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Illusion [34]
3 years ago
7

Car 1 drives 35 mph to the east, and car 2 drives 50 mph to the west. From the frame of reference of car 1, what is the velocity

of car 2?
A. 85 mph west
B. 15 mph east
C. 85 mph east
D. 15 mph west
Physics
2 answers:
Sergio039 [100]3 years ago
4 0
The distance between the two cars is increasing at the rate of 85 mph.

A passenger in Car-1 says that he is at rest in his own frame of reference,
and Car-2 is moving away from him at 85 mph, toward the west.
Radda [10]3 years ago
3 0

Answer:

A. 85 mph west

Explanation:

Velocity of car 1 is given as

v_1 = 35 mph towards East

\vec v_1 = 35(-\hat i) mph

Velocity of car 2 is given as

v_2 = 50 mph towards West

\vec v_2 = 50 (\hat i) mph

Now we need to find the speed of car 2 with respect to the frame of car 1

so it is given as

\vec v_{21} = \vec v_2 - \vec v_1

now we will have

\vec v_{21} = 50 (\hat i) - 35 (-\hat i) = 85 \hat i

so the relative speed will be 85 mph towards West

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A car is traveling south is 200 kilometers from it’s starting point after 2 hours. What is the average velocity of the car
Lesechka [4]

Answer:

100

Explanation:

take note that v=d/t (velocity is distance over(divided by) time, so in this case it would be 200 (distance) divided by 2 (time) = 100

6 0
3 years ago
Two coils of wire are placed close together. Initially, a current of 3.04 A exists in one of the coils, but there is no current
Alchen [17]

Answer:

M=0.0247H

Explanation:

Given data

V_{voltage}=4.29V\\I_{current}=3.04A\\t_{time}=1.75*10^{-2}s

To find

Mutual inductance of the two-coil system

Solution

The mutual inductance given as:

M= (-VΔt)/ΔI

Substitute the given values

So

M=-\frac{4.29V*1.75*10^{-2}s}{(0-3.04A)}\\ M=0.0247H

4 0
3 years ago
A less than youthful 82.6 kg physics professor decides to run the 26.2 mile (42.195 km) Los Angeles Marathon. During his months
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Mem me e m even have. Jags. Shah. Shiv side esicjm is n meh dish so do indbbd
4 0
2 years ago
g A loop circuit has a resistance of R1 and a current of 2 A. The current is reduced to 1.4 A when an additional 2.2 Ω resistor
Mrrafil [7]

Answer:

R1 = 5.13 Ω

Explanation:

From Ohm's law,

V = IR............... Equation 1

Where V = Voltage, I = current, R = resistance.

From the question,

I = 2 A, R = R1

Substitute into equation 1

V = 2R1................ Equation 2

When a resistance of 2.2Ω is added in series with R1,

assuming the voltage source remain constant

R = 2.2+R1,  and I = 1.4 A

V = 1.4(2.2+R1)................. Equation 3

Substitute the value of V into equation 3

2R1 = 1.4(2.2+R1)

2R1 = 3.08+1.4R1

2R1-1.4R1 = 3.08

0.6R1 = 3.08

R1 = 3.08/0.6

R1 = 5.13 Ω

6 0
2 years ago
Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10
11111nata11111 [884]

Answer:

1.62\times 10^{-8}\ \text{s}

Explanation:

\epsilon_0 = Vacuum permittivity = 8.854\times 10^{-12}\ \text{F/m}

A = Area = 10\times 2\times 10^{-4}\ \text{m}^2

d = Distance between plates = 1 mm

V_c = Changed voltage = 60 V

V = Initial voltage = 100 V

R = Resistance = 1000\ \Omega

Capacitance is given by

C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}

We have the relation

V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}

The time taken for the potential difference to reach the required level is 1.62\times 10^{-8}\ \text{s}.

5 0
2 years ago
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