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Harman [31]
4 years ago
14

While in empty space, an astronaut throws a ball at a velocity of 11 m/s. what will the velocity of the ball be after it has tra

veled 3 meters?
Physics
1 answer:
Kisachek [45]4 years ago
5 0
It will be traveling at 11m/s, because there is no friction in empty space.
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Which of the following career roadblocks is completely out of your control
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C.

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An object of mass 2kg is on an incline where there is an applied force of 15N. The
seraphim [82]

a) See free-body diagram in attachment

b) The acceleration is 2.46 m/s^2

Explanation:

a)

The free-body diagram of an object is a diagram representing all the forces acting on the object. Each force is represented by a vector of length proportional to the magnitude of the force, pointing in the same direction as the force.

The free-body diagram for this object is shown in the figure in attachment.

There are three forces acting on the object:

  • The weight of the object, labelled as mg (where m is the mass of the object and g is the acceleration of gravity), acting downward
  • The applied force, F_a, acting up along the plane
  • The force of friction, F_f, acting down along the plane

b)

In order to find the acceleration of the object, we need to write the equation of the forces acting along the direction parallel to the incline. We have:

F_a - F_f - mg sin \theta = ma

where:

F_a = 15 N is the applied force, pushing forward

F_f = 5 N is the frictional force, acting backward

mg sin \theta is the component of the weight parallel to the incline, acting backward, where

m = 2 kg is the mass of the object

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

\theta=15^{\circ} is the angle between the horizontal and the incline (it is not given in the problem, so I assumed this value)

a is the acceleration

Solving for a, we find:

a=\frac{F_a - F_f - mg sin \theta}{m}=\frac{15-5-(2)(9.8)(sin 15^{\circ})}{2}=2.46 m/s^2

Learn more about inclined planes:

brainly.com/question/5884009

#LearnwithBrainly

8 0
3 years ago
A car is moving in the positive direction along a straight highway and accelerates at a constant rate while going from point A t
rusak2 [61]

Answer:

The time where the avergae speed equals the instaneous speed is T/2

Explanation:

The velocity of the car is:

v(t) = v0 + at

Where v0 is the initial speed and a is the constant acceleration.

Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:

v_{ave} = \frac{1}{T}\int\limits^T_0 {v(t)} \, dt

v_ave = v0+a(T/2)

We can esaily note that when <u><em>t=T/2</em></u><u><em> </em></u>

v(T/2)=v_ave

Now we want to know where the car should be, the osition of the car is:

x(t) = x_A + v_0 t + \frac{1}{2}at^2

Where x_A is the position of point A. Therefore, the car will be at:

<u><em>x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2</em></u>

8 0
3 years ago
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