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2 years ago

12

A soccer ball initially at rest rolls down a hill with an acceleration of 3.0 m /s2 .it accelerates for 5 seconds before melissa

is able to catch it how far did it roll ?

1 answer:

slavikrds [6]2 years ago

6 0

The correct answer is the Soccer ball is 3.2 plus 5 that’s your answer.

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What are the two most important traits between a whale and a shark?

nordsb [41]

Answer:

Both are aquatic animals and are hunters

Explanation:

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2 years ago

What is the force of gravity for a 12 kg turkey?<br><br> Please help asap

pogonyaev

Answer: 117.6N

Explanation:

By the second Newton's law, we know that:

F = m*a

F = force

m = mass

a = acceleration

We know that in the surface of the Earth, the gravitational acceleration is g = 9.8m/s^2.

Then we just can input that acceleration in the above equation, and also replace m by 12kg, and find that the force due the gravity is:

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2 years ago

Do mirrors reflect light

777dan777 [17]

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Yes, Mirror are a surface that reflects light more perfectly than ordinary objects.

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2 years ago

Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has a magnitude of 3.7m/s2. We want

lbvjy [14]

Explanation:

It is given that, Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m/s².

The second equation of kinematics gives the relationship between the height reached and time taken by it.

Here, the ball is droped under the action of gravity. The value of acceleration due to gravity on Mars is positive.

We want to know how many seconds the basketball is in the air before it hits the ground. So, the formula is :

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3 years ago

You build a grandfather clock, whose timing is based on a pendulum. You measure its period to be 2s on Earth. You then travel wi

Elenna [48]

Answer:

$\frac{g_{2}}{g_{1}} = \frac{1}{4}$

Explanation:

The period of the simple pendulum is:

$T = 2\pi\cdot \sqrt{\frac{l}{g} }$

Where:

$l$ - Cord length, in m.

$g$ - Gravity constant, in $\frac{m}{s^{2}}$ .

Given that the same pendulum is test on each planet, the following relation is formed:

$T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}$

The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:

$\frac{g_{2}}{g_{1}} = \left(\frac{T_{1}}{T_{2}} \right)^{2}$

$\frac{g_{2}}{g_{1}} = \left(\frac{2\,s}{4\,s} \right)^{2}$

$\frac{g_{2}}{g_{1}} = \frac{1}{4}$

5 0

3 years ago