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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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3 years ago
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Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y
WITCHER [35]

Question:

A point charge of -2.14uC  is located in the center of a spherical cavity of radius 6.55cm  inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm  from the center of the cavity.  

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 3.65\times 10^5N/C

Explanation:

A point charge ,q = -2.14\times 10^{-6} C is located in the center of a spherical cavity of radius , r =6.55\times 10^{-2}  m inside an insulating spherical charged solid.  

The charge density in the solid , d = 7.35 \times 10^{-4}C/m^3.

Distance from the center of the cavity,R =9.5\times 10^{-2 }m

Volume of shell of charge= V  =(\frac{4\pi}{3})[ R^3 - r^3 ]

Charge on the shell ,Q = V \times d'

Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d

Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}

Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}

Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}

Q =1.7745 \times 10^{-6 }C

Electric field at 9.5\times 10^{-2}m due to shellE1  = \frac{k Q}{R^2}

E1 =  \frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}

E1 =1.769\times 10^6 N/C

Electric field at  9.5\times 10^{-2} due to 'q' at center E2 = \frac{kq}{R^2}

E2 =\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}

E2 =2.134\times 10^6 N/C

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

=[  2.134  - 1.769 ]\times 10^6

= 3.65\times 10^5 N/C

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Answer:

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