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Lilit [14]
3 years ago
13

When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not n

oticeable?​
Physics
2 answers:
Anarel [89]3 years ago
6 0

Answer:

True, because unlike the apple we don't have a large as$ refrence point (the earth is too big to notice being pushed)

Explanation:

Katena32 [7]3 years ago
4 0

Yes, this is true.

-- While the apple is falling, the same gravitational force acts on both the apple and the Earth.

-- The mass of the apple is somewhere in the neighborhood of 1/4 kg.

-- The mass of the Earth is about 5.972 x 10²⁴ kg.

-- Since the Earth has roughly 2.389 x 10²⁶ times as much mass as the apple has, the apple has roughly 2.389 x 10²⁶ greater acceleration than the Earth has, and moves roughly 2.389 x 10²⁶ times as far down as the Earth moves up, before they smack together.

-- That's why you don't notice the Earth's motion.

-- Also, you're standing on the Earth, moving up with it, toward the apple.  Maybe it would be different if you were sitting on the apple, riding it down to the ground, and you were able to notice the motion of the ground coming up to meet you at a speed that's 0.00000000000000000000000000419 of YOUR speed.  

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A balloon contains 2.3 mol of helium at 1.0 atm , initially at 240 ∘C. What's the initial volume? What's the volume after the ga
pashok25 [27]
A) initial volume
We can calculate the initial volume of the gas by using the ideal gas law:
p_i V_i = nRT_i
where
p_i=1.0 atm=1.01 \cdot 10^5 Pa is the initial pressure of the gas
V_i is the initial volume of the gas
n=2.3 mol is the number of moles
R=8.31 J/K mol is the gas constant
T_i=240^{\circ}C=513 K is the initial temperature of the gas

By re-arranging this equation, we can find V_i:
V_i =  \frac{nRT_i}{p_i} = \frac{(2.3 mol)(8.31 J/mol K)(513 K)}{1.01 \cdot 10^5 Pa}=0.097 m^3

2) Now the gas cools down to a temperature of
T_f = 14^{\circ}C=287 K
while the pressure is kept constant: p_f = p_i = 1.01 \cdot 10^5 Pa, so we can use again the ideal gas law to find the new volume of the gas
V_f =  \frac{nRT_f}{p_f}= \frac{(2.3 mol)(8.31 J/molK)(287 K)}{1.01 \cdot 10^5 Pa} = 0.054 m^3

3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:
W=p \Delta V= p(V_f -V_i)
by using the data we found at point 1) and 2), we find
W=p(V_f -V_i)=(1.01 \cdot 10^5 Pa)(0.054 m^3-0.097 m^3)=-4343 J
where the negative sign means the work is done by the surrounding on the gas.
5 0
3 years ago
What happens to the force between two charges if one of the charges are doubled?
Julli [10]

Answer:

The new force between the charges becomes double of the initial force.

Explanation:

The force acting between charge particles is given by :

F=k\dfrac{q_1q_2}{r^2}

k is electrostatic constant

r is distance between charges

If one of the charges are doubled, then, q₁ = 2q₁

The new force becomes,

F'=\dfrac{2kq_1q_2}{r^2}\\\\F'=2F

So, the new force between the charges becomes double of the initial force.

7 0
3 years ago
When a wave has λ=3 m and f=15 Hz, what is the speed of the wave?
Tamiku [17]

Wave speed = (wavelength) x (frequency)

Wave speed = (3 m) x (15 Hz)

<em>Wave speed = 45 m/s</em>

5 0
2 years ago
What is the effect on the force of gravity between two objects if the mass of one object remains unchanged while the distance to
Vadim26 [7]

Answer:

The correct answer to the question is

B. It always decreases

Explanation:

To solve the question, we note that the foce of gravity is given by

F_G=\frac{Gm_1m_2}{r^2} where

G= Gravitational constant

m₁ = mass of first object

m₂ = mass of second object

r = the distance between both objects

If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have

F_{G2} =\frac{Gm_1(2m_2)}{(2r)^2} = \frac{2}{4} \frac{Gm_1m_2}{r^2}

Therefore the gravitational force is halved. That is it will always decrease

4 0
3 years ago
If I were a geologist working in South America along one of those plate boundaries, what might I see if I were looking at sedime
faust18 [17]

Answer:

you'll see rocks that have been there for years

4 0
2 years ago
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