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3 years ago

A box has a mass of 3 kg. The box is lifted from the floor and placed

Physics

1 answer:

Lera25 [3.4K]3 years ago

4 0

Explanation:

Potential energy=MGH

100=3×10×H.

H=100/30

H=3.40m

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A student is holding a marble in his hand. What are the equal and opposite forces that are acting on it?

Nesterboy [21]

Answer:

Gravity and his hand

Explanation:

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3 years ago

Determine the speed of a falling object 7.8 sec after it dropped.

butalik [34]

Answer:

76.4m/s

Explanation:

Given parameter:

Time taken = 7.8sec

Unknown:

Speed after it dropped = ?

Solution:

To solve this problem, we use one of the kinematics equation:

V = U + gt

V is the final speed

U is the initial speed = 0m/s

g is the acceleration due to gravity

t is the time taken

V = 0 + 9.8 x 7.8 = 76.4m/s

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3 years ago

An underground cannon launches a cannonball from ground level at a 35 degree angle. the cannonball is shot with an initial veloc

user100 [1]

Answer:

Time = 1.75[s]; Distance traveled = 21.5 [m]; Max height = 15 [m]

Explanation:

First, we have to break down the velocity vector into the X & y components.

$(v_{x})_{0} = 15 * cos( 35)= 12.28[m/s]\\(v_{y})_{0} = 15 * sin( 35)= 8.6[m/s]\\\\$

To find the time t that lasts the ball of cannon in the air we must use the following equation of kinematics, in this equation the value of y is equal to zero because it will be proposed that the ball lands at the same level that was fired.

$y=(v_{y} )_{0}-\frac{1}{2}*g*t^{2} \\where:\\g=9.81[m/s^2]\\t = time[s]\\y=0[m]$

$0=8.6*t-\frac{1}{2}*9.81*t^{2} \\4.905*t^{2}=8.6*t\\ t=1.75[s]$

In order to find the distance traveled horizontally from the cannonball, we must use the speed kinematics equation in the X coordinate.

$x = (v_{x})_{0} *t\\x=12.28*1.75\\x=21.5 [m]$

In order to find the last value, we must bear in mind that when the cannonball reaches the maximum height, the velocity in the component y is equal to zero, and we can find the value of and with the following kinematic equation

$y = (v_{y})_{0} *t+\frac{1}{2} *g*(t)^{2} \\y = 0*t+\frac{1}{2} *9.81*(1.75)^{2}\\ y=15 [m]$

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3 years ago

Two children sit on different sides of a seesaw. The first child of mass 27 kg sits 1.5 m from the center. How far must the seco

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Answer:

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Explanation:

jsjszhxhbssbzbbzbdfnndsndnndndnsndnfndndnsmsnsxnbdbsndndndnxbcbwnnbfndbdxbbdbvh go do some gehehhsdhehndnsbxhdbdj

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3 years ago

How much force is required to cause an object with a mass of 850 kg to accelerate at a rate of 2 meters per second squared (m/s^

arlik [135]

Answer:

How much force is required to cause an object with a mass of 850 kg to accelerate at a rate of 2 meters per second squared (m/s^2)?

Explanation:

<em> Mass multiplied by acceleration gives you the amount of force needed for it.</em>

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3 years ago

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