Science, asked by sj5206026, 1 month ago

5. Solve the following examples.
a. An electric pump has 2 kW power.
How much water will the pump lift
every minute to a height of 10 m?
(Ans : 1224.5 kg)
b. If a 1200 W electric iron is used
daily for 30 minutes, how much
total electricity is consumed in the
month of April? (Ans:18 Unit)
c. If the energy of a ball falling from a
height of 10 metres is reduced by
40%, how high will it rebound?
(Ans: 6 m)
d. The velocity of a car increase from
54 km/hr to 72 km/hr. How much is
the work done if the mass of the car
is 1500 kg ?
(Ans. : 131250 J)
e. Ravi applied a force of 10 N and
moved a book 30 cm in the direction
of the force. How much was the
work done by Ravi?
(Ans: 3 J)​

Answers

Answered by Anonymous
0

Answer:

a. Power , P=2 kW

Power , P=

TimeTaken(T)

Workdone(W)

⇒W=P×T

For every minute i.e. 60 s, work done by the pump is W=2×1000×60=120000J

Now, this work done is stored in water as it potential energy. Thus, mgh = 120000 J

⇒m=

9.8×10

120000

=1224.5 kg

Hence, 1224.5 kg of water is lifted by the pump every minute to a height of 10 m.

b. total energy from 10m height is

mgh = m*g*10 = 10mg

and when it is reduced by 40%

then amount of energy left = (100-40)% = 0.6*(10mg ) = 6mg

now mgh' = 6mg

so, h' = 6m

c. Given,

Initial height of ball, h=10m

Potential energy=mgh

Reduce potential energy = mgh

new

mgh

new

=(

100

100−40

)mgh

h

new

=0.6×10=6m

Hence, height after rebound is 6m.

d. Initial Velocity, u=54km/hr=15m/s

Final velocity , v=72km/hr=20m/s

Mass =1500kg

Initial kinetic energy

2

1

mu

2

2

1

1500×15×15=168750J

Final kinetic energy

2

1

mv

2

2

1

1500×20×20=300000J

So, change in kinetic energy = work done=300000−168750=131250J.

e. Work done ,

W=F×S×cosθ

Here, θ=0

F=10NS=30cm=0.03m

∴W=10×0.03×1=3J.

Answered by shettysachi5
0

Answer:

A) In the image

B) Power =1200W=1.2kWh

Time = 30 min / day × 30 days =0.5h×30= 15 h

Energy =Power×Time=1.2×15= 18kWh.

C) Initial height of ball, h=10m

Potential energy=mgh

Reduce potential energy = mghnew​

mghnew​=(100100−40​)mgh

hnew​=0.6×10=6m

Hence, height after rebound is 6m

D) Initial Velocity, u=54km/hr=15m/s

Final velocity , v=72km/hr=20m/s

Mass =1500kg

Initial kinetic energy 21​mu2

⇒21​1500×15×15=168750J

Final kinetic energy 21​mv2

⇒21​1500×20×20=300000J

So, change in kinetic energy = work done=300000−168750=131250J

E) Work done ,  

W=F×S×cosθ

Here, θ=0∘

F=10NS=30cm=0.03m

∴W=10×0.03×1=3J

Explanation:

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