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dybincka [34]
3 years ago
6

Suppose you know only the position of an object at instants A and B. Is more than one displacement possible? Is more than one av

erage velocity possible? Is more than one average speed possible?
Physics
1 answer:
Burka [1]3 years ago
7 0
<span>Is more than one displacement possible?
No, displacement only considers the initial and final position.
Is more than one average velocity possible?
No, the average velocity is defined as displacement per time and since there is only one displacement possible, there is only one average velocity possible as well.
Is more than one average speed possible?
Yes, the average speed considers the total distance traveled and this distance may not be the same as the total displacement.
</span>
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<u>Principal</u><u> </u><u>focus</u><u> </u><u>of</u><u> </u><u>concav</u><u>e</u><u> </u><u>lens</u><u> </u><u>-</u><u> </u>

★ The point at which rays parallel to principal axis coming from infinity appear to converge after being refracted from concave lens is called the principal focus of concave lens.

<em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em>

• <u>Additional</u><u> information</u><u> </u><u>-</u><u> </u>

★ Principal focus - A number of rays parallel to the principal axis after reflection from a concave mirror meet at a point on the principal axis or appear to come from a point after reflection from a convex mirror on the principal axis. This is called principal focus.

8 0
2 years ago
Which are characteristics of light? Check all that apply
Schach [20]

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wavelength, frequency, energy and speed

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7 0
2 years ago
A 2 kg ball is moving 3 m/s when it starts rolling up a hill.
AURORKA [14]

Answer:

the height reached is = 0.458 [m]

Explanation:

We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:

Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]

Replacing the values on the equation we have:

Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\

This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.

Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\

Now we have:

h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]

In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]

6 0
3 years ago
You travel from Point A to Point B to Point C. What is your
Luba_88 [7]

Answer:

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3 0
3 years ago
Read 2 more answers
Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu
11111nata11111 [884]

Answer:

velocity = 62.89 m/s  in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $ 10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $ 10^5$  kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $ 10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $ 10^5$

So, $ V_x = \frac{10^5}{3000} $  = $ \frac{100}{3} $  m/s

and $ V_y=\frac{160}{3}$  m/s

Therefore, velocity is = $ \sqrt{V_x^2 + V_y^2} $

                                   = $ \sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2} $

                                   = 62.89 m/s

And direction is

tan θ = $ \frac{V_y}{V_x}$     = 1.6

therefore, $ \theta = \tan^{-1}1.6 $

                   = $ 58 ^{\circ}$  from x-axis

4 0
3 years ago
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