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3 years ago

Can I write a narrative writing how the first time I fell in love with blue eyes?

Physics

1 answer:

STatiana [176]3 years ago

4 0

Yes you can. Fiction or real. You can almost write any topic you want...

You might be interested in

. Which of the following is NOT a property of water? A. Ability to dissolve many substances B. Constant volume upon freezing C.

ss7ja [257]

Answer:

Explanation:

water is only able to desolve two substances , salt , sugar,

7 0

3 years ago

Read 2 more answers

A ball is thrown off a cliff at a speed of 10 m/s in a horizontally direction. The ball reaches the ground 1.5 seconds. If the b

Tems11 [23]

I am pretty sure it is d

5 0

2 years ago

Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected b

zmey [24]

Answer:

The maximum amount of work is $W = 1563.289 \ J$

Explanation:

From the question we are told that

The temperature of the environment is $T = 280\ K$

The volume of container A is $V_A = 2 m^3$

Initially the number of moles is $n = 1.2 \ moles$

The volume of container B is $V_B = 3.5 \ m^3$

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

$W = P_A V_A ln[ \frac{V_B}{V_A} ]$

Now from the Ideal gas law

$P_A V_A = nRT$

So substituting for $P_A V_A$ in the equation above

$W = nRT ln [\frac{V_B}{V_A} ]$

Where R is the gas constant with a values of $R = 8.314 \ J/mol$

Substituting values we have that

$W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]$

$W = 1563.289 \ J$

8 0

2 years ago

A student claims "everything falls at the same acceleration rate on the Moon, where there is no air or friction," how would you

Alexeev081 [22]

Now let’s say you’re on the Moon. If you were to drop a hammer and a feather from the same height, which would hit the ground first?

Trick Question! On the moon both objects would hit the ground at the same time. On Earth, the hammer lands first.

So yeah, the student is right. Galileo gave us this theory long ago.

5 0

2 years ago

The length of the track itself is 500m. The vehicles (each of mass 15kg) are dragged

gogolik [260]

Answer:

W = 100000 J = 100 KJ

Explanation:

Here we will use the most basic and general formula of work, which is as follows:

$W = Fd$

where,

W = Work Done = ?

F = Force Required = 200 N

d = Length of Track = 500 m

Therefore,

$W = (200\ N)(500\ m)\\$

5 0

2 years ago