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nikklg [1K]
3 years ago
5

14

Physics
1 answer:
romanna [79]3 years ago
6 0

Answer: 0.00068 N

Explanation: Universal gravitational constant=6.674 *10^(-11)

Force=Gm1m2/(r^2)

Force= 6.67*25000*40000*10^(-11)/(10^2)

Force=0.00068 N

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Please someone help me !!
sergij07 [2.7K]

Answer:

D

Explanation:

because if the solvent is more than the solvent then we can't resolve it.

so our product will be suspended

6 0
3 years ago
How much power is used by a hair dryer if it does 40,000 J of work in 40 seconds?
bearhunter [10]

Answer:

40000÷40=1000 joules is required to work in 40 seconds

6 0
3 years ago
A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result
Over [174]
B) law of conservation of momentum

It states that the total momentum of a system before impact is the same as the total momentum of the system after impact.

In this case total momentum before impact:

10kg*5m/s  + 5kg * 0m/s = 50 kg m/s

After Impact:

10kg*0m/s + 5kg*10m/s = 50 kg m/s

You can see the momentum before and after impact is same as 50 kg m/s  

Of course we assumed that the first cart stopped after the impact, and there are no energy losses.
7 0
3 years ago
What must be the units for the gravitational constant G in order for gravitational force to have units of newtons?
babunello [35]

Answer:

m³/(kg⋅s²)

Explanation:

Hello.

In this case, since the involved formula is:

F=G*\frac{m_1m_2}{r^2}

By writing a dimensional analysis with the proper algebra handling, we obtain:

N[=]G*\frac{kg*kg}{m^2}\\ \\kg*\frac{m}{s^2}[=]G *\frac{kg*kg}{m^2}\\\\G[=]\frac{kg*m*m^2}{kg^2*s^2}\\ \\G[=]\frac{m^3}{kg*s^2}

Thus, answer is:

m³/(kg⋅s²)

Note that the [=] is used to indicate the units of G.

Best regards

4 0
3 years ago
Air flows into a jet engine at 70 lbm/s, and fuel also enters the engine at a steady rate. The exhaust gases, having a density o
Andrews [41]

Answer:

1387908 lbm/h

Explanation:

Air flowing into jet engine = 70 lbm/s

ρ = Exhaust gas density = 0.1 lbm/ft³

r = Radius of exit with a circular cross section = 1 ft

v = Exhaust gas velocity = 1450 ft/s

Exhaust gas mass (flow rate)= Air flowing into jet engine + Fuel

Q = (70+x) lbm/s

Area of exit with a circular cross section = π×r² = π×1²= π m²

Now from energy balance

Q = ρ×A×v

⇒70+x = 0.1×π×1450

⇒70+x = 455.53

⇒ x = 455.53-70

⇒ x = 385.53 lbm/s

∴ Mass of fuel which is supplied to the engine each minute is 1387908 lbm/h

8 0
3 years ago
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