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2 years ago

The amount of WORK done is determined by 2 factors.

Physics

1 answer:

RUDIKE [14]2 years ago

6 0

Explanation :

Work is done when a force is applied to create a displacement on an object.

Thus, the work done depends on the two factors i.e.

(1) Applied force (F)

(2) Distance or displacement (d)

Mathematically, work done is $W= F.s$

It also depends on the angle between the force and the displacement.

$W=Fs\ cos\ \theta$

For example,

A person carries a weight of 20 kg and lifts it on his head 1.5 m above the surface. So, the work done by him on the luggage will be:

$W= F\times s$

$W=m\times g \times s$

$W=20\ kg\times 9.8\ m/s^2\times 1.5\ m$

So, $W=294\ Joules$

Hence, the work done by him on the luggage is 294 Joules.

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A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1 2 and t = 1 s? Use G

Romashka-Z-Leto [24]

There is one mistake in the question.The Correct question is here

A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.

Answer:

y(1s) - y(1/2s) = - 3.675 m

The cat falls 3.675 m between time 1/2 s and 1 s.

Explanation:

Given data

time=1/2 sec to 1 sec

v(t)=-9.8t m/s

To find

Distance

Solution

As the acceleration as first derivative of velocity with respect to time

acceleration(-g)= dv/dt

Solve it

dv = a dt

dv = -g dt

v - v₀ = -gt

v= dy/dt

dy = v dt

dy = ( v₀ - gt ) dt

y(1s) - y(1/2s) = ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]

y(1s) - y(1/2s) = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]

y1s - y1/2s = ( - 4.9 m/s² ) ( 3/4 s² )

y(1s) - y(1/2s) = - 3.675 m

The cat falls 3.675 m between time 1/2 s and 1 s.

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2 years ago

A jet airliner moving initially at 406 mph (with respect to the ground) to the east moves into a region where the wind is blowin

astraxan [27]

Answer:

966 mph

Explanation:

Using as convention:

- East --> positive x-direction

- North --> Positive y-direction

The x- and y- components of the initial velocity of the jet can be written as

$v_{1x} = 406 mph\\v_{1y} = 0$

While the components of the velocity of the wind are

$v_{2x} = (568)(cos 15^{\circ})=548.6 mph\\v_{2y} = (568)(sin 15^{\circ})=147.0 mph$

So the components of the resultant velocity of the jet are

$v_x = v_{1x}+v_{2x}=406+548.6=954.6 mph\\v_y = v_{1y}+v_{2y}=0+147.0=147.0 mph$

And the new speed is the magnitude of the resultant velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(954.6)^2+(147.0)^2}=965.8 mph \sim 966 mph$

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2 years ago

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faltersainse [42]

The correct answer is magnetic field, electric field, and charges.

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morpeh [17]

Your answer will be Radio Waves .

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Answer:

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