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2 years ago

What is the density of an object with mass of 60 g and a volume of 2cm3

Physics

1 answer:

Artemon [7]2 years ago

4 0

Answer:

D=mass /volume

=60/0.002

=30000

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What is the wavelength of light that is deviated in the first order through an angle of 13.1 ∘ by a transmission grating having

faltersainse [42]

Answer:

The wavelength of light is $4.53\times10^{-7}\ m$

Explanation:

Given that,

Angle = 13.1°

Number of slits = 5000

We need to calculate the wavelength of light

Diffraction of first order is defined as,

$d \sin\theta=n\lambda$ .....(I)

The separation of the slits

$d = \dfrac{1}{N}$

$d=\dfrac{1}{5000}$

$d=2\times10^{-6}\ m$

Now put the value in equation (I)

$2\times10^{-6}\sin13.1^{\circ}=\lambda$

Here, n = 1

$\lambda=4.53\times10^{-7}\ m$

Hence, The wavelength of light is $4.53\times10^{-7}\ m$

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3 years ago

What is the name of the line drawn perpendicular to the surface where a light ray strikes?

AnnyKZ [126]

It is the normal line

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3 years ago

A satellite with a mass of 2,000 ㎏ is inserted into an orbit that is twice the Earth' s radius. W hat is the force of gravity on

dolphi86 [110]

Answer:

option (b) 4900 N

Explanation:

m = 2000 kg, R = 6380 km = 6380 x 10^3 m, Me = 5.98 x 10^24 kg, h = R

F = G Me x m / (R + h)^2

F = G Me x m / 2R^2

F = 6.67 x 10^-11 x 5.98 x 10^24 x 2000 / (2 x 6380 x 10^3)^2

F = 4900 N

4 0

3 years ago

which type of wave spreading do you think causes faster energy loss-two-dimensional or three-dimensional? explain.

sdas [7]

Three dimensional would loose faster

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3 years ago

Read 2 more answers

An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane

lubasha [3.4K]

Answer:

$\rho = 12580.7 kg/m^3$

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

$F = \frac{GMm}{(r + h)^2}$

here we have

$F =\frac {mv^2}{(r+ h)}$

$\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}$

here we have

$v = \sqrt{\frac{GM}{(r + h)}}$

now we can find time period as

$T = \frac{2\pi (r + h)}{v}$

$T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}$

$1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}$

$M = 4.54 \times 10^{23} kg$

Now the density is given as

$\rho = \frac{M}{\frac{4}{3}\pi r^3}$

$\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}$

$\rho = 12580.7 kg/m^3$

8 0

3 years ago