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3 years ago

What is the mass of an object that requires 100N of force in order to accelerate it at 10m/s2 ?

Physics

2 answers:

Debora [2.8K]3 years ago

7 0

Answer:

We know that f=ma

Rearranging

M=f÷a so m=10kg

Explanation:

jeka57 [31]3 years ago

3 0

<u>Answer:</u> The mass of an object is 10 kg.

<u>Explanation:</u>

Force is defined as the push or pull on an object with some mass that causes change in its velocity.

It is also defined as the mass multiplied by the acceleration of the object.

Mathematically,

$F=ma$

where,

F = force exerted on the object = 100 N

m = mass of the object = ?

a = acceleration of the object = $10m/s^2$

Putting values in above equation, we get:

$100kg.m/s^2=m\times 10m/s^2\\\\m=\frac{100}{10}=10kg$

Hence, the mass of an object is 10 kg.

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Answer:

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3 years ago

earnstyle [38]

Answer:

0.0312J

Explanation:

Let x be the distance the staple moves:

$x=0.150m-0.115m=0.035m$

And spring constant is $k=51.0N/m$

$PE=0.5kx^2\\=0.5\times 51.0\times 0.035^2\\\\=0.312$

Hence, the potential energy is 0.0312J

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3 years ago

Imagine that a loudspeaker is producing a quiet tone with a low pitch. How will its vibrations change:

iVinArrow [24]

I believe it is A :) hope this helped

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3 years ago

A young child is playing with a very flexible hose. She is moving her hand back and forth making transverse type waves. She move

aev [14]

Answer:

The waves will increase in frequency

Explanation:

As the young girl moves her hand back and forth faster, it will be observed that number of back and forth motions increase every second. Also the distance between crest and trough of the wave (wavelength) will be reduced as she moves her hand back and forth faster.

Frequency = number of turns (moves) per second

The waves will increase in frequency since there will be more number of back and forth motions in every second.

Also,

The distance between crest and trough will be reduced, which implies that there will be decrease in waves wavelength.

This can also be verified using wave equation;

V = Fλ

At constant velocity,

F ∝ ¹/λ

Thus, decrease in wavelength will cause increase in frequency of the waves.

The right answer is : The waves will increase in frequency

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3 years ago

In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir

Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be $S_{1}$

and the distance covered by Sir Alfred be $S_{2}$

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

$S = ut + \frac{1}{2}at^{2}$

Where $S$ is the distance traveled

$u$ is the initial velocity

$a$ is the acceleration

and $t$ is the time

For Sir George,

$S = S_{1}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.21 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{1} = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1} = 0.105 t^{2}\\$

$t^{2} = \frac{S_{1}}{0.105}$

Now, for Sir Alfred

$S = S_{2}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.26 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{2} = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2} = 0.13 t^{2}\\$

$t^{2} = \frac{S_{2}}{0.13}$

Since, they traveled for the same time, $t$ just before collision, we can write

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

$S_{1} + S_{2} = 86$ m

Then, $S_{2} = 86 - S_{1}$

Then,

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$ becomes

$\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}$

$0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}$

$S_{1} = 38.43$ m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

6 0

3 years ago