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3 years ago

I need help with all of these questions

Physics

1 answer:

Sedbober [7]3 years ago

7 0

Answer:

This was my best estimation of the answers

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Two physics students shoot a water-bottle rocket from a 1 m tall stand. They calculate that rocket will travel with the given fo

Kitty [74]

Answer:

2 secs; 65 feet

Explanation:

The function guiding the water bottle is given as:

f(x) = -16t² + 64t + 1

The bottle will reach maximum height when velocity, df/dt (velocity is the first derivative of distance) = 0.

Therefore:

df/dt = 0 = -32t + 64

=> 32t = 64

t = 64/32 = 2 seconds

This is the time it will take to reach the maximum height.

To find this height, we insert t = 2 into the function f(x):

f = -16(2)² + 64(2) + 1

f = -(16 * 4) + 128 + 1

f = -64 + 128 + 1

f = 65 ft

Its maximum height is 65 ft.

7 0

3 years ago

Read 2 more answers

The hypotenuse of a right triangle is 29.0 centimeters. The length of one of its legs is 20.0 centimeters. What is the length of

Kisachek [45]

square root (29 squared - 20 squared)

Pythagoras' theorem

5 0

3 years ago

Read 2 more answers

Two pieces of clay are moving directly toward each other. When they collide, they stick together and move as one piece. One piec

Wittaler [7]

Answer:

The fraction of the total initial kinetic energy is lost during the collision is $\dfrac{11}{17}\ J$

Explanation:

Given that,

Mass of one piece = 300 g

Speed of one piece = 1 m/s

Mass of other piece = 600 g

Speed of other piece = 0.75 m/s

We need to calculate the final velocity

Using conservation of energy

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v$

Put the value intro the formula

$300\times10^{-3}\times1+600\times10^{-3}\times(0.75)=(300\times10^{-3}+600\times10^{-3})v$

$v=\dfrac{00\times10^{-3}\times1+600\times10^{-3}\times(-0.75)}{(300\times10^{-3}+600\times10^{-3})}$

$v=-0.5\ m/s$

We need to calculate the total initial kinetic energy

Using formula of kinetic energy

$K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2$

Put the value into the formula

$K.E_{i}=\dfrac{1}{2}\times300\times10^{-3}\times1^2+\dfrac{1}{2}\times600\times10^{-3}\times(0.75)^2$

$K.E_{i}=0.31875\ J$

We need to calculate the total final kinetic energy

Using formula of kinetic energy

$K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2$

Put the value into the formula

$K.E_{f}=\dfrac{1}{2}\times(300\times10^{-3}+600\times10^{-3})\times(-0.5)^2$

$K.E_{f}=0.1125\ J$

We need to calculate the energy lost during the collision

Using formula of energy lost

$energy\ lost=\dfrac{0.31875-0.1125}{0.31875}$

$energy\ lost=\dfrac{11}{17}\ J$

Hence, The fraction of the total initial kinetic energy is lost during the collision is $\dfrac{11}{17}\ J$

3 0

3 years ago

Помогите!!!

ycow [4]

Answer:

Explanation:

UIGRDE

3 0

2 years ago

Any clue on this one

aniked [119]

3rd one I believe so

6 0

4 years ago