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3 years ago

Why does 30 kg mass at the top of a hill have more potential energy?

1 answer:

6 0

So when it comes to Gravitational Potential Energy the higher an object of mass is the more energy it has as the equation for EPG is = MGH so the M is Mass of the object , the G is the gravitational constant which on earth is roughly 9.8 m/s and the H is the height of the object. So the greater the H value is the more energy you'll have. An example would be using your 30kg mass say at a height of 10 meters would have the EPG of 2940 Joules of energy since EPG= MGH so EPG= (30)(9.8)(10) . Now if the object was at a height of 20 meters the EPG would be greater as EPG=(30)(9.8)(20) which would be 5880 joules of energy

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Just before the ball leaves her hand, what is its centripetal acceleration?

erica [24]

A = $\frac{ v^{2} }{r}$ = ω²r

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3 years ago

Which is usually greater for a pair of surfaces, the coefficient of static friction or the coefficient of sliding friction? why

lorasvet [3.4K]

Explanation :

Static friction is the frictional force between two objects that are at rest. While sliding friction is the frictional force between two objects in contact and are sliding w.r.t each other.

Static friction is usually greater than sliding friction because in static friction the contact forces is more and the interlocking between objects is tight as compared to sliding friction.

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3 years ago

A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with

gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm$

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s$

The block's speed is 31.422 cm/s

4 0

3 years ago

-A 180 kg hippo is riding a bicycle at a speed of 6.0

Vlad1618 [11]

Answer:

0.0675 seconds

Explanation:

From the question,

We apply newton's second law of motion

F = m(v-u)/t.................... Equation 1

Where F = force exert by the brake, v = final speed, u = initial speed m = mass of the bicycle, t = time.

make t the subject of the equation

t = m(v-u)/F................... Equation 2

Given: m = 180 kg, u = 6.0 m/s, v = 0 m/s (comes to stop), F = -1600 N ( agianst the dirction of motion)

Substitute these value into equation 2

t = 180(0-6.0)/-1600

t = -1080/-1600

t = 0.0675 seconds.

8 0

3 years ago

If a student is sitting in their chair and the leg of the chair breaks the student will __________ toward the floor because the

USPshnik [31]

Answer:

theyll accelerate towards the floor :)

5 0

3 years ago

Read 2 more answers