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3 years ago

What is force? in Newton's Law

Physics

1 answer:

Arte-miy333 [17]3 years ago

5 0

Answer:

That situation is described by Newton's Second Law of Motion. According to NASA, this law states, "Force is equal to the change in momentum per change in time. For a constant mass, force equals mass times acceleration.

Explanation:

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In a tug of war game, team A is very strong and is pulling team B with a speed of 1 foot per second. In this situation… *

Anastasy [175]

Answer:

Explanation:

4 0

3 years ago

Read 2 more answers

An object is projected at 25m/s from the top of a building of height 50m. At the same instant,another object is projected from t

docker41 [41]

A) The objects have the same vertical position after 2 seconds

B) The objects have same vertical position at y = 30.4 m (but they do not collide since they have different x-position)

Explanation:

The motion of the first object along the vertical direction is a uniformly accelerated motion, so we can write its position at time t using the following equation:

$y_1(t)=h+u_1 t + \frac{1}{2}gt^2$

where:

h = 50 m is the initial height

$u_1=0$ is the initial vertical velocity (the object is projected horizontally, so the vertical velocity is zero at the beginning)

$g=-9.8 m/s^2$ is the acceleration of gravity

So, its vertical position can be rewritten as

$y_1(t)=50-4.9t^2$

The position of object 2 instead can be written as

$y_2(t)=(u_2 sin \theta)t + \frac{1}{2}gt^2$

where

$u_2 sin \theta$ is the initial vertical velocity, where

$u_2 = 50 m/s$ is the initial velocity

$\theta=30^{\circ}$ is the angle of projection

Substituting, we get:

$y_2(t)=(50)(sin 30^{\circ})t+\frac{1}{2}(-9.8)t^2=25t-4.9t^2$

The two objects collide when their vertical position is the same, so:

$y_1(t)=y_2(t)\\50-4.9t^2 = 25t-4.9t^2$

And solving for t, we find:

$50=25t\\t= 2 s$

Note that this means that the two object at t = 2 s have the same vertical position: however, this is not true for the horizontal position.

In order to find the point where they collide, we have to substitute the time of the collision that we found in part A into one of the expressions of the vertical position.

Substituting into the expression of object 2, we find:

$y_2(t) = 25t-4.9t^2=25(2.0)-4.9(2.0)^2=30.4 m$

We can verify that at the same time, the vertical position of object 1 is the same:

$y_1(t)=50-4.9t^2=50-4.9(2.0)^2=30.4 m$

This means that the two objects have the same vertical position at 30.4 m.

However, in reality, the two objects do not collide. In fact, object 1 is moving in the horizontal direction with constant velocity

$v_{1x}=25 m/s$

So its horizontal position at t = 2.0 s is

$x_1(2.0)=v_{1x}t=(25)(2.0)=50 m$

While object 2 is moving in the horizontal plane with velocity

$v_{2x}=u_2 cos \theta=(50)(cos 30^{\circ})=43.3 m/s$

So its horizontal position at t = 2.0 s is

$x_2(2.0)=v_{2x}t=(43.3)(2.0)=86.6 m$

So in reality, the two objects do not collide, if they start from the same x-position.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0

3 years ago

When viewing an object, we see its color according to the wavelength of light that the object

Fantom [35]

Objects appear to us to have the color that
they reflect (paint) or transmit (jello).

6 0

4 years ago

Read 2 more answers

How do you determine true displacement value of an object that has moved?

WITCHER [35]

Diceplacement is the distance an object has traveled in a certain direction
for example, if you were to walk North for 20m, then east for 40m, the <u>distance</u> you have traveled is 60m however your displacement is the distance between your starting position and your end position;
sqrt(20^2+40^2) = 44.7m
and because displacement is a vector, there needs to be a direction;
sin(theta)=40/44.7
theta=63.4 degrees East of North

therefore the true displacement is 44.7m at 63.4 degrees East of North

3 0

3 years ago

A three-wheeled car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed

DiKsa [7]

Answer:

(a) Total time = 105.5 sec

(b)average speed = 31.26 m/sec

Explanation:

We have given that three-wheeled car starts from rest that initial velocity u = 0 m/sec

Final velocity = 35 m/sec

Acceleration $a=2m/sec^2$