If the net force on an object stays the same but the mass of the object is doubled, what will happen to the

according to newton's second law of motion of the net force acting on the object increases while the mass of the object remains

Licemer1 [7]

Answer:

The Acceleration will increase

Explanation:

Newton's Second Law of motion: It states that the rate of change of momentum is directly proportional to the applied force and takes places along the direction of the force.

It can be expressed mathematically as,

F ∝ m(v-u)/t

Where (v-u)/t = a

F = kma.

F = force, m = mass of the body, a = acceleration, k = constant of proportionality which tend to unity for a unit force, a unit mass, and a unit acceleration.

Therefore,

F = ma.

From the equation above,

If the net force acting on a body increase, while the mass of the body remains constant, the acceleration will also increase.

4 0

3 years ago

1-A car moves toward east 12km is represented as A and it turns towards south 16km is represented as B. What is the resultant ve

hjlf

1. A-20 km south east

The car's displacement consists of two components into two different directions. Using a system of coordinates in which x represents the east direction and y represents the south direction, the two displacements are:

$d_x = 12 km$ east

$d_y = 16 km$ south

Since the two components are orthogonal to each other, we can find the resultant displacement by using Pythagorean's theorem:

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(12 km)^2+(16 km)^2}=\sqrt{400}=20 km$

and the direction is between the two original directions, so south-east.

2. D. 10 m/s

First of all, we need to calculate the total time the stone took to hit the ground. Since the vertical distance covered is S = 78.4 m, and since the motion is an accelerated motion with constant acceleration g=9.8 m/s^2, we have

$S=\frac{1}{2}gt^2$

From which we find the total time of the fall, t:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(78.4 m)}{9.8 m/s^2}}=4 s$

Now we can consider the horizontal motion of the stone: we know that the stone travels for d = 40 m in a time of t = 4 s, therefore the horizontal velocity of the stone is

$v=\frac{d}{t}=\frac{40 m}{4 s}=10 m/s$

3. B=32.32 m

As in the previous problem, we have to calculate the total time it takes for the stone to reach the river first. Since the vertical distance covered is S = 20 m, we have

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(20 m)}{9.8 m/s^2}}=2.0 s$