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Svet_ta [14]
2 years ago
9

Provide three examples of situations in which mass is the main factor determining an object's momentum

Physics
1 answer:
Ratling [72]2 years ago
3 0

The three examples of situations in which mass is the main factor determining an object's momentum are, a moving truck, a moving trailer and a rolling block.

<h3>What is momentum?</h3>
  • Momentum is the product of mass and velocity of an object

P = mv

where;

  • m is the mass of the object
  • v is the velocity of the object

The three examples of situations in which mass is the main factor determining an object's momentum is below;

  • A moving truck.
  • A moving trailer.
  • A rolling block.

The three examples given above shows three objects with heavy mass.

Learn more about momentum here: brainly.com/question/7538238

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Answer:

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Explanation:

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3 years ago
A 5.93 kg ball is attached to the top of a vertical pole with a 2.35 m length of massless string. The ball is struck, causing it
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Answer

given,

mass of ball = 5.93 kg

length of the string = 2.35 m

revolve with velocity of 4.75 m/s

acceleration due to gravity = 9.81 m/s²

T cos θ = mg

T cos θ = 5.93\times 9.81

T cos θ = 58.17

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T sin^2 \theta =56.93

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T (1 - cos^2 \theta) =56.93

T (1 - (\dfrac{58.17}{T})^2) =56.93

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T =  93.22 N

cos \theta = \dfrac{58.17}{93.22}

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3 years ago
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 168 cm , but its circumference is decreasi
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Answer:

103.1 V

Explanation:

We are given that

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\frac{dC}{dt}=-15cm/s

Magnetic field,B=0.9 T

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Magnetic flux=\phi=BA=B(\pi r^2)

Circumference,C=2\pi r

r=\frac{C}{2\pi}

r=\frac{168}{2\pi} cm

\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}(-15)=-\frac{15}{2\pi} cm/s

\int dr=-\int \frac{15}{2\pi}dt

r=-\frac{15}{2\pi}t+C

When t=0

r=\frac{168}{2\pi}

\frac{168}{2\pi}=C

r=-\frac{15}{2\pi}t+\frac{168}{2\pi}

E=-\frac{d\phi}{dt}=-\frac{d(B\pi r^2)}{dt}=-2\pi rB\frac{dr}{dt}

E=-2\pi(-\frac{5}{2\pi}t+\frac{168}{2\pi})B\times -\frac{15}{2\pi}

t=8 s

B=0.9

E=2\pi\times \frac{15}{2\pi}\times 0.9(-\frac{15}{2\pi}(8)+\frac{168}{2\pi})

E=103.1 V

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