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3 years ago

HELP QUICK numbers 2-6

Physics

2 answers:

Sindrei [870]3 years ago

5 0

Answer:

2. B

3. A

4. C

5. B

6. A

Explanation:

2. Sound waves use vibrating molecules to move. With this being said, sound waves would travel best through dense materials, and would travel worst in empty materials, like space. Using this information, you can infer that B would be the answer because water is denser than air and water is the densest material on that list.

3. The speed of the wave is equal to the frequency times the wavelength. Since it has a wavelength of 100 meter and 20 Hertz, the speed would 2000 m/s

4. Like noted in question 2, sound waves travel best in dense materials. Rock is the densest material in that last.

5. By looking at the chart, you can see that the speed increases by 4 m/s when the temperature increases by 10. Using simple division, you could determine that the speed would go up by 2 m/s since the jump between 15 degrees and 20 degrees is 5, which is half of 10, making the speed half of 4.

6. The speed would decrease because, like stated above, sound travels best through dense materials. Since rock is denser than air, the speed would decrease as it leaves.

White raven [17]3 years ago

3 0

B,A,C,B,A Just took the test myself

You might be interested in

(A) The figure shows the setup which is used to observe an image formed wen a lighted candle is kept in front of a bi convex len

GREYUIT [131]

Answer:

<h3>1. In the case of candle flame , the object is placed beyond c , that means the image is formed or focused between c and f .in second case the object or sun is at infinity , so the image will be </h3><h3>formed at focus.this means the distance between image and lens has decreased in the second </h3><h3>case. either we have to move the screen towards the lens or the lens towards the screen.</h3>

Explanation:

I have posted the drawing.

in the second image

it is project c and characteristics

hope it helps u :)

6 0

3 years ago

You and your friend Peter are putting new shingles on a roof pitched at 21∘. You're sitting on the very top of the roof when Pet

Marat540 [252]

Answer:

3.69 m/s

Explanation:

Forces :

mgsin Θ - mumgcosΘ = ma

g x sinΘ - mu x g x cosΘ = a

9.8 x sin 21 - 0.53 x 9.8 x cos 21 = a

a = -1.337 m/s²

so you have final velocity = 0 m/s

initial velocity = ? m/s

Given d = 5.1 m

By kinematics

vf² = vo² + 2ad

0 = vo² + 2 x -1.337*5.1

vo = 3.69 m/s

8 0

3 years ago

In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.

Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement of the object

t is the time elapsed

$\omega$ is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

$\theta=2\pi$ rad

And the time taken is

$t=95 min \cdot 60 =5700 s$

Therefore, the angular velocity of the telescope is

$\omega=\frac{2\pi}{5700}=0.0011 rad/s$

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

$v=\omega r$

where

v is the linear velocity

$\omega$ is the angular velocity

r is the radius of the circular orbit

In this problem:

$\omega=0.0011 rad/s$ is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

$r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m$

Therefore, the linear velocity of the telescope is:

$v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s$

4 0

3 years ago

A long line of charge with uniform linear charge density λ1 is located on the x-axis and another long line of charge with unifor

Dahasolnce [82]

Answer:

A.The positive z-direction

Explanation:

We are given that

Linear charge density of long line which is located on the x-axis= $\lambda_1$

Linear charge density of another long line which is located on the y-axis= $\lambda_2$

We have to find the direction of electric field at z=a on the positive z-axis if $\lambda_1$ and $\lambda_2$ are positive.

The direction of electric field at z=a on the positive z-axis is positive z-direction .

Because $\lambda_1$ and $\lambda_2$ are positive and the electric field is applied away from the positive charge.

Hence, option A is true.

A.The positive z-direction

6 0

3 years ago

The charges Q1=Q and Q2=4Q that are a distance d apart, repel each other with a force of 1.60 N. What would be the force between

gladu [14]

Answer:

50.4 N

Explanation:

Q1 = Q

Q2 = 4 Q

Distance = d

The force is given by

$F = \frac{KQ_{1}Q_{2}}{d^{2}}$

$1.60 = \frac{4KQ^{2}}{d^{2}}$ .... (1)

Now,

Q3 = 2 Q

Q4 = 7 Q

distance = d/3

$F' = \frac{9KQ_{3}Q_{4}}{d^{2}}$

$F' = \frac{126KQ^{2}}{d^{2}}$ .... (2)

Divide equation (2) by equation (1), we get

F' / 1.60 = 126 / 4

F' = 50.4 N

Thus, the force is 50.4 N.

7 0

3 years ago