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Liono4ka [1.6K]

3 years ago

7

The charge of an electronic is A.-2 B.-1 C.0 D.+1

1 answer:

NISA [10]3 years ago

6 0

It is given that the charge of an electron is -1.602x10^-19. When expressed in atomic units, the elementary charge (<span>1.602x10^-19</span>) is equal to the value of unity (1). Since an electron carries a negative charge, the charge should be equal to negative unity, or -1.

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The acceleration due to gravity on the Moon's surface is

Molodets [167]

Answer:

50 lb

Explanation:

Given,

The weight of astronaut's life support backpack on Earth (w) = 300 lb

Acceleration due to gravity on Earth (g) = 9.8 m/s²

Acceleration due to gravity on Moon = g'

$g'=\frac{g}{6}$

We know that weight of an object on Earth is,

$w = m\times g$

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

Similarly, weight on Moon will be

$w' = m\times g'$

$w' = \frac{w}{g}\times\frac{g}{6}$

$w' = \frac{300}{6}$

$w' = 50$

Thus the astronaut's life support backpack will weigh 50 lb on Moon.

7 0

3 years ago

What is the centripetal acceleration of the earth as it travels around the sun when Earth has an orbital radius of 1.5 x 10^11 m

masya89 [10]

Answer: $0.0058 m/s^{2}$

Explanation:

Centripetal acceleration $a_{C}$ is calculated by the following equation:

$a_{C}=\frac{V^{2}}{r}$

Where:

$V=29.7 \frac{km}{h} \frac{1000 m}{1 km}=29700 m/s$ is the Earth's orbital speed

$r=1.5(10)^{11} m$ is the orbital radius

$a_{C}=\frac{(29700 m/s)^{2}}{1.5(10)^{11} m}$

$a_{C}=0.0058 m/s^{2}$

4 0

3 years ago

A hockey puck with a mass of 0.175 kg slides over the ice. The puck initially slides with a speed of 5.25 m/s, but it comes to a

Neko [114]

Answer:

1.70 J

Explanation:

The heat dissipated is the difference in the kinetic energies.

This is given by

$E = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

$v_i$ and $v_f$ are the initial and final velocities.

With <em>m</em> = 0.175 kg,

$E = \frac{1}{2}\times0.175(2.85^2 - 5.25^2) = -1.701\text{ J}$

The negative sign appears because energy is lost.

7 0

3 years ago

The Greeks noticed that some stars were larger and moved haphazardly. What were these stars?

Varvara68 [4.7K]

Answer:

i guess it's shooting star

5 0

3 years ago

At what height h above the ground does the projectile have a speed of 0.5v?

dedylja [7]

Answer:height above ground at which projectile have velocity

0.5v is (0.0375v^2)

Explanation:

Using Vf = Vi - gt

Where Vf is final velocity

Vi is initial velocity

g is the acceleration due to gravity

t is the time taken

So, 0.5v = v - gt

t = 0.05v

Therefore height h = vt - 0.5gt^2

Subtitute t

h = 0.05v^2 - 0.0125v^2

h = 0.0375v^2

3 0

3 years ago