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Jet001 [13]
3 years ago
6

3. Two cars, both with a mass of 500 kg, are traveling down a road. The first car has a velocity of 65 m/s east and the second c

ar has a velocity of 85 m/s west. (Chapter 2 – Pages 54-55) a. Calculate the momentum of both cars showing the appropriate equation from your textbook and your work with units in order to receive credit. b. Which car has the larger momentum? Explain how you know.
Physics
1 answer:
-BARSIC- [3]3 years ago
6 0

Answer:

32500 kg m/s east, 42500 kg m/s west. Second car has larger momentum

Explanation:

The momentum of an object is given by

p = mv

where

m is the mass

v is the velocity

For the first car, m = 500 kg and v = 65 m/s east, so the momentum is

p_1 = (500)(65)=32500 kg m/s east

For the second car, m = 500 kg and v = 85 m/s west, so the momentum is

p_2 = (500)(85)=42500 kg m/s west

By comparing the two momentum, we see that the second car has larger momentum.

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A car of m = 1200. kg collides with a tree while traveling 60.0 mph. The collision occurs over a time period of 0.0500 seconds.
MrRissso [65]
You may know linear momentum is given by P= mass.velocity. Initially car is moving with some velocity so you know initial momentum of the car. Finally it comes to rest i.e final momentum of the car is 0. According to Newton's second law : Force = change in momentum /time. Applying this you'll get answer as 642840N. Hope it helped you. Revert back to me if you have any questions. Please check out the calculation it might be wrong!
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An object is dropped from rest from a 70.6 m tower. Air resistance is negligible. After 0.32 seconds, what is magnitude and dire
dem82 [27]

Answer:

<em>1,378.9ms²</em>

Explanation:

Given the following

Distance S = 70.6m

Time t = 0.32secs

Initial velocity = 0m/s

Required

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Using the equation of motion

S = ut+1/2at²

Substitute

70.6 = 0+1/2a(0.32)²

70.6 = 0.0512a

a = 70.6/0.0512

a = 1,378.9

<em>Hence the acceleration is 1,378.9ms²</em>

7 0
3 years ago
A point charge of 9.00 × 10−9 C is located at the origin of a coordinate system. A positive charge of 3.00 × 10−9 C is brought i
dlinn [17]

A point charge is located at the origin of a coordinate system. A positive charge is brought in from infinity to a point. The charges are at distance for given electrical potential energy is 3.34 x  10⁷ m.

<h3>What is electric potential energy?</h3>

The electric potential energy is the work done by a test charge to bring it from infinity to a particular location.

The electric potential energy is given by the relation,

V = kQ/r

where k = 9 x 10⁹ J.m/C ,Q = 3 x 10⁻⁹ C, V =8.09 × 10⁻⁷ J.

Substitute the values into the expression to get the distance between the charges.

8.09 × 10⁻⁷ =  9 x 10⁹ x  3 x 10⁻⁹ / r

r =3.34 x  10⁷ m

Thus, the distance between the charges will be 3.34 x  10⁷ m.

Learn more about  electric potential energy.

brainly.com/question/12645463

#SPJ1

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Which the answer It's science
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The answer is A. ive done a 5-k race, so its for sure 3 miles. 
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