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2 years ago

7

Which "sphere" does an astronomer most directly study?

2 answers:

Allushta [10]2 years ago

7 0

Answer:

exosphere

Explanation:

exosphere is space

astronomy is space

Lisa [10]2 years ago

5 0

I would choose B. exosphere. Astronomers study outer space, planets and ect.

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John wants a new video tape and a new shirt, but he only has $12.00. He buys the video tape and gets $0.52 change. What was his

mestny [16]

Answer:

You have the answer in your comments. I will be copying it so your question doesn't get deleted.

The answers is $0.58

$11.48

the video tape

the new shirt

7 0

2 years ago

How do I find frictional force with force applied?

lions [1.4K]

Answer: Choose the normal force acting between the object and the ground. Let's assume a normal force of 250 N.

Determine the friction coefficient.

Multiply these values by each other: 250 N * 0.13 = 32.5 N .

You just found the force of friction!

Explanation:

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2 years ago

dem82 [27]

Answer:

Explanation:

28 / 70 = 0.3857142... = 0.39 hr

280 / 100 = 2.8 hrs.

(100 - 0) / 10 = 10 m/s²

(60 - 20) / 4 = 10 m/s²

3 0

2 years ago

The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How

Vlada [557]

Answer:

The value is $w = 7.54 \ m$

Explanation:

From the question we are told that

The length of the crack is $a = 0.3 \ m$

The frequency is $f = 30.0 \ kHz = 30 *10^{3} \ Hz$

The distance outside the cave that is being consider is $D = 100 \ m$

The speed of sound is $v_s = 340 \ m/s$

Generally the wavelength of the wave is mathematically represented as

$\lambda = \frac{v}f}$

=> $\lambda = \frac{340 }{30*10^{3}}$

=> $\lambda = 0.0113 \ m/s$

Generally for a single slit the path difference between the interference patterns of the sound wave and the center is mathematically represented as

$y = \frac{ n * \lambda * D}{a}$

=> $y = \frac{ 1 * 0.0113 * 100}{0.3}$

=> $y = 3.77 \ m$

Generally the width of the sound beam is mathematically represented as

$w = 2 * y$

=> $w = 2 * 3.77$

=> $w = 7.54 \ m$

4 0

2 years ago

Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 20

Genrish500 [490]

Answer:

The time it can operate between chargins in minutes is

$t=102.8 minutes$

Explanation:

Given: $m=500kg$ , $r=1.0m$ , $w=200\pi rad/s$

a). The rotational kinetic energy

$K_R=\frac{1}{2}*I*w^2$

$I=\frac{1}{2}*m*r^2$

$I=\frac{1}{2}*500kg*(1.0m)^2$

$I=250 kg*m^2$

$K_R=\frac{1}{2}*250kg*m^2*(200\pi rad/s)^2$

$K_R=49.348x10^6J$

b). The power average 0.8kW un range time can be find

$P=\frac{K_R}{t'}$

Solve to t'

$t=\frac{K_R}{P}$

$t=\frac{49.348x10^6}{0.8x10^3w}=6168.5s$

$t=6168.5s\frac{1minute}{60s}=102.8 minutes$

3 0

2 years ago