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Nonamiya [84]
3 years ago
11

Alex, parked by the side of an east-west road, is watching car P, which is moving in a westerly direction. Barbara, driving east

at a speed 52 km/h, watches the same car. Take the easterly direction as positive. If Alex measures a speed of 78 km/h for car P, what velocity will Barbara measure?
Physics
1 answer:
Alexxandr [17]3 years ago
6 0

Answer:

v_{PB} = 130\ km/h

Explanation:

Since, Alex is at rest. Therefore, the speed measured by him will be the absolute speed of car P. Therefore, taking easterly direction as positive:

Absolute\ Velocity\ of\ Car\ P = v_{P} = -78\ km/h

And the absolute velocity of Barbara's Car is given as:Absolute\ Velocity\ of\ Barbara's\ Car = v_{B} = 52\ km/h

Now, for the velocity of Car p with respect to the velocity of Barbara's Car can be given s follows:

Velocity\ of\ Car\ P\ measured\ by\ Barbara = v_{PB} = v_{B}-v_{P}\\\\v_{PB} = 52\ km/h-(-78\ km/h)

v_{PB} = 130\ km/h

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Answer:

h = 81.63 m

Explanation:

Given that,

The speed of the car, v = 40 m/s

We need to find the height when the car comes to rest. We can use the conservation of energy to find it i.e.

mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}\\\\h=\dfrac{(40)^2}{2\times 9.8}\\\\h=81.63\ m

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3 years ago
Suppose high tide is at midnight, the water level at midnight is 3 m, and the water level at low tide is 0.5 m. Assuming the nex
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We have that the time, to the nearest minute, when the water level is at 1.125 m for the second time after midnight is

t=10.0hours

From the Question we are told that

Maximum height h_{max}=3m

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Time for  next high tide will occurT=12 hours =>720 min

Generally Average Height

h_{avg}=\frac{3+0.5}{2}\\\\h_{avg}=1.75

Therefore determine Amplitude to be

A=h_{max}=j_{avg}\\\\A=3-1.75\\\\A=1.25

Generally, the equation for Time is mathematically given by

At t=0

h(x)=Acos(Bx)+h_{avg}

Where

B=\frac{2\pi}{P}\\\\B=\frac{2\pi}{720}\\\\B=8.73*10^{-3}

Therefore

h(t)=Acos8.73*10^{-3}(t)+h_{avg}

Hence the Time at T=1.125 is

1.125(t)=1.25cos(8.73*10^{-3})(t)+1.75

-0.1249t=1.75

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2 years ago
A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

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The displacement is

d = 4.0 m

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b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

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