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1 year ago

5. A nail contains trillions of electrons. Given that electrons repel from each other, why do they not then fly out of the nail?

Physics

1 answer:

diamong [38]1 year ago

4 0

Answer:

Nails are made of iron. Iron consists of 26 protons and 26 electrons. protons are positively charged and electrons are negatively charged, so this force of attraction keeps the electrons together.

If electrons repel from each other, the positively charge protons and nucleus allow them to move in a definite orbit and prevent them flying out of the nail.

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Essam is abseiling down a steep cliff. How much gravitational potential energy does he lose for every metre he descends? His mas

Dafna11 [192]

Answer:

720 J

Explanation:

The gravitational potential energy that Essam loses for every metre is given by:

$\Delta U=mg \Delta h$

where

m=72 kg is Essam's mass

$g=10 N/kg$ is the gravitational field strength

$\Delta h=1 m$ is the difference in height

By substituting the numbers into the formula, we find

$\Delta U=(72 kg)(10 N/kg)(1 m)=720 J$

5 0

2 years ago

Read 2 more answers

The spectrum of Star A has an absorption line of hydrogen at 660.0 nm. The spectrum of Star B has an absorption line at 666 nm.

rjkz [21]

Answer:

The stars are moving away from us.

Explanation:

The observed wavelengths of hydrogen transition for stars A and B (660.0 nm and 666 nm respectively) are greater than that observed in the laboratory (656.2 nm). The observed long wavelengths for the stars means that the light from the stars is red-shifted.

According to the Doppler effect, red-shifted light means that the source is moving a way from the observer; therefore, we arrive at the conclusion that the stars A and B are moving away from us.

6 0

2 years ago

A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington

Roman55 [17]

Let $\theta$ be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

$\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

The current has velocity vector (relative to the Earth)

$\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath$

The swimmer's resultant velocity (her velocity relative to the Earth) is then

$\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}$

$\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath$

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

$\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ$

which is approximately 41º west of north.

6 0

2 years ago

A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheri

MaRussiya [10]

Answer:

$dz=19217687.07\ m$

Explanation:

Given:

initial gauge pressure in the container, $P_0=2.02\times 10^{5}\ Pa$

atmospheric pressure at sea level, $P_a=1.01\times 10^5\ Pa$

initial volume, $V_0=4.4\times 10^{-4}\ m^3$

maximum pressure difference bearable by the container, $dP_{max}=2.26\times 10^{5}\ Pa$

density of the air, $\rho_a=1.2\ kg.m^{-3}$

density of sea water, $\rho_s=1.2\ kg.m^{-3}$

The relation between the change in pressure with height is given as:

$\frac{dP_{max}}{dz} =\rho_a.g_n$

where:

dz = height in the atmosphere

$g_n$ = standard value of gravity

Now putting the respective values:

$\frac{2.26\times 10^{5}}{dz} =1.2\times 9.8$

$dz=19217.687\ km$

$dz=19217687.07\ m$

Is the maximum height above the ground that the container can be lifted before bursting. (Since the density of air and the density of sea water are assumed to be constant.)

7 0

2 years ago

A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'

lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3

= 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

$Y =\frac{mgL}{A\Delta L}$

$\Delta L =\frac{mgL}{A\times Y}$

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0

2 years ago