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VARVARA [1.3K]
3 years ago
13

Describe the effect of the mass on the net force.

Physics
2 answers:
MariettaO [177]3 years ago
3 0

Answer:

F = m a

it means as m increase force increase also and acceleration is constant

the best example of this case is free fall regardless of the mass of something the free fall acceleration is constant - 9.81 m/s^2, because as mass increase gravitational force increase also

podryga [215]3 years ago
3 0

Answer:

hlw I'm jess bregoli

your answer is here (◕ᴗ◕✿)

mass is the amount of matter..Gravity affects weight, it does not affect mass.

Masses always remain the same.

Explanation:

hope it may help you

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5 0
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A fully loaded cart with a mass of 2200 kg starts from the top of a 12-meter hill on a roller coaster.
Salsk061 [2.6K]

Answer:

A. potential energy is 258720 Joule

Explanation:

A.Gravitational potential energy is: PE = m × g × h

velocity =  15.33 m/s when the car reaches the bottom of the hill.

where, m = mass

            g = acceleration due to gravity

            h = height from the bottom of hill.

The potential energy is : m×g×h

                                     =(2200×9.8×12)

                                     =258720 Joule

B. at the bottom of the hill, the potential energy is converted into kinetic energy so PE at top = KE at bottom

                    kinetic energy= \frac{1}{2}(m*v^{2})

where v = velocity

          m= mass

therefore,               v=\sqrt\frac{2*K.E}{m} {}

                         or,  v=\sqrt{\frac{2*258720}{2200} }

                         or,   v=15.33 m/s

7 0
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5.1C A fluid flows steadily through a pipe with a uniform crosssectional area. The density of the fluid decreases to half its in
prisoha [69]

The options are;

a. V2 equals 2V1.

b. V2 equals (V1)/2.

c. V2 equals V1.

d. V2 equals (V1)/4.

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Answer:

Option A: V2 equals 2V1

Explanation:

Since the flow is steady, then we can say;

mass flow rate at input = mass flow rate at output.

Formula for mass flow rate is;

m' = ρVA

Thus;

At input;

m'1 = ρ1•V1•A1

At output;

m'2 = ρ2•V2•A2

So, m'1 = m'2

Now, we are told that the density of the fluid decreases to half its initial value.

Thus; ρ2 = (ρ1)/2

Since m'1 = m'2, then;

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Now, the pipe is uniform and thus the cross section doesn't change. Thus;

A1 = A2

We now have;

ρ1•V1•A1 = (ρ1)/2•V2•A1

A1 and ρ1 will cancel out to give;

V1 = (V2)/2

Thus, V2 = 2V1

5 0
3 years ago
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