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Inga [223]
3 years ago
8

What is the mass of a cannonball if the force a force of 2500 N gives the cannonball an acceleration of 200 m/s^2??

Physics
1 answer:
vampirchik [111]3 years ago
5 0

The answer is a.12.5kg because i just did the test and it was correct.

hope this helps


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How reactive is an atom of Sodium(Na) and why?
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Sodium (Na) is very reactive because it does not have a full valence shell.
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3 years ago
A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 19.0 m/s when the
german

Answer: 4 s

Explanation:

Given

The ball leaves the hand of student with a speed of u=19\ m/s

When the hand is h=2.5\ m above the ground

Using the equation of motion we can write

h=ut+\dfrac{1}{2}at^2

Substitute the values

\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t]

Thus, the ball will take 4 s to hit the ground.

5 0
2 years ago
A 1000 kg box is being pushed with a force of 3500 N. What acceleration is the
WARRIOR [948]
Net Force = mass x acceleration
3500=1,000a
So a= 3500/1000
a=35/10
a=3.5 m/s^2
4 0
3 years ago
A window air conditioner that consumes 1.4 kW of electricity when running and has a coefficient of performance of 3 is placed in
Shtirlitz [24]

Answer:

1.4 kJ/s

Explanation:

See attached photo

7 0
3 years ago
A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h
Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

                    \tau = I \times \alpha

and,       I = \sum mr^{2}

So,      I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}

                       = 4 kg m^{2}

      \tau_{1} = 4 kg m^{2} \times \alpha_{1}

     \tau_{2} = I_{2} \alpha_{2}

So,      I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}

                     = 1 kg m^{2}

 as \tau_{2} = I_{2} \alpha_{2}

                   = 1 kg m^{2} \times \alpha_{2}        

Hence,     \tau_{1} = \tau_{2}

                  4 \alpha_{1} = \alpha_{2}

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Thus, we can conclude that the new rotation is \frac{1}{4} times that of the first rotation rate.

8 0
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