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stellarik [79]
3 years ago
15

On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo

und lane moving with a speed of 11 m/s, and Car B is in the westbound lane moving with a speed of 31 m/s.
(a) What is the magnitude and direction of the velocity of car A as seen by the driver of car B? magnitude m/s direction
(b) What is the magnitude and direction of the velocity of car B as seen by the driver of car A? magnitude m/s direction
Physics
1 answer:
ludmilkaskok [199]3 years ago
5 0

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i

42 m/s, negative direction (to the west).

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Xplain the difference between aerobic and anaerobic exercise and give 1 example of each?
Lilit [14]

Answer:

The term "aerobic" refers to the body's ability to provide energy through the use of oxygen.  The term anaerobic refers to the body's ability to provide energy without the use of oxygen.

An example of an aerobic exercise is swimming. An example of an anaerobic exercise is sprinting.

Explanation:

6 0
3 years ago
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A coil with an inductance of 2.3 H and a resistance of 14 Ω is suddenly connected to an ideal battery with ε = 100 V. At 0.13 s
klemol [59]

Given Information:

Resistance = R = 14 Ω

Inductance = L = 2.3 H

voltage = V = 100 V

time = t = 0.13 s

Required Information:

(a) energy is being stored in the magnetic field

(b) thermal energy is appearing in the resistance

(c) energy is being delivered by the battery?

Answer:

(a) energy is being stored in the magnetic field ≈ 219 watts

(b) thermal energy is appearing in the resistance ≈ 267 watts

(c) energy is being delivered by the battery ≈ 481 watts

Explanation:

The energy stored in the inductor is given by

U = \frac{1}{2} Li^{2}

The rate at which the energy is being stored in the inductor is given by

\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1

The current through the RL circuit is given by

i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })

Where τ is the the time constant and is given by

\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16

i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}

Therefore, eq. 1 becomes

\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})

At t = 0.13 seconds

\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts

(b) thermal energy is appearing in the resistance

The thermal energy is given by

P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts

(c) energy is being delivered by the battery?

The energy delivered by battery is

P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts

4 0
3 years ago
Underground, carbon can be stored in ____, which humans ___.
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Underground, carbon can be stored in ____, which humans ___.

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coal, oil, fossil fuels ... humans burn for energy

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2 years ago
Which factors could be potential sources of error in the experiment? Check all that apply.
mote1985 [20]

Answer:

1. energy lost in the lever due to friction

3. visual estimation of height of the beanbag

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Explanation:

Edge 2021

8 0
3 years ago
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A ball rolls with a speed of 2.0m/s across a level table that is 1.0m above the floor. Upon reaching the edge of the table it fo
Mekhanik [1.2K]

Answer:0.58 m

Explanation:

The initial velocity of the ball is u = 2.0 m/s

The height of the table is, h = 1.0 m

The ball falls in vertical direction under acceleration due to gravity.

Time taken for ball to hit the floor:

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Solving this for t,

t = 0.29 s ( we have neglected the negative value of t)

In the same time, the ball would cover a horizontal distance of :

s = u t

⇒s = 2.0 m/s×0.29 s = 0.58 m

Thus, the landing spot is 0.58 m away from the table.

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