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3 years ago

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A ball is dropped from a height h and falls the last half of its distance in 4 seconds. How long does the ball fall? From what h

eight is the ball originally dropped?

1 answer:

dlinn [17]3 years ago

8 0

Answer:

How long does the ball fall is t_2 = 13.66 (s).

From what height is the ball originally dropped is h= 913.90 (m).

Explanation:

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If youre walking from point a to b, the magnitude of your displacement will always be equal or less than or greater than your di

xenn [34]

The magnitude of your displacement can be equal to the distance you covered, or it can be less than the distance you covered. But it can never be greater than the distance you covered.

This is because displacement is a straight line, whereas distance can be a straight line, a squiggly line, a zig-zag line, a line with loops in it, a line with a bunch of back-and-forths in it, or any other kind of line.

The straight line is always the shortest path between two points.

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3 years ago

The weight of an object is the force generated by Earth's gravity accelerating the object's _________. distance mass time destin

julsineya [31]

The weight of an object is the force of gravity between Earth's
mass and the object's mass.

The forces of gravity always come in equal, opposite pairs.
The Earth's weight on the object is the same as the object's
weight on the Earth, and when the object falls to Earth, Earth
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3 years ago

A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

$E_{A}=E_{B}$

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

$E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}$

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

$E_{B}=m*g*h+\frac{1}{2}*m*v^{2}$

Therefore we will have the following equation:

$(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]$

The kinetic energy can be easily calculated by means of the kinetic energy equation.

$KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]$

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

$E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]$

5 0

3 years ago

Use the galvanometers to determine the amount and direction of the induced current. Which galvanometer is

docker41 [41]

Answer:

Option B

Explanation:

Looking at the 3 galvanometer readings given above, for galvanometer A, the reading is -2 mA.

For galvanometer B, the reading is 4 mA.

While for galvanometer C, the reading is -5 MA

Thus, option B is correct.

4 0

3 years ago

Distance measurements based on the speed of light; used for objects in space

Vika [28.1K]

Answer : Yes, distance measurements based on the speed of light used for objects in space.

Explanation : A light year is measurement of distance that light travel in a one year.

In a one year light travels 9460000000000 kilometer.

We know that, speed of light is $3\times10^{8}\ m/s$

and time is 31536000 seconds in 1 year

so, distance = speed of light X time

Now, the light year is $9.4608\times10^{15} meter$

Example : The nearest star to earth is about 4.3 light year away.

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3 years ago