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xxTIMURxx [149]
3 years ago
9

How does mass and force affect the motion of an object?

Physics
2 answers:
liberstina [14]3 years ago
8 0
My teacher always said in science The greater the mass of an object, the more inertia it has so this is the awnser
djyliett [7]3 years ago
4 0
The greater the mass of an object, the more inertia it has, so the object resists changes to its motion better.
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Answer:

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A ball droped from a building. How fast is it traveling after falling 3.55s
algol [13]

Answer:

d = 61.75 m

Explanation:

Given that,

A ball droped from a building.

We need to find how fast is it traveling after falling 3.55 s.

As it is dropped, its initial velocity is equal to 0.

Let d is the distance it covers after falling 3.55 s.

We can use second equation of motion to find d.

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

Here, u = 0 and a =g

d=\dfrac{1}{2}gt^2\\\\d=\dfrac{1}{2}\times 9.8\times (3.55)^2\\\\d=61.75\ m

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4 0
3 years ago
. Starting from a stop at a traffic signal, a car speeds up to 20 m/s in 5 seconds. Calculate the acceleration of the car.
Dimas [21]

Answer: 4m/s²

Explanation:

At stop signal car speed iz 0 ; V1=0m/s

After t=5s  , car has speed is V2=20m/s

using:

a=(V2-V1)/t

a=(20m/s-0m/s)/5s

a=20m/s/5s

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4 0
3 years ago
A 240 N sphere 0.20 m in radius rolls, without slipping 6.0 m downa
Shalnov [3]

Answer:

The angular speed of the sphere at the bottom of the hill is 31.39 rad/s.

Explanation:

It is given that,

Weight of the sphere, W = 240 N

Radius of the sphere, r = 0.2 m

Angle with the horizontal, \theta=28^{\circ}

We need to find the angular speed of the sphere at the bottom of the hill if it starts  from rest.

As per the law of conservation of energy, the total energy at the top is equal to the energy at the bottom.

Gravitational energy = translational energy + rotational energy

So,

mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

I is the moment of inertia of the sphere, I=\dfrac{2}{5}mr^2

Also, v=r\omega

h is the height of the ramp, h=l\ sin\theta

mgl\ sin\theta=\dfrac{1}{2}m(r\omega)^2+\dfrac{1}{2}I\omega^2

On solving the above equation we get :

\omega=\sqrt{\dfrac{10gl\ sin\theta}{7r^2}}

\omega=\sqrt{\dfrac{10\times 9.8\times 6\ sin(28)}{7(0.2)^2}}

\omega=31.39\ rad/s

So, the angular speed of the sphere at the bottom of the hill is 31.39 rad/s. Hence, this is the required solution.

6 0
3 years ago
Which object is most efficient?
Radda [10]

Answer:

It is b

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