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4 years ago

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Suppose an astrophotographer hands you a picture with star trails taken looking toward the north celestial pole. If the star tra

ils are 1/6 of a complete circle, about how many hours was the picture exposed? Hint: How many hours would it take the stars to make a full circle? Report your answer to the nearest whole number

1 answer:

Reil [10]4 years ago

4 0

Answer:

Our answer will be 4 hours.

Explanation:

It takes 24 hours to complete a circle(360° rotation)

Hence to complete 1/6th of complete circle it will take 24/6= 4 hours

Exact in integers= 4 hrs. Hence our answer will be 4 hours.

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What should not be used to support a scaffold because they could collapse, sending the structure to the ground?

uranmaximum [27]

Answer:

Cinder blocks

Explanation:

From safety training guides, whenever there is inadequate support, poor

construction, or a movement in the components of the scaffold such as the coupler as well as the base upon which the scaffold structure is built, there will be a great danger of collapse.

In light of this safety rule, Cinder blocks such as sandcrete blocks, bricks e.t.c shouldn't be used to support scaffold because they have a high tendency of being crushed under load

5 0

3 years ago

Please help me this is a test and it needs to be done in a few

Amanda [17]

Answer:

I think A because astroids are stronger than a crash. Hope I helped! :)

Explanation:

7 0

3 years ago

Read 2 more answers

Convert 3402kgm/s to 20000Newtons

oee [108]

The 3,402 has units of kg-m/s. That's momentum. The 20,000 has units of Newtons. That's force. Momentum and force are different physical things, and you can't convert them from one to the other.

The best I can do for you is something like this:

Let's say you have a moving object with 3,402 kg-m/s of momentum, and you want to STOP it completely. You want to stand in front of it and push back on it, hard enough and for long enough to CHANGE its momentum from 3,402 kg-m/s to zero.

Also ... there's a limit to how hard you can push. The most force you can exert is 20,000 Newtons.

The amount you'll change its momentum is called the <u><em>impulse</em></u> you give it. The quantity of impulse is (force) x (length of time you push on it).

So you need to keep pushing it back for (T seconds) long enough so that

(20,000 Newtons of force) x (T seconds) = 3,402 kg-m/s of momentum .

Divide each side of that equation by (20,000 Newtons). Then it says:

(T seconds) = (3,402 kg-m/s) / (20,000 Newtons)

<em>T = 0.1701 second</em>

And that's how you provide just enough impulse to stop the flying object ... push on it with 20,000 Newtons of force for exactly 0.1701 second, and it loses all its momentum and falls out of the air onto the ground at your feet.

This story is the closest I can come to anything that looks like "convert"ing momentum into force.

3 0

3 years ago

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler

ki77a [65]

Answer:

a) $t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

b) $\theta_1=\frac{w_1^2}{2\alpha}rad$

c) $t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

Explanation:

1) Basic concepts

Angular displacement is defined as the angle changed by an object. The units are rad/s.

Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:

$w=\frac{\Delat \theta}{\Delta t}$

Angular acceleration is the rate of change of the angular velocity respect to the time

$\alpha=\frac{dw}{dt}$

2) Part a

We can define some notation

$w_o=0\frac{rad}{s}$ ,represent the initial angular velocity of the wheel

$w_1=?\frac{rad}{s}$ , represent the final angular velocity of the wheel

$\alpha$ , represent the angular acceleration of the flywheel

$t_1$ time taken in order to reach the final angular velocity

So we can apply this formula from kinematics:

$w_1=w_o +\alpha t_1$

And solving for t1 we got:

$t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

3) Part b

We can use other formula from kinematics in order to find the angular displacement, on this case the following:

$\Delta \theta=wt+\frac{1}{2}\alpha t^2$

Replacing the values for our case we got:

$\Delta \theta=w_o t+\frac{1}{2}\alpha t_1^2$

And we can replace $t_1$ from the result for part a, like this:

$\theta_1-\theta_o=w_o t+\frac{1}{2}\alpha (\frac{w_1}{\alpha})^2$

Since $\theta_o=0$ and $w_o=0$ then we have:

$\theta_1=\frac{1}{2}\alpha \frac{w_1^2}{\alpha^2}$

And simplifying:

$\theta_1=\frac{w_1^2}{2\alpha}rad$

4) Part c

For this case we can assume that the angular acceleration in order to stop applied on the wheel is $\alpha_1 =-5\alpha \frac{rad}{s}$

We have an initial angular velocity $w_1$ , and since at the end stops we have that $w_2 =0$

Assuming that $t_2$ represent the time in order to stop the wheel, we cna use the following formula

$w_2 =w_1 +\alpha_1 t_2$

Since $w_2=0$ if we solve for $t_2$ we got

$t_2=\frac{0-w_1}{\alpha_1}=\frac{-w_1}{-5\alpha}$

And from part a) we can see that $w_1=\alpha t_1$ , and replacing into the last equation we got:

$t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

5 0

3 years ago

Suppose you wanted to use a non-reflecting layer for radar waves to make an aircraft invisible. What would the thickness of the

Xelga [282]

Answer:

the thickness of the film for destructive interference is 1 cm

Explanation:

We can assume that the radar wave penetrates the layer and is reflected in the inner part of it, giving rise to an interference phenomenon of the two reflected rays, we must be careful that the ray has a phase change when

* the wave passes from the air to the film with a higher refractive index

* the wavelength inside the film changes by the refractive index

λ = λ₀ / n

so the ratio for destructive interference is

2 n t = m λ

t = m λ / 2n

indicate that the wavelength λ = 2 cm, suppose that the interference occurs for m = 1, therefore it is thickness

t = 1 2/2 n

t = 1 / n

where n is the index of refraction of the anti-reflective layer. As they tell us not to take into account the change in wavelength when penetrating the film n = 1

t = 1 cm

So the thickness of the film for destructive interference is 1 cm

8 0

3 years ago