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3 years ago

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the earth's moon has a gravitational field strength of about 1.6 n/kg near its surface. the moon has a mass of 7.35x10^22 kg. wh

at is the radius of the moon? please show steps on how to do it please.

1 answer:

Andrew [12]3 years ago

8 0

Answer:

1750km

Explanation:

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A balloon filled with helium gas at 20°C occupies 4.91 L at 1.00 atm. The balloon is immersed in liquid nitrogen at -196°C, whil

Marizza181 [45]

Answer:

V = 0.248 L

Explanation:

To do this, use the following equation:

P1*V1/T1 = P2*V2/T2

This equation is used to find a relation between two differents conditions of a same gas, which is this case. From this equation we can solve for V2.

Solving for V2:

V2 = P1*V1*T2/T1*P2

Temperature must be at Kelvin, so, we have to sum the temperature 273 to convert it in K.

Replacing the data we have:

V2 = 1 * 4.91 * (-196+273) / 5.2 * (20+273)

V2 = 378.07 / 1523.6

V2 = 0.248 L

7 0

3 years ago

What do you know about tides?

Neporo4naja [7]

Tides are the rise and fall of sea levels caused by the combined effects of the gravitational forces exerted by the Moon and the Sun, and the rotation of the Earth. Tide tables can be used to find the predicted times and amplitude (or "tidal range") of tides at any given locale.

For further expiation please contact me at 678-987-2411. Disclaimer(This is not a real number)

8 0

3 years ago

Assume a change at the source of sound reduces the wavelength of a sound wave in air by a factor of 3.

noname [10]

Explanation:

The speed of a wave is given by :

$v=f\lambda$ ......(1)

(i) Here, the wavelength of a sound wave in air reduces by a factor of 3. Equation (1) becomes :

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

Wavelength and frequency are inversely proportional to each other. So, if wavelength of a sound wave in air reduces by a factor of 3, then the frequency will increases by a factor of 3.

(ii) It remains the same.

8 0

3 years ago

A purse at radius 2.30 m and a wallet at radius 3.45 m travel in uniform circular motion on the floor of a merry-go-round as the

ivolga24 [154]

Answer:

The acceleration of the wallet is $3\hat{i}+6\hat{j}$

Explanation:

Given that,

Radius of purse r= 2.30 m

Radius of wallet r'= 3.45 m

Acceleration of the purse $a=2\hat{i}+4.00\hat{j}$

We need to calculate the acceleration of the wallet

Using formula of acceleration

$a=r\omega^2$

Both the purse and wallet have same angular velocity

$\omega=\omega'$

$\sqrt{\dfrac{a}{r}}=\sqrt{\dfrac{a'}{r'}}$

$\dfrac{a}{r}=\dfrac{a'}{r'}$

$\dfrac{a'}{a}=\dfrac{r'}{r}$

$\dfrac{a'}{a}=\dfrac{3.45}{2.30}$

$\dfrac{a'}{a}=\dfrac{3}{2}$

$a'=\dfrac{3}{2}\times(2\hat{i}+4.00\hat{j})$

$a'=3\hat{i}+6\hat{j}$

Hence, The acceleration of the wallet is $3\hat{i}+6\hat{j}$

4 0

3 years ago

A 150 kg safe on frictionless casters is being raised at 1.20m to the bed of a truck using planks 4m long. The force needed to p

Vesnalui [34]

The only thing you need to know in order to solve this task is that <span>plank length (which is force x), should equal the increase in potential energy, so what we have now : (mass)* g * (height).
It has to look like that: </span>
<span>F * 3.0 = 150 x 9.81 x 1.20
Then solve for F, the result should be in newtones = 588N

Do hope it makes sense.</span>

3 0

2 years ago