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3 years ago

Where does the heat come from that creates the motion for continental drift?

Physics

1 answer:

poizon [28]3 years ago

4 0

Answer:

The heat doesn't come from any of those the heat comes from the Sun

Explanation:

But the thing that holds the heat from the Sun is the Earth's atmosphere.

When he gets hot the carbon dioxide holds in the heat but when the when oxygen it meets the heat in the air it makes it become cooler

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A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed

Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

$\int EdS=\frac{Q_{in}}{\epsilon_o}$ (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

$\int EdS=ES=E(4\pi r^2)$ (2)

Qin is calculate by using the charge density:

$Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho$ (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

$\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}$

Next, you use the results of (3), (2) and (1):

$E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})$

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

$E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}$

hence, the electric field is 1580594.95 N/C

7 0

3 years ago

Two power lines run parallel for a distance of 220 m and are separated by a distance of 40.0 cm. If the current in each of the t

daser333 [38]

Answer:

The magnitude of force is 1.86 N and the direction of force is towards the other wire.

Explanation:

Given:

Current flowing through each power line, I = 130 A

Distance between the two power lines, d = 40 cm = 0.4 m

Length of power lines, L = 220 m

The force exerted by the power lines on each other is given by the relation:

$F = \frac{\mu_{0}LII }{2\pi d}$

Substitute the suitable values in the above equation.

$F = \frac{4\pi\times10^{-7}\times220\times130\times130 }{2\pi\times0.4}$

F = 1.86 N

Since the direction of current flowing through the power lines are opposite to each other, so the force is attractive in nature. Hence, the direction of force experienced by the power lines on each other is towards the each other.

5 0

3 years ago

jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh

Greeley [361]

Answer:

C. 590 mph

$\vert v_{cj}\vert=589.49\ mph$

Explanation:

Given:

velocity of jet, $v_j=500\ mph$
direction of velocity of jet, east relative to the ground
velocity of Cessna, $v_c=150\ mph$
direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

$\vec v_j=500\ \hat i\ mph$

$\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)$

$\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph$

Now the vector of relative velocity of Cessna with respect to jet:

$\vec v_{cj}=\vec v_j-\vec v_c$

$\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )$

$\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph$

Now the magnitude of this velocity:

$\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}$

$\vert v_{cj}\vert=589.49\ mph$ is the relative velocity of Cessna with respect to the jet.

8 0

3 years ago

Q3. You throw a ball into the air, it reaches a certain height and then comes back to you.

Grace [21]

The If a car is going round a curve , there is an acceleration because the direction of the velocity changes.

<h3>What is the direction of the velocity?</h3>

Now we know that if you throw the ball upwards, the motion is in opposite direction to gravity thus the ball is experiencing deceleration and the speed decreases. The velocity decreases and the acceleration is negative.

If the ball is coming down, then the ball is accelerated thus it speeds up and the direction of the acceleration is positive.

If a car is going round a curve, the vehicle is accelerating because the direction of the velocity changes even if its amount remains constant.

When a board is moving down a hill at 2 ms-1, it is experiencing an acceleration because the motion is in the same direction as gravity.

If a car is coming to a stop at a point, it experiences a deceleration and not an acceleration since the change of velocity with time is negative as the car comes to rest.

Learn more about acceleration:brainly.com/question/12550364

#SPJ1

5 0

2 years ago

A 5.0 kg object moving at 5.0 m/s. KE = mv2 times 1/2

steposvetlana [31]

Answer: KE = 62.5J

Explanation:

Given that

Mass of object = 5kg

kinetic energy KE = ?

velocity of object = 5m/s

Since kinetic energy is the energy possessed by a moving object, and it depends on the mass (m) of the object and the velocity (v) by which it moves. Therefore, the object has kinetic energy.

i.e K.E = 1/2mv^2

KE = 1/2 x 5kg x (5m/s)^2

KE = 0.5 x 5 x 25

KE = 62.5J

Thus, the object has 62.5 joules of kinetic energy.

5 0

3 years ago