How does the interstellar medium affect our view of most of the galaxy?

in physics lab, a cube slides down a frictionless incline as shown in the figure below, and elastically strikes another cube at

Tema [17]

<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:

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3 0

3 years ago

A ball with a mass of 3.1 kg is moving in a uniform circular motion upon a horizontal surface. The ball is attached at the cente

inessss [21]

Answer:

3.46 seconds

Explanation:

Since the ball is moving in circular motion thus centripetal force will be acting there along the rope.

The equation for the centripetal force is as follows -

$F=\frac{mv^2}{r}$

Where, $m$ is the mass of the ball, $v$ is the speed and $r$ is the radius of the circular path which will be equal to the length of the rope.

This centripetal force will be equal to the tension in the string and thus we can write,

$20.4 = \frac{3.1\times v^2}{2}$

and, $v^2 = 13.16$

Thus, $v = 3.63$ m/s.

Now, the total length of circular path = circumference of the circle

Thus, total path length = 2πr = 2 × 3.14 × 2 = 12.56 m

Time taken to complete one revolution = $\frac{\text {Path length} }{\text {Speed}}$ = $\frac{12.56}{3.63}$ = 3.46 seconds.

Thus, the mass will complete one revolution in 3.46 seconds.

4 0

3 years ago

A 0.10 g honeybee acquires a charge of +23 pC while flying.

kari74 [83]

Answer:

a) $\frac{F}{w} =2.347\times 10^{-6}\ N$

b) $E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, $m=10^{-4}\ kg$

charge acquired by the bee, $q_2=23\times 10^{-12}\ C$

a.

Electrical field near the earth surface, $E=100\ N.C^{-1}$

Now the electric force on the bee:

we know:

$F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}$

$F=E.q_2$

$F=100\times 23\times 10^{-12}$

$F=23\times 10^{-10}\ N$

The weight of the bee:

$w=m.g$

$w=10^{-4}\times 9.8$

$w=9.8\times10^{-4}\ N$

Therefore the ratio :

$\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}$

$\frac{F}{w} =2.347\times 10^{-6}\ N$

b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

$F=9.8\times10^{-4}\ N$

$E.q_2=9.8\times10^{-4}\ N$

$E\times 23\times 10^{-12}=9.8\times10^{-4}\ N$

$E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

3 0

3 years ago

Use the graph below to answer the following question: if average acceleration is calculated using the equation, “ change in velo

sergiy2304 [10]

Answer:

$a=9\ cm/s^2$

Explanation:

<u>Average Acceleration </u>

Acceleration is a physical magnitude defined as the change of velocity over time. When we have experimental data, we can compute it by calculating the slope of the line in velocity vs time graph.

Note: <em>We cannot see if the time axis is numbered in increments of 1 second, and we'll assume that. </em>

When $t_2=4\ sec$ , the graph shows a value of $v_2=36\ cm/s$