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3 years ago

What is the

Physics

1 answer:

Len [333]3 years ago

6 0

Answer:

$d=5\ g/cm^3$

Explanation:

Given that,

Mass of the object, m = 100 grams

Volume of the object, V = 20 cm³

We need to find the density of the object. We know that, density is equal to mass per unit volume. So,

$d=\dfrac{m}{V}\\\\d=\dfrac{100\ g}{20\ cm^3}\\\\d=5\ g/cm^3$

So, the density of the object is equal to $5\ g/cm^3$ .

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Find the angular momentum of a hollow spinning sphere when its angular speed is 20rad/s. The rotational inertia of the sphere is

zysi [14]

Answer:

L= 0.4 kgm²/s

Explanation:

The angular momentum of a hallow spinning sphere is

L = Iω

L = (0.02kgm²) × (20rad/s)

L= 0.4kgm²/s

5 0

3 years ago

Y=1/2 at^2 solve for t

Aloiza [94]

Explanation:

y = ½ at²

Multiply both sides by 2:

2y = at²

Divide both sides by a:

t² = 2y/a

Take the square root of both sides:

t = √(2y/a)

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3 years ago

The wavelength of violet light is about 425 nm (1 nanometer = 1 × 10−9 m). what are the frequency and period of the light waves?

BigorU [14]

1. Frequency: $7.06\cdot 10^{14} Hz$

The frequency of a light wave is given by:

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

where

$c=3\cdot 10^{-8} m/s$ is the speed of light

$\lambda$ is the wavelength of the wave

In this problem, we have light with wavelength

$\lambda=425 nm=425\cdot 10^{-9} m$

Substituting into the equation, we find the frequency:

$f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz$

2. Period: $1.42 \cdot 10^{-15}s$

The period of a wave is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

The frequency of this light wave is $7.06\cdot 10^{14} Hz$ (found in the previous exercise), so the period is:

$T=\frac{1}{f}=\frac{1}{7.06\cdot 10^{14} Hz}=1.42\cdot 10^{-15} s$

4 0

3 years ago

Read 2 more answers

platform diving in the olympic games takes place at two heights: 5 meters and 10 meters. What is the velocity of a diver enterin

posledela

1) Velocity: 9.9 m/s and 14 m/s

The motion of the diver is a free-fall motion, so it is a uniform accelerated motion. Choosing downward as positive direction, the final velocity can be found by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u = 0 is the initial velocity (the diver starts from rest)

$a=g=9.8 m/s^2$ is the acceleration of gravity

s is the displacement

For the diver jumping from 5 m, s = 5 m, so

$v=\sqrt{2as}=\sqrt{2(9.8)(5)}=9.9 m/s$

For the diver jumping from 10 m, s = 10 m, so

$v=\sqrt{2as}=\sqrt{2(9.8)(10)}=14 m/s$

2) Time: 1.01 s and 1.43 s

The time of flight of each diver can be found by using the other suvat equation

$s=ut+\frac{1}{2}at^2$

And since u = 0, it can be reduced to

$s=\frac{1}{2}at^2$

For the diver jumping from 5 m, s = 5 m, so we find

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(5)}{9.8}}=1.01 s$

For the diver jumping from 10 m, s = 10 m, so we find

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(10)}{9.8}}=1.43 s$

5 0

3 years ago

The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able

vovangra [49]

Answer:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

Explanation:

<u>Charge of an Electron</u>

Since Robert Millikan determined the charge of a single electron is

$q_e=-1.6\cdot 10^{-19}\ C$

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is $5\times -1.6\cdot 10^{-19}\ C=-8 \cdot 10^{-19}\ C$

Let's test the possible charges listed in the question:

$-8.0 \times 10 ^{-19 }$ . We have just found it's a possible charge of a particle

$-3.2 \times 10 ^{-19 }$ . Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

$-1.2 \times 10 ^{-19 }$ this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge

$-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 }$ cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

Finally, the charge $-4.8 \times 10 ^{-19 }\ C$ is four times the charge of the electron, so it is a possible value for the charge of an oil droplet

Summarizing, the following are the possible values for the charge of an oil droplet:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

5 0

3 years ago