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3 years ago

How would the moon appear from Earth if the moon did not rotate?

2 answers:

andrey2020 [161]3 years ago

6 0

It would mean that only one side of earth would be light and the other dark all the time also we would only see the sun on one side and on the other we see the moon

n200080 [17]3 years ago

4 0

As the Earth rotates, the Moon rotates the same, hence why we only see one side of the Moon at any time, but if it stopped rotating we would see different parts of it as we rotate...basically it wouldn't look the same all the time (source: my brain, or 7th grade science class: either or)

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A stone was dropped off a cliff and hit the ground with a speed of 88 ft/s. What is the height (in feet) of the cliff

igor_vitrenko [27]

Answer:

the height (in feet) of the cliff is 121 ft

Explanation:

A stone hit the cliff with

speed, v = 88 ft/s

Acceleration, a= 32 ft/s^2

initial speed, u = 0 ft/s

height is h.

To solve this problem we will apply the linear motion kinematic equations, Equation of motion describes change in velocity, depending on the acceleration and the distance traveled

so, writing the formula of Equation of motion:

v^2 - u^2 = 2*a*h

substituting the appropriate values,

(88)^2 - 0 = 2*32* h

h=(88)^2 / 64

h= 121 ft

hence

the height (in feet) of the cliff is 121 ft

learn more about height of the cliff here:

<u>brainly.com/question/24130198</u>

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3 0

1 year ago

Which quantity is measured in newton seconds (Ns)?

pshichka [43]

Answer:

Impulse

Explanation:

Impulse is force times time

7 0

2 years ago

How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium

Minchanka [31]

The height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

<h3>Pressure and temperature at equilibrium </h3>

The relationship between pressure and temperature can be used to determine the height risen by the water.

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where;

V₁ = AL
V₂ = A(L - y)
P₁ = Pa
P₂ = Pa + ρgh
T₁ = 20⁰C = 293 K
T₂ = 10⁰ C = 283 k

$\frac{PaAL}{T_1} = \frac{(P_a + \rho gh)A(L-y)}{T_2} \\\\\frac{PaL}{T_1} = \frac{(P_a + \rho gh)(L-y)}{T_2} \\\\L-y = \frac{PaLT_2}{T_1(P_a + \rho gh)} \\\\y = L (1 - \frac{PaT_2}{T_1(P_a + \rho gh)})\\\\y = 4.2(1 - \frac{101325 \times 283}{293(101325\ +\ 1000 \times 9.8 \times 100)} )\\\\y = 3.8 \ m$

Thus, the height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

The complete question is below:

A diving bell is a 4.2 m -tall cylinder closed at the upper end but open at the lower end. The temperature of the air in the bell is 20 °C. The bell is lowered into the ocean until its lower end is 100 m deep. The temperature at that depth is 10°C. How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium?

Learn more about thermal equilibrium here: brainly.com/question/9459470

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3 0

1 year ago

What type of animals does Dr. Grant study?

Rasek [7]

Dr. Alan Grant is the main protagonist in Jurassic Park, with the book written primarily from his perspective. He is a paleontology professor at the University of Denver and receives research funding from the Hammond Foundation. He became a world-renowned paleontologist after discovering dinosaur nest fossils in Montana. Billionaire John Hammond chooses Dr. Grant to evaluate his dinosaur amusement park because of his professional expertise and unbiased opinion on dinosaurs.

Idk if this is related to what you ask but it might help.

7 0

3 years ago

Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy

Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

$P_{1}V_{1}=nRT$

$V_{1}=\dfrac{nRT}{P_{1}}$

Put the value into the formula

$V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}$

$V_{1}=2.62\times10^{-3}\ m^3$

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

$T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}$

Put the value into the formula

$T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}$

$T_{2}=233.7\ K$

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

$P_{1}V_{1}=P_{2}V_{2}$

$V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}$

Put the value into the formula

$V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}$

$V_{2}=0.00786\ m^3$

$V_{2}=7.86\times10^{-3}\ m^3$

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

$V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}$

$V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}$

$V_{2}=0.0057\ m^3$

$V_{2}=5.7\times10^{-3}\ m^3$

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

6 0

3 years ago