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Dimas [21]
3 years ago
14

A 5 N force pushes on the right side of a box. At the same time, a 10 N force pushes on the left side of the box. What happens t

o the box?
Physics
2 answers:
notka56 [123]3 years ago
8 0
The resultant force is 5N.So the box moves to right with constant acceleration.
Mrac [35]3 years ago
8 0

Answer: The box will displace towards right side.

Explanation:

In the given problem, a 5 N force pushes on the right side of a box. At the same time, a 10 N force pushes on the left side of the box. The net force acting on the force is 5 N.

The magnitude of the force acting on the box from the left side is more than the force acting on the box from the right side. The body will displace in that direction in which the magnitude of the applied force is more.

The box is displaced to right by a net force of 5 N.

Therefore, the box will displace towards right side.

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How is the motion of an object affected when a force acts on it?
Arada [10]

Newton's first law of motion is sometimes called the law of inertia. When the forces acting on an object are balanced, the object is either at rest or moving with a constant velocity. Unbalanced forces can cause an object to accelerate or decelerate. Unbalanced forces can also cause an object to change direction.

Hope that helps!

3 0
3 years ago
Define si system of unit
Zepler [3.9K]
Good good good 7777777
8 0
2 years ago
A passenger on your boat falls overboard. what should you do first?
yarga [219]
The very first thing that you should do when a passenger on your boat falls overboard is to throw a PFD or also known as a Personal Flotation Device. This would include anything that can help the passenger to float. But this step would differ, only if the passenger is not wearing a lifevest. 
5 0
3 years ago
When Woods hits a 0.04593 kg golf ball, the golf ball is usually traveling around 281 kilometers per hour. What average force do
Semenov [28]

Answer:

128.9 N

Explanation:

The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

F=\frac{\Delta p}{\Delta t}

where

F is the force

\Delta p is the change in momentum

\Delta t=0.030 s is the time interval

The change in momentum can be written as

\Delta p = m(v-u)

where

m = 0.04593 kg is the mass of the ball

u = 0 is the initial velocity of the ball

v=281 km/h =78.1 m/s is the final velocity of the ball

Substituting into the original equation, we find the force exerted on the golf ball:

F=\frac{m(v-u)}{\Delta t}=\frac{(0.04593)(78.1-0)}{0.030}=128.9 N

7 0
2 years ago
Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius
Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2

The centripetal acceleration for the second radius; 4.0 m

a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2

The centripetal acceleration for the third radius; 6.0 m

a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2

The centripetal acceleration for the fourth radius; 8.0 m

a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2

The centripetal acceleration for the fifth radius; 10.0 m

a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2

6 0
3 years ago
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