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patriot [66]
3 years ago
6

A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep

the box moving at a constant velocity. What is the coefficient of kinetic friction between the box and the floor? 0.23 0.18 0.20
Physics
1 answer:
salantis [7]3 years ago
4 0
The box is kept in motion at constant velocity by a force of F=99 N. Constant velocity means there is no acceleration, so the resultant of the forces acting on the box is zero. Apart from the force F pushing the box, there is only another force acting on it in the horizontal direction: the frictional force F_f which acts in the opposite direction of the motion, so in the opposite direction of F.
Therefore, since the resultant of the two forces must be zero,
F-F_f=0
so
F=F_f

The frictional force can be rewritten as
F_f = \mu m g
where m=50 kg, g=9.81 m/s^2. Re-arranging, we can solve this equation to find \mu, the coefficient of dynamic friction:
\mu =  \frac{F}{mg}= \frac{99 N}{(50 kg)(9.81 m/s^2)}  =0.20
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If an automobile had a 100%-efficient engine, transferring all of the fuel's energy to work, would the engine be warm to your to
svetlana [45]

Answer:

The engine would be warm to touch, and the exhaust gases would be at ambient temperature. The engine would not vibrate nor make any noise. None of the fuel entering the engine would go unused.

Explanation:

In this ideal engine, none of these events would happen due to the nature of the efficiency.

We can define efficiency as the ratio between the used energy and the potential generable energy in the fuel.

n=W, total/(E, available).

However, in real engines the energy generated in the combustion of the fuel transforms into heat (which heates the exhost gases, and the engine therefore transfering some of this heat to the environment). Also, there are some mechanical energy loss due to vibrations and sound, which are also energy that comes from the fuel combustion.

5 0
3 years ago
State how much energy is transferred in each of the following cases: 2 grams of steam at 100 degrees Celsius condenses to water
dusya [7]

Answer:

Explanation:

When 2 gms of steam condenses to water at 100 degree latent heat of vaporization is releases which is calculated as follows

Heat released = mass x latent heat of vaporization

= 2 x 2260 = 4520 J

When 2 gms of water  at 100 degree is cooled to ice water at zero degree  heat  is releases which is calculated as follows

Heat released = mass x specific heat x( 100-0)

= 2 x 4.2 x 100 = 840 J

When 2 gms of water at zero degree  condenses to ice at zero degree latent heat of fusion  is releases which is calculated as follows

Heat released = mass x latent heat of fusion

= 2 x 334 = 668 J

When 2 grams of steam at 100 degrees Celsius turns to ice at 0 degrees Celsius heat released will be sum of all the heat released as mentioned above ie

4520 + 840 +668 = 6028 J

3 0
3 years ago
Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0
3 years ago
NEED IT ASAP PLEASE 3 examples of how we use physics in our everyday life. Please explain throughly.
Flauer [41]

Answer:

  1. Alarm Clock. The buzzing sound of an alarm clock helps you wake up in the morning as per your schedule. The sound is something that you can’t see, but hear or experience.
  2. Cell Phones Cellphones have become like Oxygen gas in modern social life. Hardly, anyone would have been untouched by the effects of a cell phone. Whether conveying any urgent message or doing incessant gossips, cellphones are everywhere. But do you know how does a cell phone work? It works on the principle of electricity and the electromagnetic spectrum, undulating patterns of electricity and magnetism.
  3. Walking.Now, when you get ready for your office/school, whatever medium of commutation is, you certainly have to walk up to a certain distance. You can easily walk is just because of Physics. While you have a walk in a park or on a tar road, you have a good grip without slipping because of a sort of roughness or resistance between the soles of your shoes and the surface of the road.

Explanation:

physical is related to things perceived through the senses as opposed to the mind; tangible or concrete.

3 0
3 years ago
A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal
Sati [7]

(a) The ball has a final velocity vector

\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j

with horizontal and vertical components, respectively,

v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}

v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}

The horizontal component of the ball's velocity is constant throughout its trajectory, so v_{x,i}=v_{x,f}, and the horizontal distance <em>x</em> that it covers after time <em>t</em> is

x=v_{x,i}t=v_{x,f}t

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s

The vertical component of the ball's velocity at time <em>t</em> is

v_{y,f}=v_{y,i}-gt

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}

So, the initial velocity vector is

\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j

which carries an initial speed of

\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}

and direction <em>θ</em> such that

\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height <em>y</em> at time <em>t</em> is

y=v_{y,i}t-\dfrac12gt^2

so that when it lands in the seats at <em>t</em> ≈ 6.38 s, it has a height of

y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}

6 0
3 years ago
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