Ask question

Ask question

All categories

Artyom0805 [142]

3 years ago

12

ANSWER AND I WILL BUY A CAR

1 answer:

Natali [406]3 years ago

8 0

You are exerting 100N. Since there’s no NET force, then there must be exactly 100N pushing exactly back on your 100N to cancel it to exactly zero. Newton's first law states that whether a body is at rest or travelling in a straight line at a constant speed, it will remain at rest or continue to move in a straight line at a constant speed unless acted upon by a force.

You might be interested in

List 3 examples of transition metals

STatiana [176]

Answer:

Explanation:

Titanium

Chromium.

Iron

4 0

4 years ago

If your weight is 100N and you run up a flight of stairs that is 6 m high and it takes

fenix001 [56]

Answer:

power=300watt

Explanation:

5 0

3 years ago

What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (such deceleration caused on

Wewaii [24]

The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is $252.52\ m/s^2$ .

The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.

The given data:

time, t = 1.1 s

initial speed, u = 1000 km/h = $\frac{2500}{9}\ m/s$

final speed, v = 0 m/s

So we will be using the equation of motion, that is,

v = u + at

$\therefore 0=\frac{2500}{9} + a(1.1)$

$\Rightarrow a=-\frac{2500}{9(1.1)}$

$\therefore a = - 252.52 \ m/s^2$

Hence , the deceleration of the rocket is $252.52\ m/s^2$ .

To learn more about Attention here:

brainly.com/question/28500124

#SPJ4

6 0

2 years ago

What is the mass of a neutron?

Anastaziya [24]

Answer:

1

Explanation:

8 0

3 years ago

Read 2 more answers

A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th

ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = $9.8\ m/s^2$

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

Kinetic energy K.E = $\frac{1}{2}mv^2$ .

From above we know that -

Kinetic energy at the bottom of the cliff = potential energy at a height h

$\frac{1}{2}mv^2=\ mgh$

⇒ $v^2=\ 2gh$

⇒ $h=\ \frac{v^2}{2g}$

⇒ $h=\ \frac{(24.2)^2}{2\times 9.8}$

⇒ $h=\ 29.88\ m$

Hence, the height of the cliff is 29.88 m

5 0

3 years ago