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2 years ago

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Helpppp meeeee eee!!

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jeka942 years ago

4 0

Answer: its a

Explanation: hope this help im sorry if its wrong

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The mass of a star can be determined by studying

vovikov84 [41]

The correct answer for the question that is being presented above is this one: "a. the wavelength of light emitted by the star." The mass of a star can be determined by studying <span>the wavelength of light emitted by the star. It also has something to do with the gravitational force that it exerts.</span>

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3 years ago

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In a shipping yard, a crane operator attaches a cable to a 1240-kg shipping container and then uses the crane to lift the contai

Firdavs [7]

Answer:

Explanation:

Given

mass of ship =1240 kg

height=37 m

(a)Work done by Tension in cable+work done by gravity =0

Work done by gravity =change in potential energy of mass $=-1240\times 9.8\times 37=-449.624 kJ$

Thus work done by tension=449.624 kJ

(b)work done by force of gravity $=-mg\times h=-1240\times 9.8\times 37$ =-449.624 kJ

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3 years ago

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Who sailed the Atlantic from France to explore

Masja [62]

Answer:

navigator Jacques Cartier

Explanation:

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2 years ago

Which type(s) of electromagnetic radiation do human bodies emit? Which type(s) can our senses detect?

Andrei [34K]

Our bodies emit heat, and nerve endings in our skin can detect it.
Our eyes can detect visible light, but our bodies don't emit that.

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3 years ago

A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re

fomenos

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

$PV=nRT$

For a gas

$P_{1}V_{1}=nRT_{1}$

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

$10\times V=nR\times286$ ....(I)

When the temperature of the gas is increased

Then,

$P_{2}V_{2}=\dfrac{n}{2}RT_{2}$ ....(II)

Divided equation (I) by equation (II)

$\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}$

$\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}$

$P_{2}=\dfrac{10\times368}{2\times286}$

$P_{2}= 6.433\ atm$

$P_{2}=6.4\ atm$

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

4 0

2 years ago