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3 years ago

C) Explain relative velocity with examples.

Physics

1 answer:

Mnenie [13.5K]3 years ago

5 0

Answer:

we encounter ocassion where one or more object move in the which is not stationary with respect to another example a boat is cross a river that is flowing at some rate of aeroplane encountring wind durning it motion

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What would be the current going through a 200 ohm resister that is connected across a 120 v power supply?

gavmur [86]

Here is your answer

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4 years ago

Two trains on separate tracks move toward each other. Train 1 has a speed of 145 km/h; train 2, a speed of 72.0 km/h. Train 2 bl

tekilochka [14]

Answer:

Therefore,

The frequency heard by the engineer on train 1

$f_{o}=603\ Hz$

Explanation:

Given:

Two trains on separate tracks move toward each other

For Train 1 Velocity of the observer,

$v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s$

For Train 2 Velocity of the Source,

$v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s$

Frequency of Source,

$f_{s}=500\ Hz$

To Find:

Frequency of Observer,

$f_{o}=?$ (frequency heard by the engineer on train 1)

Solution:

Here we can use the Doppler effect equation to calculate both the velocity of the source $v_{s}$ and observer $v_{o}$ , the original frequency of the sound waves $f_{s}$ and the observed frequency of the sound waves $f_{o}$ ,

The Equation is

$f_{o}=f_{s}(\dfrac{v+v_{o}}{v -v_{s}})$

Where,

v = velocity of sound in air = 343 m/s

Substituting the values we get

$f_{o}=500(\dfrac{343+40.28}{343 -25})=500\times 1.205=602.64\approx 603\ Hz$

Therefore,

The frequency heard by the engineer on train 1

$f_{o}=603\ Hz$

7 0

3 years ago

With your hand parallel to the floor and your palm upright, you raise a 3-kg book upward with an acceleration of 2 m/s2. what is

kirza4 [7]

35N i don´t know why

3 0

3 years ago

Read 2 more answers

a) Calculate the height (in m) of a cliff if it takes 2.21 s for a rock to hit the ground when it is thrown straight up from the

DedPeter [7]

Answer:

a)Height of the cliff = 6.19 m

b) t = 0.571 s

Explanation:

We need to first calculate the time it takes the rock to go up and come down to the cliff's level first.

Using the equations of motion,

g = -9.8 m/s², u₁ = Initial velocity = 8.03 m/s,

v = velocity at maximum height = 0 m/s

t₁ = time to reach maximum height = ?

a) v₁ = u₁ + gt₁

0 = 8.03 - 9.8t

t₁ = 8.03/9.8 = 0.819 s

Total time of flight for the rock to leave the cliff and reach maximum height = 0.819 s

Height that the rock attains, measured from the top of the cliff to the maximum height, y₁ = ?

y₁ = ut₁ + gt₁²/2 = 8.03(0.819) - 9.8(0.819²)/2 = 3.29m

Total time of flight of the rock, t = t₁ + t₂

Subtracting the time obtained from the first part (0.819 s) from the total time would give the time it took the rock to travel from maximum height reached to the ground.

t₂ = 2.21 - 0.819 = 1.391 s

Let y = total height fallen through = y₁ + (height of the cliff) = y₁ + H = ?

g = 9.8 m/s²

t₂ = 1.391 s

u₂ = initial velocity for 2nd phase of rock motion = velocity at maximum height = 0 m/s

y = u₂t₂ + gt₂²/2

y = 0 + 9.8(1.391²)/2

y = 9.48 m

y = y₁ + H

9.48 = 3.29 + H

H = 9.48 - 3.29 = 6.19 m

b) y = H = 6.19 m, u = 8.03 m/s, t = ?,

y = ut + gt²/2

6.19 = 8.03t + 9.8t²/2

4.9t² + 8.03t - 6.19 = 0

Solving the quadratic eqn

t = -2.21 s or 0.571 s

Since time can only be positive,

t = 0.571 s

(PROVED)

4 0

3 years ago

Read 2 more answers

You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a dis

uranmaximum [27]

Answer:

The magnitude of force exerted on the handle is 108.73 N

Explanation:

To determine the magnitude of force exerted, we will use the formula relating Power and Force.

Power is the rate at which work is done. Power can be calculated from the formula

Power = Work / Time

But, Work = Force × Distance

Hence,

Power is given by the formula

$P = \frac{F \times s}{t}$

Where P is the Power

F is the force

s is the distance

and t is the time

From $P = \frac{F \times s}{t}$ ,

Then we can write that

$F = \frac{P \times t}{s}$

From the question,

Distance, s = 1.1 m

Time, t = 1.3 s

Power, P = 92 W

Putting these values into the formula, we get

$F = \frac{92 \times 1.3}{1.1}$

$F = \frac{119.6}{1.1}$

$F = 108.73N$

Hence, the magnitude of force exerted on the handle is 108.73 N.

7 0

3 years ago

C) Explain relative velocity with examples.​

C) Explain relative velocity with examples.