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3 years ago

7

An asteroid has acquired a net negative charge of 149 C from being bombarded by the solar wind over the years, and is currently

in equilibrium whereby it expels electrons at the same rate as it acquires them. How many more electrons does it have than protons

1 answer:

bearhunter [10]3 years ago

5 0

Answer:

93.125 × 10^(19)

Explanation:

We are told the asteroid has acquired a net negative charge of 149 C.

Thus;

Q = -149 C

charge on electron has a value of:

e = -1.6 × 10^(-19) C

Now, for us to determine the excess electrons on the asteroid, we will just divide the net charge in excess on the asteroid by the charge of a single electron.

Thus;

n = Q/e

n = -149/(-1.6 × 10^(-19))

n = 93.125 × 10^(19)

Thus, it has 93.125 × 10^(19) more electrons than protons

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If a 400-mm diameter pipe with a pipe roughness coefficient of 100 flows full of pressurized water with a head loss of 0.4 ft pe

RoseWind [281]

Answer:

Q = 913.9 gpm

Explanation:

The Hazen Williams equation can be written as follows:

$P = \frac{4.52\ Q^{1.85}}{C^{1.85}d^{4.87}}$

where,

P = Friction Loss per foot of pipe = $\frac{0.4}{1000\ ft}$ = 4 x 10⁻⁴

Q = Flow Rate in gallon/min (gpm) = ?

d = pipe diameter in inches = (400 mm)(0.0393701 in/1 mm) = 15.75 in

C = roughness coefficient = 100

Therefore,

$4\ x \ 10^{-4} = \frac{4.52\ Q^{1.85}}{(100)^{1.85}(15.75)^{4.87}}\\\\Q^{1.85} = \frac{4\ x \ 10^{-4}}{1.33\ x\ 10^{-9}} \\\\Q = (300384.75)^\frac{1}{1.85}$

<u>Q = 913.9 gpm</u>

5 0

3 years ago

A Porsche challenges a Honda to a 400 m race. Because the Porsche's acceleration of 3.4 m/s2 is larger than the Honda's 3.0 m/s2

padilas [110]

Answer:

Winner wins by 0.969 s

Explanation:

For the Porche:

Given:

Displacement of Porsche s = 400 m

Acceleration of Porsche a = 3.4 m/s^2

From Newton's second equation of motion,

$s = ut + (1/2) a t^2$ (u = 0 as the car was initially at rest)

Substituting the values into the equation, we have

$t^2 = (2 * 400) / 3.4$

= 235.29 / 3.4

t = 15.33 s

For the Honda:

Displacement of Honda = 310 m

Acceleration of Honda = 3 m/s^2

Applying Newton's second equation of motion

$s = ut + (1/2) a t^2$ (u = 0 for same reason)

Substituting the values into the equation, we obtain

$t^2 = (2 * 310) / 3$

= 620 / 3

t = 14.37 s

Hence

The winner (honda) wins by a time interval of = 15.33 - 14.37

=0.969 s

8 0

3 years ago

A rock falls from the top of a hill to the bottom. The total energy of the rock at the top isn’t the same as the sum of its pote

mart [117]

Answer:

A) some of the rocks energy is transformed to thermal energy

Explanation:

If we neglect air resistance during the fall of the rock, than the mechanical energy of the rock (which is sum of its potential energy and its kinetic energy) would be constant during the entire motion, so the total energy of the rock at the top would be the same as the sum of its potential energy and kinetic energy at the bottom.

However, this not occurs, due to the presence of air resistance. In fact, air resistance acts against the fall of the rock, and because of the friction between the molecules of air and the surface of the rock, the rock loses part of its energy. This energy is converted into thermal energy of the molecules of the air.

3 0

3 years ago

A ski tow operates on a slope of angle 15.6 ∘ of length 340 m . The rope moves at a speed of 12.7 km/h and provides power for 53

GREYUIT [131]

Answer:

P = 34034.2 Watt

Explanation:

As we know that the slope angle is given as

$\theta = 15.6^0$

now the weight of the rider along the slope is given as

$W = mgsin\theta$

$W = 69(9.81)sin15.6$

$W = 182 N$

now total weight of all 53 riders along the slope is given as

$F = 53 W$

$F = 9647.5 N$

now the speed of the rope is given as

$v = 12.7 km/h = 3.53 m/s$

now the power required is given as

$P = F.v$

$P = 9647.5(3.53)$

$P = 34034.2 Watt$

3 0

3 years ago

A solid sphere of mass 4.0 kg and radius 0.12 m starts from rest at the top of a ramp inclined 15°, and rolls to the bottom. The

cluponka [151]

Answer:

v = 4.1 m/s

Explanation:

As per mechanical energy conservation law we can say that initial total gravitational potential energy of the sphere is equal to final kinetic energy of rolling

so we will say

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

now for pure rolling condition we know that

$v = R\omega$

so we will have

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2})$

$mgh = \frac{7}{10}mv^2$

now we will have

$v^2 = \sqrt{\frac{10}{7}gh}$

$v^2 = \sqrt{\frac{10}{7}(9.8)(1.2)}$

$v = 4.1 m/s$

7 0

3 years ago