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3 years ago

A 62kg box is lifted 12 meters off the ground. How much work is done?

Physics

1 answer:

Temka [501]3 years ago

4 0

Answer: 7291.2 joules

Explanation:

Work is done when force is applied on an object over a distance.

Thus, Workdone = Force X distance

Since Distance moved by box = 12 metres

mass of box = 62kg

Acceleration due to gravity when box was lifted is represented by g = 9.8m/s^2

Recall that Force = Mass x acceleration due to gravity

i.e Force = 62kg x 9.8m/s^2

= 607.6 Newton

So, Workdone = Force X Distance

Workdone = 607.6 Newton X 12 metres

Workdone = 7291.2 joules

Thus, 7291.2 joules of work was done.

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The mcb of rupa's room is tripped and keeps on tripping again and again . if it is a domestic circuit, what could be the reason

Tamiku [17]

The MCB of a rupas room is tripped and keeps on tripping again and again, and if it is a domestic circuit, what could be the reason for this phenomenon?

The reason could be a short circuit which is resulting in higher level of currents to pass through the MCB which is resulting in trip every time.

The MCB is faulty and might need a replacement.

To Diagnose the problem further more.

Turn off all the switches in rupas room and then try turning on the MCB. If it trips again then MCB is faulty (Subjective to the fact there everything was normal before this issue and no signs of short circuit or spark in wiring were observed)

If MCB does not trip in point 1 then Turn ON all the switches one by one. This shall give you the cause of problem.

4 0

2 years ago

A motorcyclist travels with an average speed of 6 m/s . If the cyclist is going to a friend's house 36 m away , how much time wi

vodka [1.7K]

Answer:

Explanation:

The time taken by the cyclist can be found by using the formula

$t = \frac{d}{v} \\$

v is the velocity

d is the distance

From the question we have

$t = \frac{36}{6} \\ = 6$

We have the final answer as

Hope this helps you

5 0

2 years ago

An observer in frame S sees lightning simultaneously strike two points 100 m apart. The first strike occurs at xx1 = yy1 = zz1 =

Veseljchak [2.6K]

Answer:

a) 0, = -0.33 us

b) 140m

c) No, The event are not simultaneous i.e they did not occur at the same time, the second even (-0.33 usec) occurs 0.33 usec earlier than the first event.

Explanation:

the lorentz factor expression is written as;

y = 1₀ / √(1 - (v²/c²))

where v is the relative speed of an observer and c is the speed of light

so we were given that relative speed to be o.7c

therefore

y = 1 / √(1 - ((0.7c)² / c²))

y = 1 / √(1 - (0.49c² / c²))

y = 1 / √(1 - 0.49)

y = 1 / 0.7141

y = 1.4

1 - the coordinates of the first event, the s' frame of reference is,

x1 ' = y(x1 - vt1) = 0

y1 ' = y1, z1' = z1 and

t1 ' = y [t1 - v/c²x1]

= 0

2 - the coordinates of the second event, the s ' frame of reference is'

x2 ' = y(x2-vt2)

= 1.4(100m - 0)

= 140m

y2 ' = y2, z2 ' = z2

t2 ' = y [ t2 - v/c²x2 ]

= 1.4 [ 0 - 0.7c/c²(100) ]

using speed of light c as 3*10^8

1.4 [ 0 - (0.7*3*10^8) / (3*10^8)²(100) ]

= -0.33 us

distance between

delltaX' = X2' - X1'

= 140m - 0

= 140m

No, The event are not simultaneous i.e they did not occur at the same time.

the second even (-0.33 us) occurs 0.33 us earlier than the first event.

4 0

3 years ago

At what point does the comet experience the strongest force of gravity?

spin [16.1K]

I think the answer may be the letter B

5 0

2 years ago

An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

andrew-mc [135]

Answer:

The object will travel 675 m during that time.

Explanation:

A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.

In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.

In this case, the position is calculated using the expression:

x = xo + vo*t + ½*a*t²

where:

x0 is the initial position.
v0 is the initial velocity.
a is the acceleration.
t is the time interval in which the motion is studied.

In this case:

x0= 0
v0= 0 because the object is initially stationary
a= 6 $\frac{m}{s^{2} }$
t= 15 s

Replacing:

x= 0 + 0*15 s + ½*6 $\frac{m}{s^{2} }$ *(15s)²

Solving:

x=½*6 $\frac{m}{s^{2} }$ *(15s)²

x=½*6 $\frac{m}{s^{2} }$ *225 s²

x= 675 m

<u><em> The object will travel 675 m during that time.</em></u>

5 0

2 years ago