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9966 [12]
4 years ago
15

The relationship between mass and inertia is described by newton's second law of motion. true or false

Physics
1 answer:
Mariana [72]4 years ago
4 0
False. Inertia and mass is not described in Newton’s second law of motion but in Newton’s first law of motion. Newton’s first law of motion or sometimes referred to as the law of inertia. In Newton’s first law indicates that an object at rest will remain at rest unless acted by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
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A ball was rolling downhill at 2 m/s. After 5s, it was rolling at 90 m/s. What is its acceleration?
olga55 [171]

Answer:

17.6 m/s²

Explanation:

Given:

v_{f} = 90 m/s (final velocity)

v_{i} = 2 m/s (initial velocity)

Δt = 5s (change in time)

The formula for acceleration is:

a_{avg} = Δv / Δt

We can find Δv by doing

Δv = v_{f} - v_{i}

Replace the values

Δv = 90m/s - 2m/s

Δv= 88m/s

Using the equation from earlier, we can find the acceleration by dividing the average velocity by time.

a_{avg} = Δv / Δt

a_{avg} = \frac{88m/s}{5/s}

acceleration = 17.6 m/s^{2}

4 0
3 years ago
A car with a mass of 1,500 kg requires a centripetal force of 640 N to safely follow a circular curve in the road at 13.4 m/s. W
Tamiku [17]
<h2>Answer</h2>

Radius is 421 m.

<u>Explanation </u>

A car with a mass of 1,500 kg requires a centripetal force of 640 N to safely follow a circular curve in the road at 13.4 m/s. Therefore, for radius we use formula which is Fc = mv ^ 2 / r,

As mass = m = 1500kg,

Centripetal force = Fc = 640N,

Velocity = v = 13.4 m / s

By putting values, Fc = mv ^ 2 / r,

r = mv ^ 2 / Fc,

=> r = ( 1500kg ) . ( 13.4 m / s) ^ 2 / 640,

=> r = ( 1500kg ) . ( 179.56 ) / 640,

r = 269340 / 640,

=> r = 420.84 m.

Radius is 421 m.

8 0
3 years ago
Read 2 more answers
Which two of the following actions would
Ganezh [65]
Answer is d using a heavier string
5 0
3 years ago
10) The closest star to the earth (other than
Reptile [31]

Answer:

It takes a little, be patient.

8 0
4 years ago
A fillet weld has a cross-sectional area of 25.0 mm2and is 300 mm long. (a) What quantity of heat (in joules) is required to acc
HACTEHA [7]

Answer:

77362.56 J

163730.28571 J

Explanation:

A = Area = 25 mm²

l = Length = 300 mm

K = Constant = 3.33\times 10^{-6}

\eta = Heat transfer factor = 0.75

f_m = Melting factor = 0.63

T = Melting point of low carbon steel = 1760 K

Volume of the fillet would be

V=Al\\\Rightarrow V=25\times 300\\\Rightarrow V=7500\ mm^3=7500\times 10^{-9}\ m^3

The unit energy for melting is given by

U_m=KT^2\\\Rightarrow U_m=3.33\times 10^{-6}\times 1760^2\\\Rightarrow U_m=10.315008\ J/mm^3

Heat would be

Q=U_mV\\\Rightarrow Q=10.315008\times 7500\\\Rightarrow Q=77362.56\ J

Heat required to weld is 77362.56 J

Amount of heat generation is given by

Q_g=\dfrac{Q}{\eta f_m}\\\Rightarrow Q_g=\dfrac{77362.56}{0.75\times 0.63}\\\Rightarrow Q_g=163730.28571\ J

The heat generated at the welding source is 163730.28571 J

7 0
4 years ago
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