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Damm [24]
3 years ago
12

Make the following conversion. 0.00432 km = _____ mm

Physics
2 answers:
lina2011 [118]3 years ago
8 0

Answer:

0.00432 km = 4320 mm

Explanation:

e-lub [12.9K]3 years ago
5 0
0.00432 km = 4320 mm
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photoshop1234 [79]

Answer:

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Explanation:

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2 years ago
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If a car travels 30 mi. north for 30 min., 60 mi. east for 1.0 hour, and 30 mi. south for 30 min., what is the average speed
Ainat [17]

Answer: 60mph

Explanation:

Given the following :

First leg travel:

Distance = 30 miles

Time of travel= 30 minutes = 0.5 hour

Second leg travel:

Distance = 60 miles

Time of travel = one hour

Average speed :

Speed = total Distance / time of travel

Total distance in miles = (30 + 60) miles = 90 miles

Total time of travel = 1 hour + 0.5 hour = 1.5 hours

Average speed = total distance traveled / total travel time

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Average speed = 60 miles / hour

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3 0
4 years ago
What is the impulse of a 3 kg object that starts from rest and moves to 20 m/s?
IrinaK [193]

Answer:

The impulse on the object is 60Ns.

Explanation:

Impulse is defined as the product of the force applied on an object and the time at which it acts. It is also the change in the momentum of a body.

F = m a

F = m(\frac{v_{2}  - v_{1} }{t})

⇒     Ft = m(v_{2} - v_{1})

where: F is the dorce on the object, t is the time at which it acts, m is the mass of the object, v_{1} is its initialvelocity and v_{2} is the final velocity of the object.

Therefore,

impulse = Ft = m(v_{2} - v_{1})

From the question, m = 3kg, v_{1} = 0m/s and v_{2} = 20m/s.

So that,

Impulse = 3 (20 - 0)

             = 3(20)

             = 60Ns

The impulse on the object is 60Ns.

8 0
3 years ago
A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck
Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

Substituting M_{truck} as 25000 Kg, V_{common} as 22.3 m/s, M_{standing} as 2000 Kg we obtain

V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

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3 years ago
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Answer:

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3 years ago
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