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Juli2301 [7.4K]
3 years ago
9

A 50 kg go cart is located at the top of a 10 m tall hill….sitting motionless. It goes down the hill and rises to the top of the

second hill while have a speed of 1 m/s. How tall is the second hill?
Physics
1 answer:
dexar [7]3 years ago
6 0

Answer:

Explanation:

This is a Law of Momentum Conservbation problem, where the total energy of the system cannot increase or decrease, only change form. The total energy equation for this situation is

TE = PE + KE where TE is total energy, PE is potential energy, and KE is kinetic energy. We begin by realizing that the go-kart is motionless at the top of a hill. If the kart isn't moving, then it has no KE, but if it is up off the ground and has the potential to fall to a point lower than it is curremtly, it has potential energy. That means that the total energy available to this go-kart is found in its potential energy and will not change throughout the trip's entirety. Thus,

TE = PE + 0 and

TE = (50.0)(9.8)(10.0) so

TE = 4900 J and since that's the total energy available throughout the trip, and we are looking to find the height of the next hill where this is both potential and kinetic energy, then

4900 = PE + KE and

4900 = (50.0)(9.8)(h) + \frac{1}{2}(50.0)(1.0)^2 and

4900 = 490h + 25.0 and

4875 = 490h so

h = 9.9 m (I kinda ignored the rules for significant digits at the end, which goes against every teacher's bone in my body, but nonetheless, there's your answer!)

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A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that
VMariaS [17]

Answer:

E = 9.66\times 10^{-6} N/C

direction is Horizontal

Explanation:

As we know that the string is horizontal here

so the tension force in the string is due to electrostatic force on it

now we will have

F = qE

so here the force is tension force on it

F = 6.57 \times 10^{-2} N

Q = 6.80 \times 10^3 C

now we have

6.57 \times 10^{-2} = (6.80 \times 10^3)E

E = 9.66\times 10^{-6} N/C

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4 0
4 years ago
A skater rotates with her arms crossed at an angular speed of 8.0 rad/s and she has a moment of inertia of 100 kg/m2. She extend
solong [7]

Answer:

When her hands extends, her momen of inertia is 4.28\ kg-m^2.

Explanation:

Given that,

Initial angular speed, \omega_i=8\ rad/s

Initial moment of inertia, I_1=100\ kg-m^2

Final angular speed, \omega_f=7\ rad/s

Initially, a skater rotates with her arms crossed and finally she extends her arms. The momentum remains conserved. Using the conservation of momentum as :

I_1\omega_1=I_2\omega_2

I_2 is final moment of inertia

I_2=\dfrac{I_1\omega_1}{\omega_2}\\\\I_2=\dfrac{100\times 8}{7}\\\\I_2=114.28\ kg-m^2

So, when her hands extends, her momen of inertia is 4.28\ kg-m^2. Hence, this is the required solution.

7 0
3 years ago
What is the potential energy of a 12 kg box on a 5 m high shelf?
Flauer [41]

Answer:6

Explanation:$’n

7 0
3 years ago
2 questions: Are seismic waves mechanical or electromagnetic waves, and are seismic waves transverse or compressional waves?
pashok25 [27]
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4 0
4 years ago
If you have two linear polarizers whose extinction axis are 90 degrees relative to eachother, no light passes through the system
marin [14]

Answer:

The third polarizer can be placed midway between the first two polarizers with its extinction axis at 45° from either polarizer to maximize the amount of light that is transmitted (one-eight).

Explanation:

If light is incident on a polarizer, it allows only light that is parallel to its 'pass-through' axis to pass through untouched.

Light whose electric direction/vector is perpendicular to the 'pass through' axis will not pass through at all. Light whose electric direction/vector points in other directions (apart from those whose direction is parallel or perpendicular) passes through according to the magnitude of the component that is parallel to the 'pass-through' axis.

The polarizer blocks half of the incident light rays and the transmitted light is polarized in the direction of the 'pass-through' axis.

A new polarizer now place at a distance from the first polarizer with its 'pass-through' axis perpendicular to the first polarizer cancel out all the light that comes through from the first polarizer. Since the light electric vector needs to be parallel to the axis of the polarizer to pass through and all the parallelized light from the first polarizer are now incident perpendicularly to the axis of the second polarizer, no light rays pass through.

But, a third polarizer can be placed midway between the first two polarizers with its axis positioned at 45° from either polarizer. Thereby allowing exactly half of the light from the first polarizer to pass through. The explanation is just like that for the first one. (Light whose electric direction/vector points in other directions (apart from those whose direction is parallel or perpendicular) passes through according to the magnitude of the component that is parallel to the 'pass-through' axis).

Then the resultant from the middle polarizer reaches the initial second polarizer and half of the light is let through again. So that, at the end of the day, (1/2) × (1/2) × (1/2) of the initial incident ray is let through.

That is, to maximize the amount of light that is transmitted (one-eight of initial incident ray) the third Polaroid is place midway between the first two and at angle 45° to either one.

8 0
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