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2 years ago

6

3.00Kg toy falls from a height of 1.00m. What is the kinetic energy just before the ground?

1 answer:

ivanzaharov [21]2 years ago

7 0

Answer:K E = 29.4 J

Explanation:

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How does a force pumb works

ser-zykov [4K]

Answer:

A force pump can be used to raise water by a height of more than 10m, the maximum height allowed by atmospheric pressure using a common lift pump.

In a force pump, the upstroke of the piston draws water, through an inlet valve, into the cylinder. On the downstroke, the water is discharged, through an outlet valve, into the outlet pipe.

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2 years ago

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What is the frequency of a wave

MAVERICK [17]

The frequency of a wave is the number of waves that passes through a point in a certain time. The less waves that pass in a period of time the lower the frequency of the wave. The more waves that pass in a period of time the higher the frequency of the wave. When measuring wave length the time period used is usually one second.

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2 years ago

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A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len

weeeeeb [17]

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}$

$\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}$

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}$

$\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}$

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

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2 years ago

Consider a 2250-lb automobile clocked by law-enforcement radar at a speed of 85.5 mph (miles/hour). if the position of the car i

Korvikt [17]

As per law of Heisenberg uncertainty law

product of uncertainty in position and uncertainty in momentum will be constant

$\Delta x . \Delta P = \frac{h}{4\pi}$

$\Delta x . m \Delta v = \frac{h}{4\pi}$

now plug in all data

$(5\times 0.3048). (2250 \times 0.454) \Delta v = \frac{6.6 \times 10^{-34}}{4\pi}$

$\Delta v = 3.37 \times 10^{-38} m/s$

So above is the uncertainty in velocity of the object

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2 years ago

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n electromagnetic wave in vacuum has an electric field amplitude of 611 V/m. Calculate the amplitude of the corresponding magnet

enot [183]

Answer:

The corresponding magnetic field is

Explanation:

From the question we are told that

The electric field amplitude is $E_o = 611\ V/m$

Generally the magnetic field amplitude is mathematically represented as

$B_o = \frac{E_o }{c }$

Where c is the speed of light with a constant value

$c = 3.0 *0^{8} \ m/s$

So

$B_o = \frac{611 }{3.0*10^{8}}$

$B_o = 2.0 4 *10^{-6} \ Vm^{-2} s$

Since 1 T is equivalent to $V m^{-2} \cdot s$

$B_o = 2.0 4 *10^{-6} \ T$

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3 years ago