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3 years ago

The amount of gravity between 1kg of lead and Earth is ___ the amount of gravity between 1kg of marshmallows and Earth.

1 answer:

3 0

The amount of gravity between 1 kg of lead and Earth is the same as the amount of gravity between 1 kg of marshmallows and Earth.

Answer: Option C

<u>Explanation:</u>

According to universal law of gravity, the gravitational force is directly proportionate to the product of masses of the interacting objects and inversely proportionate to its distance squared between them.

Since the mass of Earth is very high compared to the mass of the other objects, be it lead or marshmallows, and the distance will be mostly same. Thus, there will be no change in the amount of gravity acting between them. Thus, the amount of gravity between 1 kg of lead and Earth is the same as the amount of gravity between 1 kg of marshmallows and Earth.

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Which vector shows the direction of the centripetal acceleration at this point

Mademuasel [1]

Answer:

It's B

Explanation:

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3 years ago

This 80 kg car is moving at 20m/sec at the top where the hills radius is 100m. What is the centrifugal force?

earnstyle [38]

100 seconds is the right thing

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3 years ago

The current in two identical light bulbs connected in series is 0. 25 A. The voltage across both bulbs is 110 V. The resistance

konstantin123 [22]

Answer:

220 ohms

Explanation:

I = V / R

0.25 = 110 / R

R = 110 / 0.25

R = 440 ohms

Equivalent resistance = 440 ohms

Resistance of single light bulb = Equivalent resistance / number of bulbs

= 440 / 2

= 220 ohms

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1 year ago

A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a

UNO [17]

Answer:

$P_{avg} = 6.283*10^{-9} \ W$

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

$\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})$

where;

$E= \frac{d \phi}{dt}$

$E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t$

$E_{rms} = \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}$

Replacing our values into above equation; we have:

$E_{rms} = \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}$

$E_{rms} = \frac {8.98909588*10^{-4} }{\sqrt{2}}$

$E_{rms} = 6.356*10^{-4} \ V$

Then the $P_{avg$ is calculated as:

$P_{avg} = \frac{E^2}{R}$

$P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}$

$P_{avg} = 6.283*10^{-9} \ W$

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3 years ago

What happens when you "crack/pop your knuckles?

Art [367]

It could result in it not being good for your joints, as well as in the long run but shouldn't cause problems when your a child. I hope this helps your question!

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3 years ago