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DanielleElmas [232]
3 years ago
7

6. A skier starts from rest at the top of a frictionless incline of height 20.0 m. At

Physics
1 answer:
n200080 [17]3 years ago
6 0

Explanation:

a. KE at bottom = PE at top

½ mv² = mgh

v = √(2gh)

v = √(2 × 9.8 m/s² × 20.0 m)

v = 19.8 m/s

b. Work by friction = PE at top

mgμ d = mgh

d = h / μ

d = 20.0 m / 0.210

d = 95.2 m

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An interstellar space probe is launched from Earth. After a brief period of acceleration it moves with a constant velocity, 70.0
sleet_krkn [62]

Answer:

22.26 years

, 15.585 light years  , 11.13 light years

Explanation:

a)

t' = t/(\sqrt{1-(v/(c*v)/c)}

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What type of acceleration does an object moving with constant speed in a circular path experience?Select one:A) constant acceler
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D) centripetal acceleration.

STEP-BY-STEP EXPLANATION:

When a body performs a uniform circular motion, the direction of the velocity vector changes at every instant. This variation is experienced by the linear vector, due to a force called centripetal, directed towards the center of the circumference that gives rise to the centripetal acceleration.

Therefore, the answer is centripetal acceleration.

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1 year ago
Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to
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(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

where

K_i is the kinetic energy at the surface, when the probe is launched

U_i is the gravitational potential energy at the surface

K_f is the final kinetic energy of the probe

U_i is the final gravitational potential energy

Here we have

K_i = 5.0 \cdot 10^7 J

at the surface, R=3.3\cdot 10^6 m (radius of the planet), M=5.3\cdot 10^{23}kg (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J

At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

so the final potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

So now we can use eq.(1) to find the final kinetic energy:

K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J

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The probe reaches a maximum distance of

r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J

So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

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