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2 years ago

10

A 5kg mass is pushed with a force of 10N for a distance of 2.5 meters. The work done is

1 answer:

Aliun [14]2 years ago

7 0

W = 25 J

Explanation:

Work done on an object is defined as

$W = Fd = (10\:\text{N})(2.5\:\text{m}) = 25\:\text{J}$

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What does the area between the line and the x-axis represent on a velocity vs. time graph?

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It has been learned in this lesson that the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period. ... Once calculated, this area represents the displacement of the object.

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What two factors determine the density of a fluid?

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3 years ago

Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has magnitude of 3.7 m/s^2. We want

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2.32 s

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Using the equation of motion,

s = ut+g't²/2............................ Equation 1

Where s = distance, u = initial velocity, g' = acceleration due to gravity of the moon, t = time.

Note: Since Onur drops the basket ball from a height, u = 0 m/s

Then,

s = g't²/2

make t the subject of the equation,

t = √(2s/g')...................... Equation 2

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Plain electromagnetic wave (in air) has a frequency of 1 MHz and its B-field amplitude is 9 nT a. What is the wavelength in air?

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Answer:

Part a)

$\lambda = 300 m$

Part b)

$E = 2.7 N/C$

Part c)

$I = 9.68 \times 10^{-3} W/m^2$

$P = 3.22 \times 10^{-11} N/m^2$

Explanation:

Part a)

As we know that frequency = 1 MHz

speed of electromagnetic wave is same as speed of light

So the wavelength is given as

$\lambda = \frac{c}{f}$

$\lambda = \frac{3\times 10^8}{1\times 10^6}$

$\lambda = 300 m$

Part b)

As we know the relation between electric field and magnetic field

$E = Bc$

$E = (9 \times 10^{-9})(3\times 10^8)$

$E = 2.7 N/C$

Part c)

Intensity of wave is given as

$I = \frac{1}{2}\epsilon_0E^2c$

$I = \frac{1}{2}(8.85 \times 10^{-12})(2.7)^2(3\times 10^8)$

$I = 9.68 \times 10^{-3} W/m^2$

Pressure is defined as ratio of intensity and speed

$P = \frac{I}{c} = \frac{9.68\times 10^{-3}}{3\times 10^8}$

$P = 3.22 \times 10^{-11} N/m^2$

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3 years ago