What is net force?

Which of the following paraphrases Hubble Law?Select one:A. The greater the distance to a galaxy, the greater its redshift. B. T

olya-2409 [2.1K]

The correct answer is:

~A. The greater the distance to a galaxy, the greater its redshift.

Hope this helps!!!

4 0

3 years ago

The area of the velocity time graph gives displacement of the body true or false?

mote1985 [20]

The area of the velocity time graph gives displacement of the body true or false?

The answer this is true.

8 0

3 years ago

Read 2 more answers

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve

Gemiola [76]

Answer:

(a) v = 5.42m/s

(b) vo = 4.64m/s

(c) a = 2874.28m/s^2

(d) Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

$v=\sqrt{2gh}$ (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

$v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}$

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

$h_{max}=\frac{v_o^2}{2g}$ (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

$v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}$

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}$

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}$

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

$\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m$

THe compression of the ball when it strikes the floor is 5.11*10^-3m

4 0

3 years ago

An alternating source drives a series RLC circuit with an emf amplitude of 6.04 V, at a phase angle of +30.3°. When the potentia

Vinvika [58]

Answer:

-8.56V

Explanation:

Our values are given by,

e = 6.04 V

Φ = 30.3

VC = 5.32

We can calculate the voltage across the circuit with the emf formula, that is,

$e(t) = e* sin(wt)$

$e(t) = 6.04 * sin(Φ + π)$

$e(t) = 6.04 * sin(32.5 + 180)$

$e(t) = -3.245 V$

Now, Using Kirchoff Voltage Law,

$e(t) - VR- VL - VC = 0$

$-3.24 - 0 - VL - 5.32 = 0$

Finally we have the potential difference across the inductor.

$VL = - 8.56 v$

5 0

3 years ago

A child stands on the edge of a merry-go-round of radius 1.63 m which is rotating at 2.13 rad/s.

Brilliant_brown [7]

Answer:

9

Explanation:

2.13 rad/s * 26.9 sec

2.13 * 26.9

57.297

3282.88 deg / 360 deg = 9.12

It makes 9 complete revolutoins

7 0

2 years ago