Answer:
[tex]a=4.2524\ m.s^{-2}[/tex]
Explanation:
Given:
two workers push a crate.
Force by first worker, [tex]F_1=543\ N[/tex]force by second worker, [tex]F_2=333\ N[/tex]mass of crate, [tex]m=206\ kg[/tex](Assuming that both the workers push in the same direction.)
We know that,
Acceleration is given as:
[tex]a=\frac{F}{m}[/tex]
[tex]a=\frac{F_1+F_2}{m}[/tex]
[tex]a=\frac{543+333}{206}[/tex]
[tex]a=4.2524\ m.s^{-2}[/tex]
A parachutist of mass 39.4 kg jumps out of an airplane at a height of 1340 m and lands on the ground with a speed of 5.78 m/s. The acceleration of gravity is 9.8 m/s². How much energy was lost to air friction during this jump?
Answer:
Approximately 500kJ
Explanation:
According to the law of conservation of energy which states that energy can neither be created nor destroyed but can be converted from one form to another.
A parachutist jumping out of the airplane covering a particular height and under the influence of gravity will possess potential energy during fall. Since PE = mass × acceleration due to gravity × height
PE = 39.4×1340×9.8
PE = 501,562Joules
If the body lands on the ground with a speed of 5.78m/s, this means the body possesses kinetic energy at the point of landing. The kinetic energy on landing is 1/2mv²
KE = 1/2×39.4×5.78²
KE = 658.14Joules
The amount of energy lost due to friction will be PE-KE
= 501,562-658.14
= 500,903Joules approximately 500kJ
Final answer:
The solution involves calculating the parachutist's initial potential energy and final kinetic energy, then finding the difference to determine the energy lost to air friction during the jump.
Explanation:
The question asks how much energy was lost to air friction during the jump of a parachutist who has a mass of 39.4 kg, jumps from a height of 1340 m, and lands with a speed of 5.78 m/s. To find the energy lost to air friction, we first calculate the potential energy at the start and the kinetic energy at the end of the jump, then find the difference.
Initial potential energy (PE_initial) is given by mgh, where m is mass, g is acceleration due to gravity (9.8 m/s²), and h is height. Therefore, PE_initial = 39.4 kg * 9.8 m/s² * 1340 m.
Final kinetic energy (KE_final) is given by 1/2 mv², where m is mass and v is velocity at landing. Thus, KE_final = 1/2 * 39.4 kg * (5.78 m/s)².
The energy lost to air friction equals the initial potential energy minus the final kinetic energy. Subtracting KE_final from PE_initial gives the amount of energy lost to air resistance.
If an astronaut goes on a space walk outside the Space Station, she will quickly float away from the station unless she has a tether holding her to the station. Part A Choose the correct explanation why does the statement make sense (or is clearly true) or does not make sense (or is clearly false). Choose the correct explanation why does the statement make sense (or is clearly true) or does not make sense (or is clearly false).
a. This statement is true. She and the Space Station have different orbits at the beginning and will move apart.
b. This statement is true. She and the Space Station cannot share the same orbit and will move apart quickly.
c. This statement is false. She and the Space Station have different orbits at the beginning but will stay together due to mutual gravity.
d. This statement is false. She and the Space Station share the same orbit and will stay together unless they are pushed apart.
Answer:
d. This statement is false. She and the Space Station share the same orbit and will stay together unless they are pushed apart.
Explanation:
In astronomy, orbit is simply a path of an object around another object in a space. That is, orbit is a path of a body that revolves around a gravitating center of mass. Examples of an orbit is are satellite around a planet, orbit around a center of galaxy, planet around the sun, and among others.
On the other hand, space station refers to a spacecraft that can support a group of human for long time in the orbit. Another names for space stations are orbital space station and orbital station.
Therefore, an astronaut goes on a space walk outside the Space Station shares the same orbit with the space station and they will stay together unless they are pushed apart.
The statement is true that an astronaut would float away during a spacewalk if not tethered. The astronaut and ISS, although influencing each other gravitationally to a small extent, they would continue along their separate paths under the influence of Earth's gravity and microgravity.
Explanation:
This statement is true. When an astronaut steps out for a spacewalk, she and the International Space Station (ISS) are both in orbit around Earth but the astronaut will not stay in place relative to the ISS without a tether holding her to the station.
The reason for this is due to microgravity and the properties of motion in space. The astronaut and the ISS are both in free-fall around the Earth, so in the absence of other forces, they would continue along their separate paths. Without a tether, even a small force (like the push off the astronaut does to get away from the airlock) can cause her to drift away from the station.
While the ISS and the astronaut do influence each other gravitationally, this effect is extremely small compared to the force of Earth's gravity. So, without a tether, the astronaut can float away from the ISS.
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A flying dragon is rising vertically at a constant speed of 6.0m/s. When the dragon is 30.0m above the ground, the rider on its back drops a small golden egg which, subsequently, is in free fall.
a) What is the maximum height above the ground reached by the egg?
b) How long after its release does the egg hit the ground?
c) What is the egg’s velocity immediately before it hits the ground?
d) Sketch, qualitatively, position, velocity, and acceleration of the egg as functions of time.
Final answer:
The detailed response covers the maximum height reached by the egg, the time it takes to hit the ground, and its velocity just before landing.a)60.0m,b)2.45 seconds,c)-6.0m/s.
Explanation:
a) Maximum Height: The maximum height above the ground reached by the egg can be calculated using the kinematic equation. It will be twice the initial height. In this case, it would be 60.0m.
b) Time to Hit Ground: You can determine the time it takes for the egg to hit the ground by using the kinematic equation for vertical motion. The time would be approximately 2.45 seconds after its release.
c) Velocity Before Landing: The velocity of the egg immediately before hitting the ground would be the same as the initial velocity when it was thrown, but in the opposite direction, which is -6.0m/s.
Water behind a dam has a certain amount of stored energy that can be released as the water falls over the top of the dam. It may be enough energy to turn a mill wheel or an electricity-generating turbine. Choose the term that best describes the type of energy stored in the water at the top of the dam.
Answer:
Gravitational potential energy
Explanation:
Gravitational potential energy is the type of energy an object has due to its position in a gravitational field. Water behind a dam possesses gravitational potential energy due to it being at a higher level than the water on the other side of the dam. When the water falls the gravitational potential energy is converted to kinetic energy, leading to the turning of the turbines to generate electricity.
Based on the data thomson collected in his experiments using cathode rays, the concept of atomicc structure was modified. What were the four things validated by his cathode ray expirement?
Answer:
Based on the data thomson collected in his experiments using cathode rays, the concept of atomicc structure was modified. What were the four things validated by his cathode ray expirement?
Explanation:
Cathode Ray tubes' Experiment of J.J. Thomson, showed that atoms contain electrons or tiny negative charged subatomic particles.
Based on the Thomson's cathode ray tubes experiment, He validated four things which are as following
1. Cathode rays have mass.
2. Matter contains positive and negative charge.
3. Particles of cathode rays are fundamental to all matter.
4. An atom is divisible.
A capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery that maintains a constant potential difference V across the plates. If the separation between the plates is doubled, the electrical energy stored in the capacitor will be:_______
a- doubled
b- unchanged
c- quadrupled
d- quartered
e- halved
Answer:e
Explanation:
Given
Area of parallel plates is A
distance between plates is d
Potential difference between Plates is V
Capacitance is given by
[tex]C=\frac{\epsilon _0A}{d}[/tex]
If separation is doubled then capacitance become half
[tex]C'=\frac{\epsilon _0A}{2d}[/tex]
[tex]C'=\frac{C}{2}[/tex]
Electrical energy stored in the capacitor is given by
[tex]E=\frac{1}{2}CV^2[/tex]
When distance is doubled
[tex]E'=\frac{1}{2}\times \frac{C}{2}\times V^2[/tex]
[tex]E'=\frac{E}{2}[/tex]
Therefore Energy is halved
If the separation between the plates is doubled, the electrical energy stored in the capacitor will be: d. The electrical energy stored in the capacitor will be quartered.
To understand why the energy stored in the capacitor is quartered when the separation between the plates is doubled, let's consider the formula for the capacitance of a parallel plate capacitor and the energy stored in a capacitor.
The capacitance C of a parallel plate capacitor is given by:
[tex]\[ C = \frac{\varepsilon_0 A}{d} \][/tex]
The energy U stored in a capacitor is given by:
[tex]\[ U = \frac{1}{2} C V^2 \][/tex]
[tex]\[ C' = \frac{\varepsilon_0 A}{2d} = \frac{1}{2} \frac{\varepsilon_0 A}{d} = \frac{1}{2} C \][/tex]
[tex]\[ U' = \frac{1}{2} C' V^2 = \frac{1}{2} \left(\frac{1}{2} C\right) V^2 = \frac{1}{4} C V^2 = \frac{1}{4} U \][/tex]
So, when the separation between the plates is doubled, the capacitance is halved, and since the energy is directly proportional to the capacitance, the energy stored in the capacitor is quartered.
A 0.500 kg pendulum bob passes through the lowest part of its path at a speed of 3.40 m/s.
(a) What is the tension in the pendulum cable at this point if the pendulum is 80.0 cm long? (in Newtons)
(b) When the pendulum reaches its highest point, what angle does the cable make with the vertical? (in Degrees)
(c) What is the tension in the pendulum cable when the pendulum reaches its highest point? (in Newtons)
Final answer:
The tension in the pendulum at the lowest point is 12.1 N. Without additional information, we cannot determine the exact angle at which the pendulum reaches its highest point. The tension in the pendulum at the highest point is 4.9 N.
Explanation:
Tension in the Pendulum Cable
To determine the tension in the pendulum cable at the lowest point, we can use the formula for circular motion.
T = mg + m(v^2/r)
Where:
g is the acceleration due to gravity (9.8 m/s²)
r is the length of the pendulum
Substituting the values given, we have:
T = 0.500 kg * 9.8 m/s² + 0.500 kg * (3.40 m/s)^2 / 0.80 m
Now, we can calculate the tension:
T = 4.9 N + 7.225 N
T ≈ 12.1 N
The Angle at the Highest Point
To find the angle at the highest point, we use the principle of conservation of energy. The kinetic energy at the lowest point equals potential energy at the highest point.
mgh = 1/2 mv^2
Solving for h gives us h = v^2/(2g), and then use trigonometry to find the angle θ.
h = r(1 - cosθ)
Without sufficient information to calculate the speed of the pendulum, we must leave the answer as an open calculation.
Tension at the Highest Point
At the highest point, the tension is equal to the weight of the pendulum alone since it is momentarily at rest, and there is no centripetal force required.
T = mg
Thus, the tension at the highest point will be:
T = 0.500 kg * 9.8 m/s² = 4.9 N.
Juanita lifts a round box and a square box to a shelf. The gravitational potential energy (GPE) for the round box increases by 50 J. The GPE for the square box increases by 100 J. On which box did Juanita do more work? Explain your reasoning.
Juanita did more work on the square box
Explanation:
According to the law of conservation of energy, the work done in lifting an object is equal to the increase in gravitational potential energy of the object. This is due to the fact that the work done on the object is converted into potential energy.
The potential energy of an object (GPE) is given by
[tex]GPE=mgh[/tex]
where
m is the mass of the object
g is the acceleration of gravity
h is the height of the object
In this problem, we have two objects:
- The roud box is lifted and its GPE increases by 50 J --> this means that the work done by Juanita on the box is 50 J
- Thr square box is lifted and its GPE increases by 100 J --> this means that the work done by Juanita on the box is 100 J
Therefore, Juanita did more work on the square box.
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The square box received more work by Juanita than the round box based on the increase in gravitational potential energy.
The square box received more work done by Juanita compared to the round box. When the gravitational potential energy increase for the square box is 100 J, which is greater than the 50 J increase in the round box, it indicates that Juanita did more work on the square box.
A localized electron has been polarized so that its spin is oriented in the positive z- direction. It is now subject to the application of a constant uniform magnetic field B = Bî along x over a period of time of duration T. After that, it is subject to the application of another magnetic field of the same magnitude B but along y: B2 = Bý, also with duration T. (a) What is the probability P that the spin-flip would occur as a result? That is, what is the probability that the spin of the electron would be found oriented in the negative z-direction after the application of the magnetic fields is over? (b) Is it possible to find such duration that the spin-flip would occur with certainty: P = 1? If yes, what would be that time r?
Answer: The answer is attached
Explanation:
Final answer:
In quantum mechanics, the probability of a spin-flip and the potential for achieving a spin flip with certainty are explored through the application of magnetic fields in different directions.
Explanation:
P(a): The probability that the spin-flip would occur can be calculated using quantum mechanics. When applying magnetic fields B along x and y, the probability of finding the spin of the electron oriented in the negative z-direction after is over is determined by the transition probabilities.
P(b): The time to ensure a spin flip with certainty (P = 1) can be calculated by adjusting the duration of the magnetic field application. By manipulating the duration, the spin can be controlled to achieve a certain outcome at a specific time.
In one contest at the county fair, a spring-loaded plunger launches a ball at a speed of 3.2m/s from one corner of a smooth, flat board that is tilted up at a 20 degree angle. To win, you must make the ball hit a small target at the adjacent corner, 2.40m away. At what angle theta should you tilt the ball launcher?
Answer:
Explanation:
Given
Speed of ball [tex]u=3.2\ m/s[/tex]
Plane is inclined at an angle [tex]20^{\circ}[/tex]
To win the Game we need to hit the target at [tex]x=2.4\ m[/tex] away
Launch angle of ball [tex]\theta [/tex]
Motion of ball can be considered in two planes i.e. Vertical to the plane and horizontal to the plane
So Net acceleration in vertical plane is [tex]g\sin 20[/tex]
Range of Projectile is given by
[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]
for [tex]R=2.4\ m[/tex]
[tex]2.4=\frac{3.2^2\times sin 2\theta }{g\sin 20}[/tex]
[tex]\sin 2\theta =\frac{2.4\times 9.8\times \sin 20}{3.2^2}[/tex]
[tex]\sin 2\theta =0.7855[/tex]
[tex]2\theta =51.77[/tex]
[tex]\theta =25.88^{\circ}[/tex]
so ball must be launched at an angle of [tex]25.88^{\circ}[/tex]
To answer this, we use projectile motion principles with an equation specifically structured for the problem. Inserting the given values into the equation derived from the horizontal and vertical equations of motion, we can find the required launching angle to hit the target.
Explanation:To solve this problem, we can apply the concepts found in projectile motion physics. Given the initial speed or velocity (3.2 m/s) and the horizontal distance of the target (2.4 m) we can find the necessary angle (theta) to hit the target. The angle θ can be given by the equation of motion for a projectile which can be derived from the horizontal and vertical equations of motion, θ = atan[(vf² ± sqrt(vf⁴ - g*(g*x² + 2*y*vf²)) / (g*x)] where g = 9.8 m/s² is the acceleration due to gravity, x = 2.4 m is the horizontal distance, y = 0 (height difference), and vf = 3.2 m/s is the final velocity, the speed at which the ball is launched.
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A particle moves along the x-axis according to x(t)=10t−2t²m. (a) What is the instantaneous velocity at t = 2 s and t = 3 s? (b) What is the instantaneous speed at these times? (c) What is the average velocity between t = 2 s and t = 3 s?
Answer:
a) v(2) = 2m/s, v(3) = -2m/s
b) speed at t = 2s is 2m/s
speed at t = 3s is 2m/s
c) 0 m/s
Explanation:
We can take the derivative of x(t) to find the equation of velocity
v(t) = x'(t) = 10 - 4t
(a) v(2) = 10 - 4*2 = 10 - 8 = 2 m/s
v(3) = 10 - 4*3 = 10 - 12 = -2 m/s
(b) The speed would be the same as velocity without the direction
speed at t = 2s is 2m/s
speed at t = 3s is 2m/s
(c) The average velocity between t = 2s and t = 3s is distance it travels over period of time
[tex]v_a = \frac{s(3) - s(2)}{\Delta t} = \frac{10*3 - 2*3^2 - (10*2 - 2*2^2)}{3 - 2}[/tex]
[tex]v_a = \frac{12 - 12}{1} = 0/1 = 0 m/s[/tex]
Final answer:
The instantaneous velocity at t = 2 s is 2 m/s and at t = 3 s is -2 m/s. The instantaneous speed at both times is 2 m/s. The average velocity between t = 2 s and t = 3 s is 12 m/s.
Explanation:
(a) To find the instantaneous velocity, we need to find the derivative of the position function x(t) with respect to time. The derivative of x(t) = 10t - 2t² is v(t) = 10 - 4t. Substituting t = 2 and t = 3 into v(t), we get v(2) = 10 - 4(2) = 2 m/s and v(3) = 10 - 4(3) = -2 m/s.
(b) The instantaneous speed is the magnitude of the instantaneous velocity. Since speed is always positive, the speed at t = 2 s and t = 3 s is 2 m/s for both.
(c) The average velocity between t = 2 s and t = 3 s is given by the change in position divided by the change in time. The change in position is x(3) - x(2) = (10(3) - 2(3)²) - (10(2) - 2(2)²) = 12 m, and the change in time is 3 s - 2 s = 1 s. Therefore, the average velocity is 12 m/1 s = 12 m/s.
write a statement that warns people about the presence of peroxide in hair dyes
Answer:
Hydrogen peroxides are known to be an active ingredient in hair dyes. Hydrogen peroxide is a damaging chemical although it has been added in diluted amounts in hair dyes. Some hair dye companies have already started looking for alternatives for hydrogen peroxide to use in their products, concerning the damaging effects of the chemical.
Hydrogen peroxide can cause hair loss, dermatitis and scalp burns.
Concerning the harmful chemicals in dyes, many people consider it is best to use henna if you really want a colour change for your hair.
A bullet is fired through a wooden board with a thickness of 9.0 cm. The bullet hits the board perpendicular to it, and with a speed of +370 m/s. The bullet then emerges on the other side of the board with a speed of +259 m/s. Assuming constant acceleration (rather, deceleration!) of the bullet while inside the wooden board, calculate the acceleration.
Remember that since the bullet is traveling in the positive direction and it is slowing down, the acceleration is in the opposite, or negative, direction. (In this case the acceleration is negative.)
Calculate also the total time the bullet is in contact with the board (in sec).
Answer:
-387883.3 m/s²
0.000286168546055 seconds
Explanation:
t = Time taken
u = Initial velocity = 370 m/s
v = Final velocity = 259 m/s
s = Displacement = 9 cm
a = Acceleration
[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{259^2-370^2}{2\times 9\times 10^{-2}}\\\Rightarrow a=-387883.3\ m/s^2[/tex]
The acceleration of the bullet is -387883.3 m/s²
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{259-370}{-387883.3}\\\Rightarrow t=0.000286168546055\ s[/tex]
The time taken is 0.000286168546055 seconds
Final answer:
The acceleration of the bullet is approximately -111 m/s^2 and the total time the bullet is in contact with the board is approximately 0.00081 seconds.
Explanation:
The acceleration of the bullet can be calculated using the equation of motion:
a = (v - u) / t
where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time. In this case, the initial velocity of the bullet is 370 m/s (positive because it is in the direction of motion) and the final velocity is 259 m/s (positive because it is also in the direction of motion). The time it takes for the bullet to decelerate can be calculated using the equation:
x = ut + (1/2)at^2
where x is the displacement, u is the initial velocity, t is the time, and a is the acceleration. In this case, the displacement is the thickness of the wooden board, which is 9.0 cm (or 0.09 m). Rearranging the equation, we get:
t = (v - u) / a
Substituting the values, we have:
t = (259 - 370) / a
Combining the equations for time and acceleration, we get:
t = (0.09) / (370 - 259) = (0.09) / (111) = 0.00081 seconds.
So, the acceleration of the bullet is approximately -111 m/s^2 (negative because it is in the opposite direction of the bullet's motion) and the total time the bullet is in contact with the board is approximately 0.00081 seconds.
What frequency (in Hz) is received by a person watching an oncoming ambulance moving at 108 km/h and emitting a steady 900 Hz sound from its siren? The speed of sound on this day is 345 m/s.
Answer:985.71 Hz
Explanation:
Given
Speed of ambulance (source) [tex]v_s=108\ kmph\approx 30\ m/s[/tex]
Original Frequency of sound wave [tex]f=900\ Hz[/tex]
speed of sound waves [tex]v=345\ m/s[/tex]
According to Doppler apparent frequency will be different to original frequency when speed of source is moving w.r.t to observer.
[tex]f'=f(\frac{v+v_o}{v-v_s})[/tex]
where f'=apparent frequency
f=original frequency
[tex]v_o[/tex]=observer speed
[tex]f'=900\times (\frac{345+0}{345-30})[/tex]
[tex]f'=900\times (\frac{345}{315})[/tex]
[tex]f'=985.71\ Hz[/tex]
An object moves 15.0 m north and then 11.0 m south. Find both the distance traveled and the magnitude of the displacement vector.
Answer:
Distance = 26.0m Displacement = 4.0m
Explanation:
Distance specifies only how far an object has traveled while displacement is the distance traveled in a specified direction.
Total distance traveled by the object will be distance travelled through north + distance travelled through south i.e 15.0m + 11.0m = 26.0m
Displacement is gotten by using the Pythagoras theorem. Since the object traveled in the same vertical direction (15.0m through north which is upward i.e positive y direction and 11.0m through south i.e in the negative y direction), the displacement will be 15.0m - 11.0m = 4.0m
The distance traveled is 26.0 m and the magnitude of the displacement vector is 4.0 m
First, we will define the terms distance and displacement
Distance is the total movement of an object without any regard to direction.
Displacement is the difference between the original and final position of a path taken by an object.
Since, the object moves 15.0 m north and then 11.0 m south,
Then,
Distance traveled = 15.0 m + 11.0 m
Distance traveled = 26.0 m
For the magnitude of the displacement,
The object moves 15.0 m north and then 11.0 m south, which is in the opposite (negative) direction
Then,
Magnitude of displacement = 15.0 m - 11.0 m
Magnitude of displacement = 4.0 m
Hence, the distance traveled is 26.0 m and the magnitude of the displacement vector is 4.0 m
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A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 20 m/s. It is hit straight back at the pitcher with a final speed of 25 m/s. Assume the direction of the initial motion of the baseball to be positive.(a) What is the impulse delivered to the ball? (b) Find the average force exerted by the bat on the ball if the two are in contact for 2.0?
Answer:
a.-6.75 kgm/s
b.[tex]3375 N[/tex]
Explanation:
We are given that
Mass of baseball=0.15 kg
Initial speed=[tex]u=20m/s[/tex]
Final speed=[tex]v=-25m/s[/tex]
a.We know that
Impulse=Change in momentum=[tex]\Delta p=mv-mu=m(v-u)[/tex]
Momentum=[tex]mass\times velocity[/tex]
Using the formula
Impulse=[tex]0.15(-25-20)=-6.75 kgm/s[/tex]
b.Time=[tex]2\times 10^{-3} s[/tex]
Force=[tex]\frac{Impulse}{time}[/tex]
Using the formula
Average force exerted by the bat on the ball=[tex]\frac{-6.75}{2\times 10^{-3}}[/tex] N
Average force exerted by the bat on the ball=[tex]3375N[/tex]
The height of a typical playground slide is about 6 ft and it rises at an angle of 30 ∘ above the horizontal.
a.)Some children like to slide down while sitting on a sheet of wax paper. This makes the friction force exerted by the slide very small. If a child starts from rest and we take the friction force to be zero, what is the speed of the child when he reaches the bottom of the slide?
b.)If the child doesn't use the wax paper, his speed at the bottom is half the value calculated in part A. What is the coefficient of kinetic friction between the child and the slide when wax paper isn't used? μk = ?
I found the answer to part a, it is 3m/s, but I need part b. Please show all steps, thank you!
Answer:
What is the coefficient of kinetic friction = 0.432
Explanation:
The detailed steps and derivation with appropriate substitution is as shown in the attached file.
This question involves the concepts of the law of conservation of energy and frictional energy.
a) The speed of the child when he reaches the bottom of the slide is "6 m/s".
b) The coefficient of kinetic friction between the child and slide when the wax paper isn't used is "0.432".
a)
According to the law of conservation of energy in this situation:
Loss in Potential Energy = Gain in Kinetic Energy
[tex]mgh = \frac{1}{2}mv^2\\\\2gh=v^2\\v=\sqrt{2gh}\\\\[/tex]
where,
v = velocity = ?
g = acceleration due to gravity = 9.81 m/s²
h = height lost = 6 ft = 1.83 m
Therefore,
[tex]v=\sqrt{2(9.81\ m/s^2)(1.83\ m)}[/tex]
v = 6 m/s
b)
Now, the velocity becomes half and the friction comes into action. So in this case the law of conservation of energy will be written as:
Loss of Potential Energy = Gain of Kinetic Energy + Frictional Energy
[tex]mgh=\frac{1}{2}m(\frac{v}{2})^2+\mu_kRl\\\\mgh=\frac{1}{2}m(\frac{v}{2})^2+\mu_kmgCos\theta l\\\\gh-\frac{1}{2}(\frac{v}{2})^2+\mu_kgCos\theta l[/tex]
where,
l = length of slide = [tex]\frac{h}{sin\theta}=\frac{1.83\ m}{sin30^o}=3.66\ m[/tex]
[tex]\mu_k[/tex] = coefficient of kinetic friction = ?
Therefore,
[tex](9.81\ m/s^2)(1.83\ m)-\frac{1}{2}(\frac{6\ m/s}{2})^2=\mu_k(9.81\ m/s^2)(Cos30^o)(3.66\ m)\\\\\mu_k=\frac{13.45\ m^2/s^2}{31.09\ m^2/s^2}\\\\\mu_k=0.432[/tex]
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The attached picture explains the law of conservation of energy.
The volume of gas in a container is 125,000 liters, and the pressure is 1.2 atmospheres. Suppose the temperature remains constant, and the pressure changes to 1.6 atmospheres. What is the new volume of the gas in liters?
Answer:
the new volume =93,750 Liters
Explanation:
From Boyles law
P1V1 = P2V2
1.2 X 125000 = 1.6 X V2
V2 = 93,750 Liters
Connor is a basketball player who makes, on average, 65% of her free throws. Asume each shot is independent. SHOW ALL WORK FOR FULL CREDIT. A) What is the probability Connor will miss three straight free throws before she makes one? (If you use your calculator to get your answer, make sure to show the key strokes in which you got your answer). Write your answer in Standard notation. B) During a season, Connor takes 150 free throws. What is the probability that he will make at least 100 out of 150 of these throws? C) What is the probability that he will make no more than 110 out of 150 free throws?
Answer:
DUNNO
Explanation:
BIT BAD FOR 65% WOULD GET -100%
Calculate the speed of a proton having a kinetic energy of 0.995 × 10−19 J and a mass of 1.673 × 10−27 kg. Answer in units of m/s.
Answer:
1.19×10²² m/s
Explanation:
Kinetic Energy: This can be defined as the the energy of a body due to motion.
The formula for kinetic energy is given as,
Ek = 1/2mv²................... Equation 1
Where Ek = Kinetic energy, m = mass of proton. v = velocity of proton.
Making v the subject of the equation,
v = √(2Ek/m).................. Equation 2.
Given: Ek = 0.995×10⁻⁵ J, m = 1.673×10⁻²⁷ kg.
Substitute into equation 2
v = √(2×0.995×10⁻⁵/1.673×10⁻²⁷ )
v = 1.19×10²² m/s.
Therefore, the velocity of proton = 1.19×10²² m/s
Technician A says dc circuits are not normally used in automotive circuits. Technician B says certain automotive computer sensors use ac. Who is right?
Answer:
A & B are correct
Explanation:
Certain automotive sensors like crank shaft position sensor uses a.c. while most automotive components require DC to work properly through the battery which is incorporated with an alternator to keep it charged through a diode while continual supply of DC is sustained.
Technician A is incorrect as DC circuits are commonly used in automobiles, and Technician B is partially correct since some automotive sensors can generate an AC signal.
Explanation:In the scenario presented, Technician A says that dc circuits are not normally used in automotive circuits, which is incorrect. DC circuits are indeed commonly used in automobiles for various functions such as starting the engine, powering lights, and running the dashboard instruments. Technician B says that certain automotive computer sensors use ac current. While most sensors in a car operate on DC current, there are some sensors, like variable reluctance sensors, that can generate an AC signal as they operate. Therefore, both statements are not entirely accurate; however, Technician B's statement has some validity regarding the existence of sensors that produce an AC signal within automotive applications.
According to the cartoon video on physics, electrons sent through 2 slits at once without measuring what slit is goes through, will produce what?
A) An interference pattern
B) A single line
C) 4 lines
D) A blank screen
Answer:
A) An interference pattern
Explanation:
the two slit experiment is key to understand the microscopic world. The wave-like properties of light were demonstrated by the famous experiment first performed by Thomas Young in the early nineteenth century. In original experiment, a point source of light illuminates two narrow adjacent slits in a screen, and the image of the light that passes through the slits is observed on a second screen.
Key Points
waves can interfere, for light this will make a series of light and dark bands matter particles, such as electrons, also produce interference patterns due to their wave-like nature so with a high flux of either photons or electrons, the characteristic interference pattern is visibleConsider a spherical Gaussian surface and three charges: q1 = 1.60 μC , q2 = -2.61 μC , and q3 = 3.67 μC . Find the electric flux through the Gaussian surface if it completely encloses (a) only charges q1 and q2, (b) only charges q2 and q3, and (c) all three charges.
Answer:
Explanation:
Guass Law: Also known as "Gauss's flux theorem" is the total of the electric flux "φ" out of a closed surface is equal to the charge "Q" enclosed divided by the permittivity εο. Solution is attached.The electric flux through a Gaussian surface can be calculated using Gauss's law. (a) Calculate electric flux for q1 and q2, (b) Calculate electric flux for q2 and q3, (c) Calculate electric flux for all three charges.
Explanation:The electric flux through a Gaussian surface can be calculated using Gauss's law. Gauss's law states that the total electric flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space.
(a) To find the electric flux through the Gaussian surface enclosing only charges q1 and q2, we need to calculate the net charge enclosed by the surface, which is the sum of the two charges. Then, we divide this sum by the permittivity of free space to obtain the electric flux.
(b) Following the same procedure, we can find the electric flux through the Gaussian surface enclosing only charges q2 and q3.
(c) To find the electric flux through the Gaussian surface enclosing all three charges, we calculate the net charge enclosed by the surface, which is the sum of all three charges. Again, we divide this sum by the permittivity of free space to obtain the electric flux.
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A transformer has a 240 V primary and a 60 V secondary. With a 5 ohm load connected, what is the primary current?
The primary current for the given transformer is 3 A which is determined by considering the principle of a transformer, the turns ratio and the law of conservation of power, and calculating secondary current using Ohm's law.
Explanation:To find the primary current of a transformer, we need to first understand the principle of a transformer. A transformer operates on the electromagnetic induction principle, and the ratio of turns in the primary and secondary coil determines how much voltage is changed. In your transformer, the primary voltage is 240 V and the secondary voltage is 60 V. Therefore, the turns ratio is 240/60 which gives a value of 4.
Now, considering the law of conservation of power in an ideal transformer, the power output is equal to the power input. Power is given by the product of voltage and current.
So, primary voltage (V1) times primary current (I1) is equal to secondary voltage (V2) times secondary current (I2). The secondary current (I2) can be calculated using Ohm's law as V2/R, where R is the resistance of the load connected. Plugging in the given values we get
I2 = 60/5 = 12 A.
Now, substituting the known variables in the power conservation equation we get
240*I1 = 60*12.
Solving for I1 gives a value of 3A.
Therefore, the primary current for the given transformer is 3 A.
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A 100-watt electric incandescent light bulb consumes __________ J of energy in 24 hours. [1 Watt (W)
Answer : The energy consumed by bulb in 24 hours is, 8.64 × 10⁶ J
Explanation :
As we are given that:
1 watt = 1 J/s
So,
100 watt = 100 J/s
Now we have to calculate the energy consumed by bulb in 24 hours.
As we know that:
1 hr = 60 min
1 min= 60 sec
So,
24 hr = 24 × 60 × 60 sec = 86400 sec
As, the energy consumed by bulb in 1 second = 100 J
So, the energy consumed by bulb in 86400 second = 86400 × 100 J
= 8.64 × 10⁶ J
Thus, the energy consumed by bulb in 24 hours is, 8.64 × 10⁶ J
To calculate the energy consumed by a 100-watt electric incandescent light bulb in 24 hours, we use the formula E = Pt, where E is the energy in joules (J), P is the power in watts (W), and t is the time in seconds (s). In this case, the energy consumed is 8640000 J.
Explanation:To calculate the electrical energy used by a 100-watt electric incandescent light bulb in 24 hours, we can use the formula E = Pt, where E is the energy in joules (J), P is the power in watts (W), and t is the time in seconds (s).
First, we need to convert the time from hours to seconds. There are 3600 seconds in an hour. So, 24 hours is equal to 24 x 3600 = 86400 seconds.
Now, we can substitute the values into the formula. E = 100W x 86400s = 8640000 J.
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When you jump from an elevated position you usually bend your knees upon reaching the ground. By doing this, you make the time of the impact about 10 times as great as for a stiff-legged landing. In this way the average force your body experiences is ________.a. less than 1/10 as great. b. more than 1/10 as great. c. about 1/10 as great.d. about 10 times as great.
Answer:
c. about 1/10 as great.
Explanation:
While jumping form a certain height when we bend our knees upon reaching the ground such that the time taken to come to complete rest is increased by 10 times then the impact force gets reduced to one-tenth of the initial value when we would not do so.
This is in accordance with the Newton's second law of motion which states that the rate of change in velocity is directly proportional to the force applied on the body.
Mathematically:
[tex]F\propto\frac{d}{dt} (p)[/tex]
[tex]\Rightarrow F=\frac{d}{dt} (m.v)[/tex]
since mass is constant
[tex]F=m\frac{d}{dt}v[/tex]
when [tex]dt=10t[/tex]
then,
[tex]F'=m.\frac{v}{10\times t}[/tex]
[tex]F'=\frac{1}{10} \times \frac{m.v}{t}[/tex]
[tex]F'=\frac{F}{10}[/tex] the body will experience the tenth part of the maximum force.
where:
[tex]\frac{d}{dt} =[/tex] represents the rate of change in dependent quantity with respect to time
[tex]p=[/tex] momentum
[tex]m=[/tex] mass of the person jumping
[tex]v=[/tex] velocity of the body while hitting the ground.
Erica (39kg \;kg ) and Danny (45kg \;kg ) are bouncing on a trampoline. Just as Erica reaches the high point of her bounce, Danny is moving upward past her at 4.8m/s \; m/s . At that instant he grabs hold of her.
What is their speed just after he grabs her?
help please
can you show formula too
Answer:
2.57 m/s
Explanation:
mass of Erica (ME) = 39 kg
velocity of Erica (VE) = 0 m/s (since she is at her high point of her jump, her velocity will be 0)
mass of Danny (MD) = 45 kg
velocity of Danny (VD) = 4.8 m/s
from the conservation of momentum total initial momentum is equal to the total final momentum
total initial momentum = final initial momentum
(ME)(VE) + (MD)(VD) = (ME + MD) V
where "V" is the velocity of Erica and Danny after Danny grabs her.
(39 x 0) + (45 x 4.8) = (39 + 45) x V
0 + 216 = 84 V
V = 216 / 84 = 2.57 m/s
The speed just after Danny grabs Erica is equal to 2.57 m/s.
Given the following data:
Mass of Erica = 39 kgMass of Danny = 45 kgVelocity of Erica = 0 m/s (since she is at the highest point)Velocity of Danny = 4.8 m/sTo determine their speed just after Danny grabs Erica, we would apply the law of conservation of momentum:
[tex]M_EV_E + M_DV_D = (M_E + M_D)V_{f}[/tex]
Where:
[tex]M_E[/tex] is the mass of Erica.[tex]M_D[/tex] is the mass of Danny.[tex]V_f[/tex] is the final speed.[tex]V_E[/tex] is the speed of the Erica.[tex]V_D[/tex] is the speed of the Danny.Substituting the given parameters into the formula, we have;
[tex]39 \times 0 + 45 \times 4.8 = (39 +45)V_f\\\\0+216=84V_f\\\\V_f = \frac{216}{84} \\\\V_f = 2.57 \;m/s[/tex]
Final speed = 2.57 m/s
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A distant star is moving toward the earth at a speed of 1/4 the speed of light. Compared to the light from a flashlight on the earth, the speed of the light from the star would be__________
Answer: THE SAME
Explanation: visible light is an electromagnetic wave, which has the properties of both magnet and electrical. The speed of light in air has Generally been estimated to be
299,792 kilometers per second. For for a distant start moving towards the Earth at a speed one-fourth (1/4th) of 299,792kilometers per second compared to a flashing light from the Earth,the speed of the light from the star is expected to be the same.
A light-year is _________. a. about 10 trillion kilometers the time it takes light to travel around the Sun b. about 300,000 kilometers per second the time it takes light to reach the nearest star
Answer:
about 10 trillion kilometers
Explanation:
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
t = Seconds in one year = [tex]365.25\times 24\times 60\times 60[/tex]
1 Light year
[tex]1\ ly=ct\\\Rightarrow 1\ ly=3\times 10^8\times 365.25\times 24\times 60\times 60=9.46728\times 10^{15}\ m\\ =9.46728\times 10^{15}\times 10^{-3}\\ =9.46728\times 10^{12}\ km\approx 10\ trillion\ km[/tex]
The answer is a. about 10 trillion kilometers
A friend of yours who takes her astronomy class very seriously challenges you to a contest to find the thinnest crescent moon you can find just after new moon? What time of day is best for looking for this very thin crescent?
Answer:
after the sun sets or just as it is setting
Explanation:
a crescent moon is thin and reflects less sunlight during the daylight sky so it becomes difficult to spot, but can be spotted when the sun is setting or just sets.
The best time to find the thinnest crescent moon just after the new moon is in the early evening, just after sunset, when the moon starts to reflect sunlight towards the earth, showing the crescent shape. The appearance of the Moon's surface can vary significantly with its phase and this can be better viewed through binoculars.
To find the thinnest crescent moon just after new moon, the best time to look is usually in the early evening just after sunset. The Moon, moving eastward each day in its 30 days cycle around the Earth, moves roughly 12° in the sky each day. A day or two after the new phase, the thin crescent first appears, as we begin to see a small part of the Moon's illuminated hemisphere reflecting a little sunlight toward us.
Because the Moon is moving eastward away from the Sun, it rises later and later each day. Therefore, after a new moon, the thin crescent will be seen in the west just after sunset.
Keep in mind that the bright crescent increases in size on successive days as the Moon moves farther and farther around the sky away from the direction of the Sun.
Bear in mind that the brighter the Moon is in the night sky, the harder it is to see the faint flashes of meteors. Furthermore, as seen through a good pair of binoculars, the appearance of the Moon's surface changes dramatically with its phase, revealing more topographic details when sunlight streams in from the side, causing topographic features to cast sharp shadows.
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