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OverLord2011 [107]
2 years ago
15

If Weight=mg and Force=ma, and the only force acting on an object is gravity, then the masses cancel each other out and you are

left with g=a. Therefore, that must mean everything on earth must fall at the same rate. Is this true Why or why not?
Physics
2 answers:
anygoal [31]2 years ago
8 0

The correct answer to question:

Explanation:

gregori [183]2 years ago
5 0

When g=a, that means everything on earth fall at the same rate.

<h3>Why does everything fall to the earth at the same rate?</h3>

As such, all objects free fall at the same rate regardless of their mass. Because the 9.8 N/kg gravitational field at Earth's surface causes a 9.8 m/s/s acceleration of any object placed there, we often call this ratio the acceleration of gravity.

<h3>Why is gravity equal to acceleration?</h3>

When objects fall to the ground, gravity causes them to accelerate. Acceleration is a change in velocity, and velocity, in turn, is a measure of the speed and direction of motion. Gravity causes an object to fall toward the ground at a faster and faster velocity the longer the object falls.

Know more about gravity here

brainly.com/question/4014727

#SPJ2

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Find the absolute value of the change of the gravitational potential energy (GPE) of the puck-Earth system from the moment the p
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3 years ago
A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta
weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq  0 \ \ \ -------> \ \ \ 1

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be  in the opposite direction

Formula for the change in specific entropy can be calculated as:

s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \  ------->  \ \ \ 2

where;

s_1, s_2 , s^0(T_2), s^0(T1) are specific entropies

R = universal gas constant

P_1 = pressure at location 1

P_2 = pressure at location 2

We obtain the specific properties of air at temperature at T_1 = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

s^0(T1) = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature T_2 = 22°C + 273) K = 295 K

From the table A- 22

s^0(T_2) = 1.68515 kJ/kg . K

R = \frac{8.314 kJ}{28.97 kg.K}

P_1 = 0.95 bar

P_2 = 0.8 bar

Now replacing our values  into equation (2) from above; we have;

s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )

s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97}  \ In (\frac{0.8}{0.95} )

s_2-s_1 = 1.68515 -1.8279+ 0.0493

s_2-s_1 =-0.0934 \  kJ/kg.K

Equating our result to equation (1)

s_2-s_1 \geq 0\\-0.0934 \leq 0

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

3 0
3 years ago
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Radda [10]

I believe the answer is a

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