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3 years ago

11

A book on a 2 meter high shelf has 20 Joules of potential energy.

1 answer:

Sever21 [200]3 years ago

4 0

Mass of this book is 1 kg.

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Three polarizing filters are shown in the figure below. The alignment of each filterâs transmission axis is shown, as measured f

lidiya [134]

Answer: B.

Explanation:

Hope help :p

5 0

2 years ago

How to find a planet’s gravitational field strength using its radius?

grin007 [14]

The gravitational field strength is approximately equal to 10 N.

<u>Explanation:</u>

Gravitational field strength is the measure of gravitational force acting on any object placed on the surface of the planet. Generally, the mass of the object is considered as 1 kg.

So the gravitational field strength will be equal to the gravitational force acting on the object.

The formula for gravitational field strength is

$g = \frac{F}{m}$

Here g is the gravitational field strength, m is the mass of the object placed on the surface and F is the gravitational force acting on the object.

Since, the mass of any object placed on the surface of earth will be negligible compared to the mass of Earth, so the mass of the object is considered as 1 kg.

Then the g = F

And $F =\frac{GMm}{r^{2} }$

Here G is the gravitational constant, M is the mass of Earth and m is the mass of the object placed on the surface, while r is the radius of the Earth.

$g = F = \frac{6 \times 10^{24} \times 6.67 \times 10^{-11} \times 1}{(6.6 \times 10^{6}) ^{2} }$

$g = 0.977 \times 10^1= 9.77\ N$

So, the gravitational field strength is approximately equal to 10 N.

5 0

2 years ago

a particle of mass m sits at rest at x = 0. At time t = 0 a force given by F = Fe^(-t/T) is applied in the +x direction; F and T

wlad13 [49]

Explanation:

Given: $F=m\ddot{x}=Fe^{-\frac{t}{T}}$

Solving for $\ddot{x}$ :

$\ddot{x}=\frac{F}{m}e^{-\sqrt{\frac{F}{m} } t}$

where:

$T=\sqrt{\frac{m}{F}}$

Integrating to get $\dot{x}$ with initial conditions $\dot{x}(0)=0$ :

$\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}} t}$

Integrating to get x with initial conditions x(0) = 0:

$x=-1+\sqrt{\frac{F}{m}} t+e^{-\sqrt{\frac{F}{m}}t}$

When t=T:

$x=-1+\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}+e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\frac{1}{e}$

$\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\sqrt{\frac{F}{m}}(1-\frac{1}{e})$

4 0

2 years ago

Monochromatic light of wavelength 385 nm is incident on a narrow slit. On a screen 3.00 m away, the distance between the second

LiRa [457]

To solve this problem it is necessary to apply the concepts related to the concept of overlap and constructive interference.

For this purpose we have that the constructive interference in waves can be expressed under the function

$a sin\theta = m\lambda$

Where

a = Width of the slit

d = Distance of slit to screen

m = Number of order which represent the number of repetition of the spectrum

$\theta =$ Angle between incident rays and scatter planes

At the same time the distance on the screen from the central point, would be

$sin\theta = \frac{y}{d}$

Where y = Represents the distance on the screen from the central point

PART A ) From the previous equation if we arrange to find the angle we have that

$\theta = sin^{-1}(\frac{y}{d})$

$\theta = sin^{-1}(\frac{1.4*10^{-2}}{3})$

$\theta = 0.2673\°$

PART B) Equation both equations we have

$a sin\theta = m\lambda$

$a \frac{y}{d} = m\lambda$

Re-arrange to find a,

$a = \frac{(2)(385*10^{-9})(3)}{(1.4*10^{-2})}$

$a = 1.65*10^{-4}m$

8 0

3 years ago

Which of the following describes the efficiency of real machines?

Leokris [45]

Choice-B is the true one.

3 0

3 years ago

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