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3 years ago

Which two statements are true of electromagnetic waves?

Physics

1 answer:

Usimov [2.4K]3 years ago

3 0

Answer:

B and C

Explanation:

Because EMWs are varying magnetic and electric radiation traveling at 90° to each other propagating energy form one place to another through vibration of these magnetic and electric fields

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The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

5 0

3 years ago

Read 2 more answers

A car traveling with an initial velocity of 12 m/s accelerates at a constant rate of 2.2 m/s2 for a time of 4 seconds.

bazaltina [42]

So what's the question?

4 0

3 years ago

Elsie is finishing second grade. If she goes to school 147 day per year and she have 10 years of school left, how many days of s

ICE Princess25 [194]

Answer:

1,323 days left

Explanation:

147 x 10 = 1,470

1470 - 147 = 1,323

Hopefully this helps you :)

pls mark brainlest ;)

6 0

3 years ago

I would like help with this physics problem

Darina [25.2K]

(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.

Recall that

$v_f=v_i+at$

We have $v_f=-v_i$ , so that

$-2v_i=at\implies-2\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)=\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=8\,\mathrm s$

(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.

(d) The position of the ball $x_f$ at time $t$ is given by

$x_f=x_i+v_it+\dfrac12at^2$

Take the starting position to be the origin, $x_i=0$ . Then after 6 seconds,

$x_f=\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2=42\,\mathrm m$

so the ball is 42 m away from where it started.

We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

$x_f-x_i=\dfrac{v_i+v_f}2t\implies42\,\mathrm m=\dfrac{8\,\frac{\mathrm m}{\mathrm s}+v_f}2(6\,\mathrm s)\implies v_f=6\,\dfrac{\mathrm m}{\mathrm s}$

Since the velocity is positive, the ball is still moving up the incline.

8 0

3 years ago

Calculate the force generated by a car that hits the wall at an

Makovka662 [10]

This is a defective question. It was WRITTEN by someone who is unclear on the concepts. DON'T try and answer it.

It's trying to get us to use Newton's second law ... F = m • a.

But that only tells us how much force must act ON THE CAR in order to accelerate it. (45 kg) • (4 m/s^2) = 180 newtons.

This is NOT the force exerted BY the car when it hits something. THAT force depends on its speed WHEN it hits, AND how long it takes for the wreckage to actually come to rest, AND how hard or soft the wall is.

DON'T try to answer this question. Your answer will be wrong, you won't understand why, and the teacher you try to argue with probably won't either.

============================================

More explanation:

Think about jumping off of a ladder in your back yard. Several times.

Your mass is the same every time. Your acceleration is the same every time . . . 9.8 m/s² down, the acceleration of Earth gravity, every time.

BUT ...

-- I'll bet you would rather land on wood than on concrete. The force of landing would be less.

-- I'll bet you would rather land on dirt than on wood. The force of landing would be less.

-- I'll bet you would rather land on grass than on dirt. The force of landing would be less.

-- I'll bet you would rather land on a pile of blankets than on dirt. The force of landing would be less.

-- I'll bet you would rather land on a trampoline than on a pile of blankets. The force of landing would be less.

-- I'll bet you would rather jump from a short ladder than from a tall one. Your speed would be less when you landed, and the force of landing would be less.

==> Your mass is the SAME every time, and your acceleration is the SAME every time. But the force when you hit is DIFFERENT every time.

The mass and acceleration of the car DON'T tell us the force of the hit when the car hits a wall.

6 0

3 years ago