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3 years ago

5

A boulder with a mass of 50 kg, sits on a cliff that is 50 m high. The boulder is pushed off of the cliff. How much kinetic ener

gy does the boulder have just before it hits the ground?

1 answer:

rosijanka [135]3 years ago

8 0

50 +50 =100 Since it’s sitting on a 50m cliff that’s high with a mass of 50 kg it would be adding because once it goes down it’s adding speed

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Bill and Nageen have each built an electric motor and they are testing them by having them lift boxes. Bill's motor lifts a box

photoshop1234 [79]

Answer:

Bill's motor power: W_B = F x S / T = F x 0.35 / 2= 0.175F

Nageen's motor power: W_N = F x S / T = F x 0.35 / 1.8 = 0.194F

=> 0.194F > 0.175F => Nageen's motor applied more power to the box than Bill's motor.

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3 years ago

3. A pendulum with a 1.0-kg weight is set in motion from a position 0.04 m above the lowest point on the path of the weight.

gavmur [86]

Answer: K.E = 0.4 J

Explanation:

Given that:

M = 1.0 kg

h = 0.04 m

K.E = ?

According to conservative of energy

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K.E = 1 × 9.81 × 0.04

K.E = 0.3924 Joule

The kinetic energy of the pendulum at the lowest point is 0.39 Joule

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3 years ago

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Akimi4 [234]

D.) White Dwarf

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Here, M = mass of the Sun.

Hope this helps!

3 0

3 years ago

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. A book is moved once around the perimeter of a table of dimensions 2.00 m by 3.00 m. If the book ends up at its initial positi

Lynna [10]

Answer:

10

Explanation:

displacement would be 10 because knowledge

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3 years ago

A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.What is the

podryga [215]

Answer:

$6.9\times 10^{28}m^{-3}$

Explanation:

We are given that

Diameter of wire=d=4.12 mm

Radius of wire=r $r=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m$

$1mm=10^{-3} m$

Current=I=8 A

Drift velocity= $v_d=5.4\times 10^{-5} m/s$

We have to find the density of free electrons in the metal

We know that

Density of electron= $n=\frac{I}{v_deA}$

Using the formula

Density of free electrons= $\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}$

By using Area of wire= $\pi r^2$

$\pi=3.14\\e=1.6\times 10^{-19} C$

Density of free electrons= $6.9\times 10^{28}m^{-3}$

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3 years ago

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