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Olin [163]
4 years ago
14

A car has momentum, p=10880 kg m/s, east. If the car is traveling east at 16 m/s, its mass must be [m] kg.

Physics
2 answers:
GuDViN [60]4 years ago
6 0
<h3><u>Given</u> :</h3>

Momentum of car = 10880 kg m/s

Velocity of car = 16 m/s

<h3><u>To Find</u> :</h3>

We. have to find mass of the car.

<h3><u>Solution</u> :</h3>

★ Momentum of body is measured as the <u>product of mass and its velocity</u>.

It is a vector quantity having both <u>magnitude</u> as well as <u>direction</u>.

SI unit : kg m/s or N s

Formula : p = m × v

  • p denotes momentum
  • m denotes mass
  • v denotes velocity

★ <u>Calculation</u> :

⭆ p = m × v

⭆ 10880 = m × 16

⭆ m = 10880/16

⭆ m = 680 kg

∴ <u>Mass of the car = 680 kg</u>

<h3>Hope It Helps!</h3>

MaRussiya [10]4 years ago
5 0

Answer:

<h3>The answer is 67.5 kg</h3>

Explanation:

To find the mass of an object given it's momentum and velocity we use the formula

m =  \frac{p}{v}  \\

where

p is the momentum

v is the velocity

From the question

p = 1080 kgm/s

v = 16 m/s

We have

m =  \frac{1080}{16}  =  \frac{135}{2}  \\

We have the final answer as

<h3>67.5 kg</h3>

Hope this helps you

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Answer:

Option B is the correct answer.

Explanation:

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            \Delta L=L\alpha \Delta T

L = 1.2 meter

ΔT = 65 - 15 = 50°C

Thermal Expansion Coefficient for aluminum, α = 24 x 10⁻⁶/°C

We have change in length

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Option B is the correct answer.

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A cruise ship sails due south at 2.00 m/s while a coast guard patrol boat heads 19.0° north of east at 5.60 m/s. What are the x-
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The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

Explanation:

Given that,

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Velocity of boat = 5.60 m/s due north

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We need to calculate the component

The velocity of the ship in term x and y coordinate

v_{s_{x}}=0

v_{s_{y}}=2.0\ m/s

The velocity of the boat in term x and y coordinate

For x component,

v_{b_{x}}=v_{b}\cos\theta

Put the value into the formula

v_{b_{x}}=5.60\cos19

v_{b_{x}}=5.29\ m/s

For y component,

v_{b_{y}}=v_{b}\sin\theta

Put the value into the formula

v_{b_{y}}=5.60\sin19

v_{b_{y}}=1.82\ m/s

We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat

For x component,

v_{sb_{x}}=v_{s_{x}}-v_{b_{x}}

Put the value into the formula

v_{sb_{x}=0-5.29

v_{sb}_{x}=-5.29\ m/s

For y component,

v_{sb_{y}}=v_{s_{y}}-v_{b_{y}}

Put the value into the formula

v_{sb_{x}=2.-1.82

v_{sb}_{x}=0.18\ m/s

Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

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