]a = \dfrac{25-0}{10}[/tex]
a = 2.5 m/s²
hence, acceleration of bike is equal to a = 2.5 m/s²
a skater increases her speed uniformly from 2.0 m/s to 10 m/s in 3.0s. what is the acceleratioon of the biker?
Answer:
a= 2.667 m/s²
Explanation:
Given that
Initial velocity ,u= 2 m/s
Final velocity ,v= 10 m/s
Time taken ,t= 3 s
As we know that
v = u + at
a=Acceleration ,u=Initial velocity ,v= Final velocity ,t=time
Now by putting the values in the above equation
v = u + at
10 = 2 + a x 3
8 = 3 a
[tex]a=\dfrac{8}{3}\ m/s^2[/tex]
a= 2.667 m/s²
Therefore the acceleration of the biker will be 2.667 m/s²
A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg·m2. If the arms are pulled in so the moment of inertia decreases to 1.80 kg·m2
Answer:
The angular velocity when the arms are pulled is ω₂= 6.25 rad/s
Explanation:
Assuming that you want to determine the final angular velocity when is the arms are pulled, then it is calculated using the principle of conservation of angular momentum. It states that:
I₁ω₁=I₂ω₂
where I = moment of inertia , ω= angular velocity , 1 and 2 denote the skater with extended hands and pulled respectively.
Thus
I₁ω₁=I₂ω₂
ω₂= I₁ω₁/I₂
replacing values
ω₂= ω₁ *(I₁/I₂) = 5 rad/s *(2.25 kg·m2/1.80 kg·m2) = 6.25 rad/s
ω₂= 6.25 rad/s
Two train whistles have identical frequencies of 1.64 102 Hz. When one train is at rest in the station and the other is moving nearby, a commuter standing on the station platform hears beats with a frequency of 4.00 beats/s when the whistles operate together. What are the two possible speeds that the moving train can have?
Answer:
Vs = 6.73 m/s or Vs = 16.3 m/s
Explanation:
frequency of the trains whistle (f) = 1.64 x 10^{2} Hz = 164 Hz
frequency of beats heard = 4 beats/s = 4 Hz
velocity of the stationary train (Vr) = 0
velocity of sound in air (V) = 343 m/s
velocity of the moving train (Vs) = ?
we can get the velocity of the moving train from the formula below
Fn = f x [tex]\frac{V + Vr}{V - Vs}[/tex] ...equation 1
where Fn = net frequency
case one - assuming the train is approaching the station Fn = 164 + 4 = 168 Hzsubstituting the known values into equation 1
168 = 164 x [tex]\frac{343 + 0}{343 - Vs}[/tex]
1.02 = [tex]\frac{343 + 0}{343 - Vs}[/tex]
Vs = [tex]343 - \frac{343 + 0}{1.02}[/tex]
Vs = 6.73 m/s
case two - assuming the train is leaving the station Fn = 164 - 4 = 160 Hzsubstituting the known values into equation 1
168 = 160 x [tex]\frac{343 + 0}{343 - Vs}[/tex]
1.05 = [tex]\frac{343 + 0}{343 - Vs}[/tex]
Vs = [tex]343 - \frac{343 + 0}{1.05}[/tex]
Vs = 16.3 m/s
A vertical spring has a spring constant of 100 newtons per meter. When an object is attached to the bottom of the spring, the spring changes from its unstretched length of 0.50 meters to a length of 0.65 meters. The magnitude of the weight of the attached object is
a. 1.1N
b. 15N
c. 50N
d. 65N
Answer:
option B
Explanation:
given,
spring constant = k = 100 N/m
distance of unstretched length = 0.5 m
distance after stretched length = 0.65 m
magnitude of weight = ?
We know,
W = K x
x is the displacement
x = 0.65 - 0.5 = 0.15 m
W = K x
W = 100 x 0.15
W = 15 N
Hence, the correct answer is option B
Final answer:
The weight of the attached object is found by multiplying the spring constant (100 N/m) by the stretch in the spring (0.15 m), which equals 15 N. Therefore, the weight of the object is 15 N.
Explanation:
The weight of the attached object can be calculated using Hooke's Law, which states that the force exerted by the spring (F) is equal to the spring constant (k) multiplied by the displacement from the rest position ( extDelta x): F = k extDelta x. In this case, the spring has a spring constant of 100 newtons per meter and the displacement from the original length of 0.50 meters to the new length of 0.65 meters is 0.15 meters.
Thus, F = 100 N/m exttimes 0.15 m, which gives us a force of 15 N. Since the weight of the object is the force exerted by the spring and these two forces are equal and opposite when the spring is at rest with the object hanging, the weight of the attached object is 15 N. Therefore, the correct answer is (b) 15N.
What is the maximum value of the resistance of the pull-up resistor, such that you avoid floating when the switch is open? pull-up
Answer:
Resistance of order of 10Kiloohms
Explanation:
Pull up resistors are essential in every microcontroller devices or digital circuits. Which helps prevent floating ( a process of not being able to determine the state of a pin) It works by allowing the state of a pin to be in either a high state or low state. A resistor is connected between the Vcc and the pin.
When the switch is open, the entire voltage flows through the resistor (V=IR). When the switch is closed the pull up resistor and the internal impedance of the pin form a voltage divider to limit the current flow.
The maximum value of a pull-up resistor to avoid floating when the switch is open is typically between 1KΩ to 10KΩ. This helps to prevent excessive current flow and quickly pulls the input pin high to avoid a floating state. Extremely high resistance values, like 1 megaohm, may cause brief floating, potentially leading to incorrect readings.
Explanation:The maximum value of the pull-up resistor, in order to avoid a floating condition when the switch is open, depends on several factors including the characteristics of the microcontroller's input pin and the voltage level of the system. However, a moderately high resistance value is often chosen in the range of 1KΩ to 10KΩ to prevent excessive current flow when the switch is closed while still pulling the input pin high quickly enough to avoid a floating state when the switch is open. Remember that a pull-up resistor's purpose is to define a state of a digital input pin when it is not otherwise driven by an external component or system.
In some cases, the maximum value of the pull-up resistor will need to be calculated based on the specifications of your particular circuit. An extremely high resistance, like 1 megaohm, can cause the input to float for a very short time, long enough that your microcontroller might read it incorrectly.
Learn more about Pull-up Resistor here:https://brainly.com/question/14307039
#SPJ12
An artery is constricted at one location to 1/2 its normal cross-sectional area.
How does the speed of blood past the constriction compare to the speed of blood flow in the rest of the artery? (Note: Assume ideal fluid flow.)
a) it is 1/4 as fast
b) it is 1/2 as fast
c) it is 2 times as fast
d) it is 4 times as fast
Answer:
v₂ = 2 v₁ blood velocity in the built zone is twice the speed of the normal zone, so the correct response is c
Explanation:
For that problem we must use the fluid continuity equation
A₁ v₁ = A₂ v₂
Where point 1 is in the normal artery and point 2 in the contraction
v₂ = v₁ A₁ / A₂
In the exercise indicates
A₂ = ½ A₁
v₂ = 2 v₁
The blood velocity in the built zone is twice the speed of the normal zone, so the correct response is c
In 2014, a space probe approached the rocky core of the comet Churyumov–Gerasimenko, which is only a few km in diameter. The probe then entered orbit around the comet at a distance of 30 km. The comet was found to have a mass of 1.0 * 10^13 kg. What was the orbital period of the probe around the comet, in earth days?
Answer: 14.62 Earth days
Explanation:
This problem can be solved by Kepler’s Third Law of Planetary motion:
[tex]T=2 \pi \sqrt{\frac{a^{3}}{GM}}[/tex]
Where:
[tex]T[/tex] is the period of the probe
[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the Gravitational Constant
[tex]M=1(10)^{13} kg[/tex] is the mass of the comet Churyumov–Gerasimenko
[tex]a=30 km \frac{1000 m}{1 km}=30000 m[/tex] is the semimajor axis of the orbit the probe described around the comet (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)
[tex]T=2 \pi \sqrt{\frac{(30000 m)^{3}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(1(10)^{13} kg)}}[/tex]
[tex]T=1,263,771.768 s \frac{1 h}{3600 s} \frac{1 Earth-day}{24 h}=14.62 Earth-days[/tex]
Hence, the orbital period of the probe is 14.62 Earth days.
The orbital period of the Rosetta space probe, in orbit around the comet Churyumov–Gerasimenko in 2014, is calculated to be approximately 0.178 Earth days using Kepler's Third Law of Planetary Motion and given values.
Explanation:This question pertains to the calculation of the orbital period of the Rosetta space probe when it entered orbit around the comet Churyumov–Gerasimenko in 2014. First, we establish the gravitational constant (G) as approximately 6.674 × 10^-11 m^3 kg^-1 s^-2. To calculate the orbital period of an object in orbit around a celestial body, we can use Kepler's Third Law of Planetary Motion which states the square of the orbital period T is proportional to the cube of the semi-major axis a. Rearranging this law and substituting, we get T = 2π sqrt(a^3/(G*M)). Applying the given values, we get T = 2π sqrt((30*10^3 m)^3/(6.674 × 10^-11 m^3 kg^-1 s^-2*1.0*10^13 kg)) = 15378 seconds or approximately 0.178 days.
Learn more about Orbital Period Calculation here:https://brainly.com/question/1766066
#SPJ3
How will W change if the initial speed of the box at Point A is increased by a factor of 2?
If the initial speed of the box at Point A is increased by a factor of 2, the final velocity, denoted as W, will also increase by a factor of 2.
Explanation:If the initial speed of the box at Point A is increased by a factor of 2, the final velocity of the box, denoted as W, will also increase by a factor of 2.
For example, if the initial speed of the box was 10 m/s, after increasing it by a factor of 2, the new speed would be 20 m/s.
Therefore, option (a), It will increase by a factor of two, is the correct answer choice.
A ray of light in air is incident upon a glass plate at an angle of 45°. The angle of refraction of the ray in the glass is 30°. What is the index of refraction of the glass?
Answer:
Explanation:
Laws of refraction:
(a) The incident ray, the refracted ray and the normal at the point of incident all lies in the same plane
(b) The ratio of the sine of incident to the sine of refraction is a constant for a given pair of media, which is the refractive index of the second medium with respect to the first medium. The is also called Snell's law
from Snell's law,
Index of refraction of the glass = sini/sinr.............. Equation 1
where i = incident angle, r = angle of refraction.
given: i = 45°, r = 30°
Substituting these values into equation 1,
Index of refraction of the glass = sin45°/sin30°
Index of refraction of the glass = (1/√2)/(1/2)
Index of refraction of the glass = 2/√2
Index of refraction of the glass = √2.
Therefore, The Index of refraction of the glass = √2.
Final answer:
To calculate the index of refraction of the glass, we apply Snell's Law, resulting in an index of refraction of approximately 1.41.
Explanation:
To find the index of refraction of the glass, we can use Snell's Law, which relates the angles of incidence and refraction to the indices of refraction of the two media. Snell's Law states:
n1 * sin(θ1) = n2 * sin(θ2),
where n1 and n2 are the indices of refraction of the first and second media (air and glass, respectively), and θ1 and θ2 are the angles of incidence and refraction. Given that the index of refraction of air is approximately 1 (since it's very close to a vacuum) and using the provided angles (45° and 30°), the calculation will look like this:
1 * sin(45°) = n2 * sin(30°)
Which simplifies to:
1 * √2/2 = n2 * 1/2
After calculating, we find:
n2 = √2
Thus, the index of refraction of the glass is √2, which is approximately 1.41.
A person holding weights sits on a spinning stool. Then the person pulls the weights in,decreasing the rotational inertia of the system of the man, the weights,and the stool by a factor of 2. The stool can rotate about its axle with negligible friction.How does the kinetic energy of the system change?
Answer:
Ķ rot = 2K rot
When the arms and weight are retracted toward the body, the energy increases by twice the initial kinetic energy because the moment of inertia has decreased coupled with the energy needed to bring the hands towards the body.
Explanation:
K rot = kinetic energy of system when hands + weight are stretched out = ½ Iw²
Where w = angular velocity when arms and weight are stretched out
I = inertia of the system = ( Ie + Ip) where Ie= moment of inertia of the extra weight carried and Ip = moment of inertia of the person.
Ķ rot = final rotational energy when arms and external weight are pulled in = ½Ìŵ² where Ì =( Ìe + Ìp)
And Ìe = inertia of weight when hands are retracted and Ìp = inertia of person when hand are retracted
Using conservation of angular momentum which is
Iw = Ìŵ
Substitute ŵ = (I/Ì)*W in Ķ rot to give
Ķ rot =½Ì *[ (I/Ì) * W ]²
Ķ rot = ½ ( I²/Ì) * w²
Since K rot = ½ I w²
So Ķ rot = K rot ( I/Ì )...eq3
From the question above system inertia reduces by a factor of 2
Ì = ½I
Sub expression into equ3
Ķ rot = 2K rot
An electron is accelerqated in the uniform field betwen two parallel charged oplates. The separation of the plates is 1.20 cm
Answer:
The speed of electron is [tex]8.7\times10^{6}\ m/s[/tex]
Explanation:
Given that,
Separation of the plate = 1.20 cm
Suppose the field is [tex]E=1.80\times10^{4}\ N/C[/tex].
If the electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate.
What the speed does it leave the hole?
We need to calculate the acceleration
Using formula of electric force
[tex]F = qE[/tex]
[tex]ma=qE[/tex]
[tex]a=\dfrac{qE}{m}[/tex]
We need to calculate the speed of electron
Using equation of motion
[tex]v^2=u^2+2as[/tex]
[tex]v^2=2as[/tex]
Put the value of acceleration in the formula
[tex]v^2=2\times\dfrac{qE}{m}\times s[/tex]
Put the value into the formula
[tex]v^2=2\times\dfrac{1.6\times10^{-19}\times1.80\times10^{4}\times1.20\times10^{-2}}{9.11\times10^{-31}}[/tex]
[tex]v=\sqrt{2\times\dfrac{1.6\times10^{-19}\times1.80\times10^{4}\times1.20\times10^{-2}}{9.11\times10^{-31}}}[/tex]
[tex]v=8.7\times10^{6}\ m/s[/tex]
Hence, The speed of electron is [tex]8.7\times10^{6}\ m/s[/tex]
ListenA person on a ledge throws a ball vertically downward, striking the ground below the ledge with 200 joules of kinetic energy. The person then throws an identical ball vertically upward at the same initial speed from the same point. What is the kinetic energy of the second ball when it hits the ground? [Neglect friction.]
A. 200 J
B. 400 J
C. less than 200 J
D.more than 400 J
Answer:
A. 200 J
Explanation:
The initial kinetic energy depends on the initial speed, while the gravitational potential energy depends on the height, both balls are thrown with the same initial speed and from the same height. Therefore, due to the law of conservation of energy, the balls must have the same mechanical energy (the sum of both energies) when both impact the ground. Since the potential energy is zero at this point, its final kinetic energy must also be the same.
The second ball, thrown vertically upward, will have kinetic energy of 200 joules when it hits the ground, making option A. the correct option.
To answer this question, let's analyze the energy transformations involved. When the ball is thrown downward, it strikes the ground with 200 joules of kinetic energy.
This energy comes from the initial throw plus the gravitational potential energy converted during its fall.
When the second ball is thrown upward, it initially gains gravitational potential energy as it rises, then this potential energy is converted back to kinetic energy as it falls back down.
The initial speed given to the ball in both scenarios is the same, meaning the total energy input into the system is identical for both throws.
Since both balls have the same initial speed and both experience the same gravitational acceleration, the second ball will also have 200 joules of kinetic energy when it hits the ground.
Therefore, the correct answer is A. 200 J.
An astronaut drops a ball off the edge of a crater on the Moon, where gravity is 1.6 m/s(s). If it takes 5 seconds for the ball to reach the bottom of the crater, what is the depth of the crater? (Need HELP, AAAAAAAAAAAAAAAAAAAAAA.)
Answer:
The depth of the crater is 20[m]
Explanation:
We can solve this problem using the kinematics equations, it should be noted that the negative sign of acceleration means that the drop of the object is down.
Data:
a = 1.6[m/s^2]
t = 5 [s]
[tex]y=y_{0} +v_{0} *t-\frac{1}{2} *g*(t)^{2}[/tex]
v0= 0, because there is not initial veocity
y = final distance = 0 the bottom of the crater
g = a = acceleration of the crater [m/s^2]
y0= Point where the ball was dropped. = Distance from the original point to the bottom of the crater.
[tex]y_{0} =\frac{1}{2}*1.6*(5)^{2}\\ y_{0} = 20 [m][/tex]
An electron is accelerated by a potential difference of 50v, what is the de broglie wavelength?
Answer:
[tex]1.73553\times 10^{-10}\ m[/tex]
Explanation:
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
K = Potential difference = 50 V
m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]
The de broglie wavelength is given by
[tex]\lambda=\dfrac{h}{\sqrt{2mK}}\\\Rightarrow \lambda=\dfrac{6.626\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 50\times 1.6\times 10^{-19}}}\\\Rightarrow \lambda=1.73553\times 10^{-10}\ m[/tex]
The wavelength is [tex]1.73553\times 10^{-10}\ m[/tex]
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hour (mph). At full power, the car can accelerate from zero to 32.0mph in time 1.10s .A)At full power, how long would it take for the car to accelerate from 0 to 64.0mph ? Neglect friction and air resistance. =4.40sPart BA more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 32.0mph in time 1.10s , how long would it take to go from zero to 64.0mph ?am not sure how to do part B
Answer:
a) 4.40 s
b) 2.20 s
Explanation:
Given parameters are:
At constant power ,
initial speed of the car, [tex]v_0=0[/tex]
final speed of the car, [tex]v=32[/tex] mph
At full power,
initial speed of the car, [tex]v_0=0[/tex]
final speed of the car, [tex]v=64[/tex] mph
a)
At constant power, [tex]KE = \frac{1}{2} mv^2[/tex]
At full power, [tex]KE = \frac{1}{2} m(2v)^2[/tex]
So [tex]KE_f = 4KE_i[/tex]
So, time to reach 64 mph speed is 4 times more than the initial time
[tex]t = 4*1.10 =4.40[/tex] s
b)
[tex]v=v_0+at\\a=\frac{v-v_0}{t}=\frac{32-0}{1.1/3600}=104727.27[/tex] [tex]miles/hours^2[/tex]
For final 64 mph speed,
[tex]v=v_0+at\\t=\frac{v-v_0}{a}=\frac{64-0}{104727.27} = 6.111*10^{-4}[/tex] [tex]hours[/tex] = [tex]6.111*10^{-4}*3600=2.20[/tex] s
Technician A says that low compression on a single cylinder will cause an engine not to start. Technician B says that low compression on a single cylinder means that the engine can be fixed with a tune-up. Who is correct?
Tech A
Tech B
Both A and B
Neither A nor B
Answer:
Technician B
Explanation:
Car engine compression refers to when air and gas are mixed together in the cylinders of an engine. This process is required for the car to move and function. If there are any problems with the compression process, then you can expect to experience all kinds of car problems.
It will be easy to tell when you have a low compression problem because you may experience a misfire when you try to start the engine. Either that or the engine will offer poor performance as you’re driving the vehicle down the road. The worst-case scenario would be the car not starting if all the cylinders have no compression.
Generally speaking, if you have low compression in one(single) cylinder, the engine will start but you’ll likely experience misfires and your vehicle will run rough. If you experience no compression in ALL cylinders, your engine simply won’t start. This rule out A statement.
An engine tune-up is a exercise for engines to undergo regularly. It is a way of making a car's work at the level and standard intended by the car manufacturer when the car was first made. All manufacturers will stipulate a schedule of when a car will required an engine tune-up to ensure that a car runs at it's most efficient
Engine tune-up are imperative to ensure that all the power and efficiency that your car is capable are being reached . This bring the answer to the question to Tech B.
Which of the following is defined as an area of the body surface that is innervated by a single spinal nerve?
A. Transverse process
B. Malar
C. Dermatome
D. Spinous process
Answer:
Dermatome. (Ans. C).
Explanation:
Dermatome is defined as the area of the human anatomy skin which is supplied by single spinal sensory nerve root. At the spinal cord these spinal sensory nerve enter the nerve root, and the branches of spinal sensory reach to the periphery of the body.
The sensory nerve which is present in the periphery of the body are the type of nerve which helps to transmit signals from sensation such as pain, temperature, etc. to the spinal cord from some specific area of the anatomy.