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Stels [109]
3 years ago
15

FAST ANSWERS I NEED HELP FASTT

Physics
2 answers:
GuDViN [60]3 years ago
6 0
The answer is Radiation
yuradex [85]3 years ago
5 0
The answer would be radiation hope it helped
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Professional assessments are the only reliable method for determining an individual's levels of health-related fitness.
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What is the average speed between the times t = 4s and t = 12 s?
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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.

B)

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
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