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OLEGan [10]
3 years ago
9

what is the net force of the box and which direction will it move (50 N left, 50N left and 100 N right)​

Physics
2 answers:
Bingel [31]3 years ago
5 0

the net force is 0 and it will not move

Serggg [28]3 years ago
5 0

The net force on the box is zero.

-- If it was already moving when the forces began, it continues moving in a straight line with constant speed.

-- If it was NOT already moving when the forces began, then it won't start moving, because the forces are 'balanced', and their effect on the box is the same as if there were no forces there at all.

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An object of mass 2 kg starts from rest and is allowed to slide down a
WARRIOR [948]

Answer: B

Explanation:

Given that an object of mass 2 kg starts from rest and is allowed to slide down a frictionless incline so that its height changes by 20 m. 

The parameters given from the question are:

Mass M = 2kg

Height h = 20m

Let g = 9.8m/s^2

At the bottom of the incline plane, the object will experience maximum kinetic energy.

From conservative of energy, maximum K.K.E = maximum P.E

Maximum P.E = mgh

Maximum P.E = 2 × 9.8 × 20 = 392 J

But

K.E = 1/2mv^2

Substitute the values of energy and mass into the formula

392 = 1/2 × 2 × V^2

V^2 = 392

V = sqrt( 392 )

V = 19.8 m/s

V = 20 m/s approximately

7 0
3 years ago
23. If a jogger runs 100 meters west and then turns around and runs 30 meters east. What
marshall27 [118]

Answer:

A

Explanation:

if he goes to the west, the east is opposite so 100-30

7 0
3 years ago
Read 2 more answers
A net force of 3000 N is accelerating a 1200 kg elevator upward. If the elevator starts from rest, how long will it take to trav
alexdok [17]

F = net force acting on the elevator in upward direction = 3000 N

m = mass of the elevator = 1200 kg

a = acceleration of the elevator = ?

Acceleration of the elevator is given as

a = F/m

a = 3000/1200

a = 2.5 m/s²

v₀ = initial velocity of the elevator = 0 m/s

Y = displacement of the elevator = 15 m

t = time taken

Using the kinematics equation

Y = v₀ t + (0.5) a t²

15 = (0) t + (0.5) (2.5) t²

t = 3.5 sec

5 0
3 years ago
A 2000 kg car experiences a braking force of 10000N and skids to a 14 m stop. What was the speed of the car just before the brak
Fudgin [204]

Answer:

V = 11.83 m/s

Explanation:

Given the following data;

Mass = 2000 kg

Force = 10000N

Distance = 14 m

To find the final velocity of the car;

First of all, we would determine the acceleration of the car;

Acceleration = force/mass

Acceleration = 10000/2000

Acceleration = 5 m/s²

Next, we would use the third equation of motion to find the final velocity;

V^{2} = U^{2} + 2aS

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement measured in meters.

Substituting into the equation, we have;

V² = 0² + 2*5*14

V² = 0 + 140

V = √140

V = 11.83 m/s

5 0
3 years ago
Which of the following statements is true?
Alborosie
My guess is A. I'm not 100% positive but i'm pretty sure.
8 0
3 years ago
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