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1 year ago

Calculate the net force acting on the box in the following. refer to the screenshot below.

Physics

1 answer:

Katarina [22]1 year ago

3 0

The net force acting on the box is 6N

The force are unbalanced

<h3>Explain the concept?</h3>

The net force on an object is the sum of all the individual forces acting on it.

If two opposing forces act on an object, the net force is the difference between the two forces.

When two forces act in the same direction on an object, the net force is the sum of the two forces.

Here the two force 3N and 3N acting on same direction, so

net force = 3N + 3N

= 6N

Forces that are balanced are those that are opposite in direction and equal in size, but here the equal forces acts in same direction so it is unbalanced

To learn more about net force refer to :

brainly.com/question/18038995

#SPJ13

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En un balde de 25 cm de altura se hace un orificio a 15 cm del suelo.calcula la velocidad de salida del agua

Contact [7]

Explanation:

En un balde de 25 cm de altura se hace un orificio a 15 cm del suelo.calcula la velocidad de salida del agua

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2 years ago

An object moves from the position +16m to the position +43m in 12s. What us the total displacement

NeX [460]

First method

initial distance = 16m

final distance= 43 m

total distance covered= final -initial

=43m -16m

=27m

Second method

Si= 16m

Sf =43 m

t= 12 s

first we will find V

V = (Sf-Si)/ t

V =( 43- 16)/ 12

V = 27/12 ⇒ V= 9/4

V= distance / time

distance= V×time

distance = (9/4) ×12

distance =27

3 0

3 years ago

What must be the acceleration of a train in order for it to stop 12 m/s in a distance if 541 m

earnstyle [38]

Answer:

The acceleration of the train must be - 0.133 m/s²

Explanation:

Lets explain how to solve the problem

A train in order for it to stop 12 m/s in a distance if 541 m

That means the initial velocity of the train is 12 m/s

Its final velocity is zero (stop)

The distance it covers is 541 m

We want to find its acceleration

The acceleration will be negative quantity because the train reduced its

velocity from 12 m/s to zero

We need rule contains velocity, acceleration and distance

So we will use ⇒<em> v² = u² + 2as</em>, where v is the final velocity, u is the

initial velocity, a is the acceleration and s is the distance

v = 0, u = 12 m/s, s = 541 m

Substitute these values in the rule

(0)² = (12)² + 2(a)(541)

0 = 144 + 1082 a

Subtract 144 from both sides

-144 = 1082 a

Divide both sides by 1082

- 0.133 = a

<em>The acceleration of the train must be - 0.133 m/s²</em>

3 0

3 years ago

A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m

Marrrta [24]

Answer:

$-1.47 m/s^2$

Explanation:

We can use the following SUVAT equation to solve the problem:

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d = 196 m is the displacement of the car before coming to a stop

Solving the equation for a, we find the acceleration:

$a=\frac{v^2-u^2}{2d}=\frac{0-(24)^2}{2(196)}=-1.47 m/s^2$

4 0

2 years ago

Can someone help me please

stealth61 [152]

B hope this helps you!!!

4 0

3 years ago