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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_x = a*cos(theta)

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Light enters air from water. The angle of refraction will be A. less than the angle of incidence. B. greater than or equal to th
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Answer:

E. greater than the angle of incidence.

Explanation:

Snell's law states that:

n_i sin \theta_i = n_r sin \theta_r (1)

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\sin \theta_r = \frac{n_i}{n_r} sin \theta_i = 1.33 sin \theta_i

which means that

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Problem 9: Suppose you wanted to charge an initially uncharged 85 pF capacitor through a 75 MΩ resistor to 90.0% of its final vo
Bingel [31]

Answer:

t=14.678\times 10^{-3}s

Explanation:

Given:

Capacitance, C = 85 pF = 85 × 10⁻¹² F

Resistance, R = 75 MΩ = 75×10⁶Ω

Charge in capacitor at any time 't' is given as:

Q=Q_o(1-e^{-\frac{t}{RC}})

where,

Q₀ = Maximum charge = CE

E = Initial voltage

t = time

also, Q = CV

V= Final voltage = 90% of E = 0.9E

thus, we have

C\times 0.9E=CE(1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})

or

0.9=1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}

or

e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}=1-0.9=0.1

taking log both sides, we get

ln(e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})=ln(0.1)=ln(\frac{1}{10})

or

-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}=-ln(10)

or

t=75\times 10^6\ \times\ 85\times 10^{-12}\times ln{10}

or

t=14.678\times 10^{-3}s

3 0
3 years ago
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