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Neko [114]
4 years ago
11

Form a hypothesis. Which will be greater, the potential energy of the car at the top of the ramp or the kinetic energy of the ca

r at the bottom of the ramp
Physics
1 answer:
AleksandrR [38]4 years ago
6 0

The potential energy of the car at the top of the ramp is greater than the kinetic energy of the car at the bottom of the ramp

Explanation:

Here we have a car moving down along a ramp. In absence of force of friction, the total mechanical energy of the car while moving down would be conserved:

E=PE+KE = const.

where

PE is the potential energy

KE is the kinetic energy

However, this is not true if friction acts on the car. When the car is still at the top of the ramp, its speed is zero, so its kinetic energy is zero, and all the energy is just potential energy:

E=PE

While the car moves down along the ramp, friction does work on it, and part of the total mechanical energy is wasted and converted into thermal energy. As a result, the car reaches the bottom of the ramp with a final energy which is less than the initial energy:

E'

Also, at the bottom of the ramp, all the energy is just kinetic energy, since the height of the car is now zero, so

E'=KE

It follows that

KE

So, the potential energy of the car at the top of the ramp is greater than the kinetic energy of the car at the bottom of the ramp.

Learn more about potential and kinetic energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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In a science museum, a 140 kg brass pendulum bob swings at the end of a 16.8 m -long wire.
Mice21 [21]

The period of the pendulum is 8.2 s

Explanation:

The period of a simple pendulum is given by the equation:

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity

T is the period

We notice that the period of a pendulum does not depend at all on its mass, but only on its length.

For the pendulum in this problem, we have

L = 16.8 m

and

g=9.8 m/s^2 (acceleration of gravity)

Therefore the period of this pendulum is

T=2\pi \sqrt{\frac{16.8}{9.8}}=8.2 s

#LearnWithBrainly

3 0
3 years ago
A 2-kg object is initially at the bottom of a long 50° inclined plane, and is beginning to slide up this inclined plane. The ini
Over [174]

Answer:

Explanation:

Given

mass of object m=2 kg

inclination \theta =50^{\circ}

\mu _k=0.3

\mu _s=0.4

initial velocity u=3 m/s

acceleration of block during upward motion

a=g\sin \theta -\mu _kg\cos \theta

a=g(\sin 50-0.3\cos 50)

a=5.617 m/s^2

using relation

v^2-u^2=2a\cdot s

where s=distance\ moved

v=final\ velocity

v=0 because block stopped after moving distance s

0-(3)^2=2\cdot (-5.617)\cdot s

s=\frac{4.5}{5.617}

s=0.801

If block stopped after s m then force acting on block is

F=mg\sin \theta =friction force f_r=\mu mg\cos \theta

F>f_r therefore block will slide back down to the bottom            

4 0
3 years ago
A 24-cm-diameter vertical cylinder is sealed at the top by a frictionless 15 kg piston. The piston is 90 cm above the bottom whe
abruzzese [7]

Answer:

(A) P_i=3249.41\ Pa

(B) V_f=0.3358\ m

Explanation:

Given:

  • diameter of the cylinder, d= 0.24\ m
  • mass of piston sealing on the top, m_p=15\ kg
  • initial temperature of the piston, T_i=315+273= 588\ K
  • initial height of piston, h_i=0.9\ m
  • atmospheric pressure on the piston, p_a=1 atm=101325\ Pa

(A)

<u>Initial pressure of gas is the pressure balanced by the weight of piston:</u>

P_i=\frac{m_p.g}{\pi.d^2\div 4}

P_i=\frac{15\times 9.8}{\pi\times (0.24^2\div 4)}

P_i=3249.41\ Pa

<em>Which is gauge pressure because it is measured with respect to the atmospheric pressure.</em>

(B)

Given:

  • Final temperature, T_f=18+273=291\ K

<u>Now, volume of air initially in the cylinder:</u>

V_i=\pi.d.h_i

V_i=\pi\times 0.24\times 0.9

V_i=0.6786\ m^3

Using gas law:

\frac{P_i V_i}{T_i}= \frac{P_f V_f}{T_f} ........................................(1)

<em>∵In every condition of equilibrium the gas pressure will be balanced by the weight of the piston so it is an </em><em>isobaric transition</em><em>.</em>

∴P_i=P_f

<u>Hence eq. (1) is reduced to:</u>

\frac{V_i}{T_i}= \frac{V_f}{T_f}

putting respective values:

\frac{0.6786}{588}= \frac{V_f}{291}

V_f=0.3358\ m

6 0
4 years ago
Can someone help me please
Vedmedyk [2.9K]
The correct answer would be

Mile
3 0
3 years ago
If the torque required to loosen a nut that is holding a flat tire in place on a car has a magnitude of 41 N · m, what minimum f
Marina86 [1]

We have that the Force  is mathematically given as

F=170.833N

<h3></h3><h3>Force </h3>

Question Parameters:

  • The torque required to loosen a nut that is holding a flat tire in place on a car has a magnitude of 41 N · m,
  • the end of a 24 cm-long  wrench to <em>loosen </em>the nut

Generally the equation for the Force   is mathematically given as

F=\frac{T}{d}

Therefore

F=41/0.24

F=170.833N

For more information on Force visit

brainly.com/question/26115859

3 0
2 years ago
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