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3 years ago

A car moves around a circular track at a constant rate. What must change?

Physics

2 answers:

suter [353]3 years ago

4 0

B. only its velocity should change

Diano4ka-milaya [45]3 years ago

4 0

The answer should be B. only it's velocity.

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A 12.0kg microwave oven is pushed 14.0m up the sloping surface of a loading ramp inclined at an angle 37 degrees above the horiz

Semenov [28]

Answer:

Answered

Explanation:

a) What is the work done on the oven by the force F?

W = F * x

W = 120 N * (14.0 cos(37))

<<<< (x component)

W = 1341.71

b) $F_f=\mu_k N$

$F_f=0.25\times12\times9.8$

= 29.4 N

$W_f= F_f\times x$

$W_f= 29.0\times 14 cos(37)$

W_f= 328.72 J = 329 J

c) increase in the internal energy

U_2 = mgh

= 12*9.81*14sin(37)

= 991 J

d) the increase in oven's kinetic energy

U_1 + K_1 + W_other = U_2 + K_2

0 + 0 + (W_F - W_f ) = U_2 + K_2

1341.71 J - 329 J - 991 J = K_2

K_2 = 21.71 J

e) F - F_f = ma

(120N - 29.4N ) / 12.0kg = a

a = 7.55m/s^2

vf^2 = v0^2 + 2ax

vf^2 = 2(7.55m/s)(14.0m)

V_f = 14.5396m/s

K = 1/2(mv^2)

K = 1/2(12.0kg)(14.5396m/s)

K = 87.238J

4 0

3 years ago

A 48.0-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocit

Kipish [7]

Answer:

F = 47.6 N

Explanation:

Newton's 2nd law can be expressed as the rate of change of the total momentum, respect of time, as follows:

$F = \frac{\Delta p}{\Delta t}$

So, in order to find the average force exerted by the skater on the wall, we can find the change in momentum due to the force exerted by the wall (which is equal and opposite to the one exerted by the skater), and divide it by the time interval , as follows:

$F_{wall} = \frac{\Delta p}{\Delta t} =\frac{(48.0 kg*(-1.06m/s)}{1.07s} = -47.6 N$

⇒ Fsk = 47.6 N (normal to the wall)

3 0

3 years ago

Which indicates that light travels in straight lines?

aleksandr82 [10.1K]

Answer:

Explanation:

8 0

3 years ago

Read 2 more answers

A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener

trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

$U=mgh$

where

m is its mass

$g=9.8 m/s^2$ is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

$U=(m)(9.8)(15)=147 m$ (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

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

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

$147 m = \frac{1}{2}mv^2$

And re-arranging for v, we find the velocity:

$v=\sqrt{2\cdot 147}=17.1 m/s$

5 0

3 years ago

Prove that the unit of area derived quantities are derived units

frozen [14]

answer: derived physical quantities are those quantities that are obtained from the basic physical quantities by multiplication or division and area is one of them

8 0

3 years ago