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2 years ago

A huge block of stone has a mass of 650kg and gains 70000j of gpe. how high has it been raised? (g = 10m/s2 )

1 answer:

5 0

The formula for calculating GPE is

GPE = m x h x g

Where m is mass, h is height raised and g is gravity.

To solve the problem we just have to substitute the values we are given

70,000J = 650 x 10 x y

70,000J = 6500 x y

70,000 / 6500 = y

10.769 = y

So, the block of stone has been raised approximately 10.769 metres.

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An object with an initial velocity of 10 m/s accelerates at a rate of 3.5 m/s2 for 8 seconds. How far will it have traveled duri

ICE Princess25 [194]

Answer:

Distance, d = 192 meters

Explanation:

We have,

Initial velocity of an object is 10 m/s

Acceleration of the object is 3.5 m/s²

Time, t = 8 s

We need to find the distance travelled by the object during that time. Second equation of motion gives the distance travelled by the object. It is given by :

$d=ut+\dfrac{1}{2}at^2$

$d=10\times 8+\dfrac{1}{2}\times 3.5\times 8^2\\\\d=192\ m$

So, the distance travelled by the object is 192 meters.

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3 years ago

imagine you are going on a rid in a spacecraft next to earth. Your trip takes one whole year. Describe earth's tilt in the north

mylen [45]

You will have to fly around the whole earth to get to your landing station

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3 years ago

Can a goalkeeper at his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will

zmey [24]

The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

Explanation:

Consider the vertical motion of ball,

We have equation of motion v = u + at

Initial velocity, u = u sin θ

Final velocity, v = 0 m/s

Acceleration = -g

Substituting

v = u + at

0 = u sin θ - g t

$t=\frac{usin\theta }{g}$

This is the time of flight.

Consider the horizontal motion of ball,

Initial velocity, u = u cos θ

Acceleration, a =0 m/s²

Time, $t=\frac{usin\theta }{g}$

Substituting

s = ut + 0.5 at²

$s=ucos\theta \times \frac{usin\theta }{g}+0.5\times 0\times (\frac{usin\theta }{g})^2\\\\s=\frac{u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{2g}$

This is the range.

In this problem

u = 30 m/s

g = 9.81 m/s²

θ = 45° - For maximum range

Substituting

$s=\frac{30^2\times sin(2\times 45)}{2\times 9.81}=45.87m$

Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.

So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

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3 years ago

Truck drivers probably cannot see your vehicle if____.

laiz [17]

Answer Is Tailgating

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3 years ago

An action potential arriving at the presynaptic terminal causes what to occur?

Ulleksa [173]

Answer:

Voltage-gated calcium ion channels open, and calcium ions diffuse into the cell

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3 years ago