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ehidna [41]
3 years ago
15

Write the equation which links gravitational field strength,gravitational potential energy,height and mass

Physics
1 answer:
Andru [333]3 years ago
8 0

Answer:

PE = mgh

Explanation:

PE is potential energy

m is mass

g is gravitational field strength

h is height

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The speed of light is greater in a vacuum than in air or water.<br><br> True or false
Dima020 [189]
True !! Hope I helped you out a bit!
7 0
3 years ago
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20 kg object travels 28 meter and stops. coefficient friction= 0.085 how much work was done by friction?
Tresset [83]
Assuming it is on a horizontal surface:
friction = μR
R = 20g (g is gravity 9.81)
so Friction = 0.085 x 20g
Work done is force x distance 
so Work done = 0.085 x 20g x 28
 = 466.956 J

7 0
3 years ago
BRAINLIEST FOR YOU IF YOU ANSWER
luda_lava [24]

Answer:

d

Explanation:

i think

5 0
2 years ago
A physics student skis down a hill, accelerating at a constant
ikadub [295]
<h3>Answer:</h3>

225 meters

<h3>Explanation:</h3>

Acceleration is the rate of change in velocity of an object in motion.

In our case we are given;

Acceleration, a = 2.0 m/s²

Time, t = 15 s

We are required to find the length of the slope;

Assuming the student started at rest, then the initial velocity, V₀ is Zero.

<h3>Step 1: Calculate the final velocity, Vf</h3>

Using the equation of linear motion;

Vf = V₀ + at

Therefore;

Vf = 0 + (2 × 15)

    = 30 m/s

Thus, the final velocity of the student is 30 m/s

<h3>Step 2: Calculate the length (displacement) of the slope </h3>

Using the other equation of linear motion;

S = 0.5 at + V₀t

We can calculate the length, S of the slope

That is;

S = (0.5 × 2 × 15² ) - (0 × 15)

= 225 m

Therefore, the length of the slope is 225 m

6 0
3 years ago
A disk-shaped grindstone of mass 3.0 kg and radius 8.0 cm is spinning at 600 rev/min. After the power is shut off, a man continu
kolbaska11 [484]

Answer:

τ=0.060 N.m

Explanation:

By kinematics:

\omega f = \omega o-\alpha*t

Solving for α:

\alpha=\frac{\omega o-\omega f}{t}

where ωo = 600*2*π/60;   ωf = 0;    t=10s

\alpha=6.283rad/s^2

The sum of torque is:

\tau=I*\alpha

\tau=M*R^2/2*\alpha

\tau=0.060 N.m

8 0
3 years ago
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