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3 years ago

If a 60-g object has a volume of 30 cm", what is its density?

Physics

1 answer:

alisha [4.7K]3 years ago

5 0

Answer:

Density = Mass/volume. D= 60/30.Divide it and you'll get ur answer as 2

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A disk with a diameter of 0.05 m is spinning with a constant velocity about an axle perpendicular to the disk and running throug

Maksim231197 [3]

Answer:

f= 7.8Hz

Explanation:

a= 12g= 120m/s²

a= ω²r

120= ω²×0.05

ω= 48.99 rad/s

2πf= 48.99

f= 7.8Hz

8 0

3 years ago

Can you hellp me please

mamaluj [8]

Answer:

i think the answer is B but im not sure

7 0

3 years ago

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The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t

Lubov Fominskaja [6]

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = $854\times 10^{3}\ Hz$

Power, P = 50 kW = $50\times 10^{3}\ W$

$I_{p} = 1\ photon/s/m^{2}$

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

$n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s$

Area of the sphere, A = $4\pi r^{2}$

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is $1\ photon/s/m^{2}$

$I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}$

$1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}$

$r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m$

Now,

We know that:

1 ly = $9.4607\times 10^{15}\ m$

Thus

$r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly$

5 0

3 years ago

What is the unit of measurement of velocity

melomori [17]

Usually the unit of measurement of velocity is meters per second or m/s

7 0

3 years ago

Read 2 more answers

A charged particle is moving in a uniform magnetic field at a speed of 8.2Ã10^3 m/s in a direction 87Â° from the direction of th

77julia77 [94]

Answer:

Magnetic field, B = 0.042 T

Explanation:

It is given that,

Speed of charged particle, $v=8.2\times 10^3\ m/s$

Angle between velocity and the magnetic field, $\theta=87$

Charge, $q=5.7\ \mu C=5.7\times 10^{-6}\ C$

Magnetic force, F = 0.002 N

The magnetic force is given by :

$F=qvB\ sin\theta$

B is the magnetic field

$B=\dfrac{F}{qv\ sin\theta}$

$B=\dfrac{0.002}{5.7\times 10^{-6}\times 8.2\times 10^3\times sin(87)}$

B = 0.042 T

So, the strength of the magnetic field is 0.042 Tesla. Hence, this is the required solution.

5 0

4 years ago