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4 years ago

11

A 60 kg box is lifted by a rope a distance of 10 meters straight up at a constant speed. How much power is required to complete

this task in 5 seconds?

1 answer:

kogti [31]4 years ago

5 0

The work done is energy which is required to lift the box,fitting the force of gravity.

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4 years ago

A 25 kg block is held against a compressed spring and then the spring is allowed to decompress giving the block a velocity. The

Alex787 [66]

Answer:

h=18.05 cm

Explanation:

Given that

m= 25 kg

K= 1300 N/m

x=26.4 cm

θ= 19.5 ∘

When the block just leave the spring then the speed of block = v m/s

From energy conservation

$\dfrac{1}{2}Kx^2=\dfrac{1}{2}mv^2$

$Kx^2=mv^2$

$v=\sqrt{\dfrac{kx^2}{m}}$

By putting the values

$v=\sqrt{\dfrac{kx^2}{m}}$

$v=\sqrt{\dfrac{1300\times 0.264^2}{25}}$

v=1.9 m/s

When block reach at the maximum height(h) position then the final speed of the block will be zero.

We know that

$V_f^2=V_i^2-2gh$

By putting the values

$0^2=1.9^2-2\times 10\times h$

h=0.1805 m

h=18.05 cm

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4 years ago

A charged cloud system produces an electric field in the air near earth's surface. a particle of charge -2.0 × 10-9 c is acted o

defon

Part a)

Magnitude of electric field is given by force per unit charge

$E = \frac{F}{q}$

$E = \frac{4.3 * 10^{-6}}{2 * 10^{-9}}$

$E = 2150 N/C$

Part b)

Electrostatic force on the proton is given as

F = qE

$F = 1.6 * 10^{-19} * 2150$

$F = 3.44 * 10^{-16} N$

PART C)

Gravitational force is given by

$F_g = mg$

$F_g = 1.6 * 10^{-27}*9.8$

$F_g = 1.57 * 10^{-26} N$

PART d)

Ratio of electric force to weight

$\frac{F_e}{F_g} = \frac{3.44 * 10^{-16}}{1.57*10^{-26}}$

$\frac{F_e}{F_g} = 2.2 * 10^{10}$

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Th answer would be 1.5

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3 years ago

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