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2 years ago

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You toss a coin straight upward in a train which is moving at a constant velocity. If the air friction is negligible, the coin w

ill land (a) In front of you (b) behind you (c) on your hand (d) at either in front of you or behind you

1 answer:

grandymaker [24]2 years ago

6 0

Answer:

option (c)

Explanation:

As you toss the coin, it will land in your hand. It is because the coin has same horizontal velocity as you and as that of train. So the horizontal distance traveled by you and the coin remains same and thus it will land in your hand.

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A charged particle is accelerated in a uniform electric field. When its velocity is 2 m/s, its electric potential energy is 100

zavuch27 [327]

Answer:

particle's potential energy = 70J

Explanation:

From conservation of energy; K1 + Ue1 = K2 + Ue2

where K1 and K2 are the kinetic energies at two positions and Ue1 and Uue2 are the electrical potential energies at two positions.

k1 = 10J, Ue1 = 100J

K2 = 40J

substitute into K1 + Ue1 = K2 + Ue2

Ue2 = K1 + Ue1 - K2

= 10 +100 - 40

Ue2 = 70J

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2 years ago

Arturiano [62]

Explanation:

A light bulb changes electrical energy into <em>heat energy and light energy .</em>

4 0

3 years ago

The magnetic field on the Sun is created by______.

Ber [7]

The answer is actually c hope this helps
( - - )
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| |

8 0

3 years ago

Verify that the linear speed of an ultracentrifuge is about 0.50 km's, and Earth in its orbit is about 30 km/s by calculating:

FrozenT [24]

Answer:

a) Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.

Explanation:

The linear speed of the particle ( $v$ ), measured in kilometers per second, rotating in a circular pattern is calculated by the following formula:

$v = R\cdot \omega$ (1)

Where:

$R$ - Radius, measured in kilometers.

$\omega$ - Angular speed, measured in radians per second.

Now we proceed to calculate the linear speed of each element:

a) Ultracentrifuge

If we know that $\omega \approx 5235.988\,\frac{rad}{s}$ and $R = 1\times 10^{-4}\,km$ , then the linear velocity is:

$v = (1\times 10^{-4}\,km)\cdot \left(5235.988\,\frac{rad}{s} \right)$

$v = 0.524\,\frac{km}{s}$

Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Earth

The Earth is 150 million kilometers away from the Sun and takes 365 days to complete one revolution around the Sun. First, we calculate angular speed of the planet:

$\omega = \frac{2\pi}{T}$ (2)

Where $T$ is the period, measured in seconds.

If we know that $T = 31536000\,s$ , then the angular speed of the Earth is:

$\omega = \frac{2\pi}{31536000\,s}$

$\omega = 1.992\times 10^{-7}\,\frac{rad}{s}$

Now, we determine the linear speed:

$v = (1.5\times 10^{8}\,km)\cdot \left(1.992\times 10^{-7}\,\frac{rad}{s} \right)$

$v = 29.88\,\frac{km}{s}$

Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.

6 0

2 years ago

The escape velocity on earth is 11.2 km/s. What fraction of the escape velocity is the rms speed of H2 at a temperature of 31.0

Sveta_85 [38]

To solve this problem it is necessary to apply the concept related to root mean square velocity, which can be expressed as

$v_{rms} = \sqrt{\frac{3RT}{n}}$

Where,

T = Temperature

R = Gas ideal constant

n = Number of moles in grams.

Our values are given as

$v_e =11.2km/s = 11200m/s$

The temperature is

$T = 30\°C = 30+273 = 303K$

Therefore the root mean square velocity would be

$v_{rms} = \sqrt{\frac{3(8.314)(303)}{0.002}}$

$v_{rms} = 1943.9m/s$

The fraction of velocity then can be calculated between the escape velocity and the root mean square velocity

$\alpha = \frac{v_{rms}}{v_e}$

$\alpha = \frac{1943.9}{11200}$

$\alpha = 0.1736$

Therefore the fraction of the scape velocity on the earth for molecula hydrogen is 0.1736

7 0

3 years ago