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3 years ago

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How do you find total displacement

1 answer:

Monica [59]3 years ago

5 0

That's actually a lot easier than finding total <u><em>distance.</em></u>

To find displacement, you only have to know where the trip started from, and where it ended. It doesn't matter what route was followed to actually travel from the start to the finish. The displacement is the straight-line distance and direction between those two points.

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Spymaster Paul, flying a constant 215km/h horizontally in a low-flying helicopter, want to drop secret documents into his contac

aleksandrvk [35]

Answer:

$\theta = 49.56 degree$

Explanation:

Relative horizontal velocity of plane with respect to the car is given as

$v_r = v_p - v_c$

so we have

$v_p = 215 km/h$

$v_c = 155 km/h$

now we have

$v_r = 215 - 155$

$v_r = 60 km/h$

$v_r = 16.67 m/s$

Now time taken by the object to drop the vertical height is given as

$y = \frac{1}{2}gt^2$

$78 = \frac{1}{2}(9.81)t^2$

$t = 3.98 s$

so the distance of the car must be

$d = v_r\times t$

$d = 16.67 \times 3.98$

$d = 66.47 m$

Angle of the car with horizontal is given as

$tan\theta = \frac{y}{x}$

$tan\theta = \frac{78}{66.47}$

$\theta = tan^{-1}(1.17)$

$\theta = 49.56 degree$

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3 years ago

At t = 0, object A is dropped from the roof of a building. At the same instant, object B is dropped from a window 10 m below the

dsp73

Answer:

Explanation:

Given two objects are dropped simultaneously

Object A is 10 m higher than object B therefore

Distance covered by object A is given by

$y_a(t)$ is given by

$y=ut+\frac{1}{2}at^2$

where y=displacement

u=initial velocity

a=acceleration

t=time

$y_a(t)=0+0.5gt^2--1$

for object B

$y_b(t)=0+0.5gt^2--2$

Subtract 1 and 2 we get

$y_a(t)-y_b(t)=0$

i.e. they will travel equal distance in equal time and distance between them remain 10 m until object B hits the ground

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3 years ago

A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d

oksano4ka [1.4K]

Answer:

x = 45 cm

Explanation:

Given that,

The length of a rod, L = 50 cm

Mass, m₁ = 0.2 kg

It is at 40cm from the left end of the rod.

We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.

The centre of mass of the rod is at 25 cm.

Taking moments of both masses such that,

$15\times 0.2=x\times 0.6\\\\x=\drac{3}{0.6}\\\\x=5\ cm$

The distance from the left end is 40+5 = 45 cm.

Hence, at a distance of 45 cm from the left end it will balance the rod.

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3 years ago