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AnnZ [28]
3 years ago
11

A force of 4 kg weight acts on a body of mass 9.8 kg calculate the acceleration

Physics
2 answers:
klio [65]3 years ago
4 0
  • Mass=9.8kg
  • Force=4N

Using newtons second law

\\ \bull\sf\dashrightarrow Force=Mass\times Acceleration

\\ \bull\sf\dashrightarrow Acceleration=\dfrac{Force}{Mass}

\\ \bull\sf\dashrightarrow Acceleration=\dfrac{9.8}{4}[tex][tex]\\ \bull\sf\dashrightarrow Acceleration=2.45m/s^2

White raven [17]3 years ago
3 0

Answer:

here given is a weight

then force becomes mg

that is F=Mg

=4*9.8

then by using the formula

F=Ma

a=F/M

=4*9.8/9.8

=4

Explanation:

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A 100 N force is applied to move an object a horizontal distance of 5 meters at constant speed in 10 seconds. How much power is
Tpy6a [65]

Answer:

50 W

Explanation:

<h3><u>Given :</u></h3>

  • Force applied = 100 N
  • Distance covered = 5 metres
  • Time = 10 seconds

<h3><u>To find :</u></h3>

Power

<h3><u>Solution :</u></h3>

For calculating power, we first need to know about the work done.

\bf \boxed{Work = Force \times displacement}

Now, substituting values in the above formula;

Work = 100 × 5

= 500 Nm or 500 J

We know that,

\bf \boxed{Power=\dfrac{Work\:done}{Time\: taken}}

Substituting values in above formula;

Power = 500/ 10

= 50 Nm/s or 50 W

Hence, power = 50 W .

5 0
3 years ago
A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,
egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

6 0
4 years ago
A person sitting in a chair with wheels stands up, causing the chair to roll backward across the floor. How would you describe t
OleMash [197]
Scenes the chair wheels are up the person is rolling backwards and if the wheels were down then the person would go forwards
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3 0
3 years ago
Learning Task 1: Determine what kind of carbon allotropes are the given pictures
kotegsom [21]

Answer:

1.C70

2.Fullerene

3.Lonsdaleite

4.Graphite

5.Diamond

6.Amorphous carbon

I hope this helps. thank you

Explanation:

4 0
3 years ago
Read 2 more answers
A 5-turn square loop (10 cm along aside, resistance = 4.0 ) is placed in a magnetic field that makes an angle of30o with the pla
andreev551 [17]

Answer:

Explanation:

Area A of the coil = .1 x .1 = .01 m²

no of turns n = 5

magnetic field B = .5 t²

Flux  Φ perpendicular to plane passing through it.= nBA sin30

rate of change of flux

dΦ/dt = nAdBsin30 / dt

= nA d/dt (.5t²x .5 )

= nA x 2 x .25 x t

At t = 4s

dΦ/dt = nA x 2

= 5x .01 x 2

= .1

current = induced emf / resistance

= .1 / 4

= .025 A

= 25 mA.

4 0
3 years ago
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