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3 years ago

An object with a mass of 2 kg is acted upon by a force of 8 N. What is the resulting acceleration of the object?

Physics

1 answer:

Tomtit [17]3 years ago

4 0

Answer:

4 meter per second square is the acceleration

Explanation:

Force=mass x acceleration

8N= 2kg x acceleration

8N/2kg = acceleration

4 meter per second square is the acceleration.

<em>Please mark me as brainliest.</em>

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A leaf floating down from a tree is an example of an object in free fall.

Nuetrik [128]

Answer:

Oi, mate its false

Explanation:

because if an leaf floats down from a tree it is not considered an object for a free-fall

7 0

4 years ago

PLEASEE HELPP

topjm [15]

Explanation:

u=166m/s, v=0(at it's highest point final velocity is zero), a=9.8m/s², t=8.6s

by the formula, S=ut+½at².

S=[166×8.6+½.×9.8×(8.6)²]. ...by calculation

S = 1427.6+362.404

S=1790.004m

hope this helps you.

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3 years ago

If you place 1 C of positive charge on Earth and 1 C of negative charge on the moon, 384,500 km away, how much force would the p

ivolga24 [154]

Answer:

6.1 x 10^-8 newtons

Explanation:

F = 8.98 *109 *1*1/3845000002

4 0

2 years ago

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at

yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

$\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)$

Therefore, the work done by the worker in lifting the bucket is given as:

$W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)$

Now, plug in the values given and solve for 'W'. This gives,

$W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J$

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

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3 years ago

Astronauts need special tools to work in space. One specially designed tool that astronauts use is a screwdriver. Using Newton’s

Softa [21]

Due to the fact that for every action there is an equal reaction for every turn on the screw the screw will exert energy equal to the amount you gave

3 0

3 years ago