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2 years ago

Between the ball and the player’s head, there are forces. Which of Newton’s laws does this represent? Support your choice.

Physics

2 answers:

Nana76 [90]2 years ago

8 0

Answer:according to newtons third law for every action force there is an equal in size and opposite in direction reaction force forces always comes in pairs

Explanation:

ira [324]2 years ago

7 0

I need help with this one to

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Olivia put a glass of water in the freezer. She left it there for three hours. When she returned, the water had turned to ice. W

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A. freezing, when water turns to ice the water is turning from a liquid to a solid.

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3 years ago

A bookcase has a base of 1 m long and 0.5 m wide. It has a mass of 300kg. Find the pressure it exerts on the floor in kPa.

Snezhnost [94]

$1m*0.5m=0.5m^2\\
g=10\frac{m}{s^2}\\
300kg*10\frac{m}{s^2}=3000N\\
3000N--0.5m^2\\
\ \ x\ \ --\ 1\\
x=6000Pa=60kPa$

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2 years ago

A giraffe is slowly walking across a field. Is it balanced or unbalanced?

vodka [1.7K]

Your answers are correct

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3 years ago

Use the drop-down menu to complete the statement. One difference between prokaryotic cells and eukaryotic cells is that a eukary

Ahat [919]

Answer:

Nucleus

Explanation: I did the quiz and got it right. Have an amazing day!

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2 years ago

A nonconducting container filled with 25 kg of water at 23°C is fitted with a stirrer, which is made to turn by gravity acting o

Paul [167]

Explanation:

Given that,

Weight of water = 25 kg

Temperature = 23°C

Weight of mass = 32 kg

Distance = 5 m

(a). We need to calculate the amount of work done on the water

Using formula of work done

$W=mgh$

$W=32\times9.8\times5$

$W=1568\ J$

The amount of work done on the water is 1568 J.

(b). We need to calculate the internal-energy change of the water

Using formula of internal energy

The change in internal energy of the water equal to the amount of the work done on the water.

$\Delta U=W$

$\Delta U=1568\ J$

The change in internal energy is 1568 J.

(c). We need to calculate the final temperature of the water

Using formula of the change internal energy

$\Delta U=mc_{p}\Delta T$

$\Delta U=mc_{p}(T_{2}-T_{1})$

$T_{2}=T_{1}+\dfrac{\Delta U}{mc_{p}}$

$T_{2}=23+\dfrac{1568}{25\times4.18\times10^{3}}$

$T_{2}=23.01^{\circ}\ C$

The final temperature of the water is 23.01°C.

(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.

The amount of heat is 1568 J.

Hence, This is the required solution.

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2 years ago