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3 years ago

What would happen to the moon if the earth stopped exerting the force of gravity onto it?

1 answer:

nataly862011 [7]3 years ago

5 0

It would drift away from the Earth and enter its own orbit around the sun. It would become the 9th planet, in an orbit very close to Earth's orbit. (Our moon is larger than Pluto, and almost as large as Mercury.)

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Temperature and pressure can change the what of a solute

igomit [66]

Temperature and pressure can change the solubility of a solute.

4 0

3 years ago

The transfer of thermal energy via electromagnetic waves is know as what

katrin2010 [14]

Answer:

Radiation

Explanation:

7 0

3 years ago

If 250 g of calcium chlorate decomposes into calcium chloride and oxygen, what mass of oxygen would be produced?

sineoko [7]

115.94g of O₂ are produced by 250g of Ca(ClO₃)₂

<u>Explanation:</u>

Ca(ClO₃)₂ -->CaCl₂ + 3O₂

One mole of calcium chlorate give three moles of oxygen.

Molar mass of calcium chlorate is

40+(35.5+16*3)*2 = 207g/mol

Molar mass of Oxygen = 32g

1 mole of calcium chlorate decomposes to form 3 moles of oxygen.

So,

207g of Ca(ClO₃)₂ produces 3 X 32g of O₂

250g of Ca(ClO₃)₂ will produce = $\frac{3 X 32}{207} X 250$

= 115.94g of O₂

Therefore, 115.94g of O₂ are produced by 250g of Ca(ClO₃)₂

7 0

3 years ago

TIMED TEST! 10 POINTS!

Airida [17]

D. 11cm
hope i helped

3 0

3 years ago

A car accelerates uniformly from rest to 24.8 m/s in 7.88 s along a level stretch of road. Ignoring friction, determine the aver

nevsk [136]

Answer:

When he weight of the car is 8.55 x $10^{3}$ N then power = 314.012 KW

When he weight of the car is 1.10 x $10^{4}$ N then power = 43.76 KW

Explanation:

Given that

Initial velocity $V_{1}$ = 0

Final velocity $V_{2}$ = 24.8 $\frac{m}{s}$

Time = 7.88 sec

We know that power required to accelerate the car is given by

P = $\frac{change \ in \ kinetic \ energy}{time}$

Change in kinetic energy Δ K.E = $\frac{1}{2} m (V_{2}^{2} - V_{1}^{2} )$

Since Initial velocity $V_{1}$ = 0

⇒ Δ K.E = $\frac{1}{2} m V_{2} ^{2}$

⇒ Power P = $\frac{1}{2} \frac{m}{t} V_{2} ^{2}$

⇒ Power P = $\frac{1}{2} \frac{W}{g\ t} V_{2} ^{2}$ -------- (1)

(a). The weight of the car is 8.55 x $10^{3}$ N = 8550 N

Put all the values in above formula

So power P = $\frac{1}{2} \frac{8550}{(9.81)\ (7.88)} (24.8) ^{2}$

P = 314.012 KW

(b). The weight of the car is 1.10 x $10^{4}$ N = 11000 N

Put all the values in equation (1) we get

P = $\frac{1}{2} \frac{11000}{(9.81)\ (7.88)} (24.8) ^{2}$

P = 43.76 KW

5 0

3 years ago