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marshall27 [118]
3 years ago
11

A boulder of mass 4 kg rolls over a cliff and reaches the beach below with a velocity of 20 metres per second minus 1.

Physics
1 answer:
Ksenya-84 [330]3 years ago
5 0

Given:-

Mass of boulder = 10kg

Final Velocity = 15m/s

Acceleration due to gravity = +10m/s

To Find:-

The kinetic energy of the boulder as it lands

The potential energy of the boulder when it was at the top of the cliff

The height of the cliff

Formulae used:-

Kinetic energy = mv²/2

Potential energy = mgh

Now,

We will first find the Kinetic enrgy,

→ K. E = ½mv²

→ K. E = ½ × 10 × (15)²

→ K. E = 5 × 225

→ K. E =1125J

But we know that when the body starts moving the Potential enrgy of the body starts Converting into Kinetic enrgy. Before reaching to the last spot it has equal kinetic energy and Potential energy.

Therefore,

→ mgh = 1125J

The Potential enrgy of the boulder when it was on top is 1125J

Now,

→ mgh = 1125J

→ 10 × 10 × h = 1125J

→ 100h = 1125J

→ h = 1125/100

→ h = 11.25m

Hence, The Height of the Cliff is 11.25m

Thanks!

Mark me brainliest!

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Calculate the energy needed to heat 4 kg of water from 25°C to 45°C.
inna [77]
(1 cal/g °C) x (4000 g) x (45 - 25)°C = 80000 cal = 80 kcal. So the answer is 80 kcal .
8 0
3 years ago
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Bowling balls are roughly the same size, but come in a variety of weights. Given its official radius of roughly 0.110 m, calcula
velikii [3]

Answer:

6.1328 kg

60.16284 N

Explanation:

r = Radius of ball = 0.11 m

\rho = Density of fluid = 1.1\times 10^3\ kg/m^3 (Assumed)

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of ball

V = Volume of ball = \frac{4}{3}\pi r^3

The weight of the bowling ball will balance the buouyant force

W=F_b\\\Rightarrow mg=V\rho g\\\Rightarrow m=\frac{V\rho g}{g}\\\Rightarrow m=V\rho\\\Rightarrow m=\frac{4}{3}\pi 0.11^3\times 1.1\times 10^3\\\Rightarrow m=6.1328\ kg

The mass of the bowling ball will be 6.1328 kg

Weight will be 6.1328\times 9.81=60.16284\ N

5 0
3 years ago
Skater begins to spend with arms held out at shoulder height. The skater wants to match the speed of the spin to the beat of the
Aleksandr [31]

Answer:

the moment of inertia with the arms extended is Io and when the arms are lowered the moment

I₀/I > 1    ⇒   w > w₀

Explanation:

The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,

           L₀ = L_f

           I₀ w₀ = I w

          w =\frac{I_o}{I} w₀

where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms

          I₀ = I_cylinder + 2 m r²

where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.

If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be

        I <I₀

        I₀/I > 1    ⇒   w > w₀

therefore the angular velocity (rotations) must increase

in this way the skater can adjust his spin speed to the musician.

7 0
3 years ago
A car travels up a hill at a constant speed of 38 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the a
mojhsa [17]

Answer:

Average speed will be 48.23 km/h

Explanation:

Let the distance up to hill is = d km

Speed when car goes to hill = 38 km/h

So time required t=\frac{distance}{speed}=\frac{d}{38}hour

Speed when car return from hill = 66 km/h

So time required to return fro hill t=\frac{d}{66}h

Total time t_{total}=\frac{t}{38}+\frac{t}{66}

Total distance = d+d =2d

So average speed=\frac{total\ distance}{total\ time}=\frac{2d}{\frac{d}{38}+\frac{d}{66}}=48.23km/h

8 0
3 years ago
A ball traveling at 15 m/s hits a bat with a force of 200N. How much force does the bat (moving at 20m/s)
just olya [345]

Answer:

200 N

Explanation:

Given that,

A ball traveling at 15 m/s hits a bat with a force of 200 N.

We need to find the force that the bat moving at 20 m/s hit the ball with.

We know that, this probelm is based on Newton's third law of motion. The force that the ball exerting on bat should be equal to the force that the bat exerting in the ball but in opposite direction.

It would mean that the ball hits the ball with a force of 200 N. Hence, the correct option is (a).

8 0
3 years ago
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