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IceJOKER [234]
3 years ago
13

How are geologists able to study continental drift by observing rock samples, volcanic activity, etc., but not when they look wi

th just their eyes?
Physics
1 answer:
Andrej [43]3 years ago
7 0

Answer:

On the Earth today, volcanoes are mainly formed by continental drift. As lighter continental rocks are pushed over heavier oceanic ones, friction melts the rock which rises and produces a volcano. ... As it sinks it melts and hot rock rises up, creating volcanoes along the coast. The Andes are being created in this way.

Explanation:

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What law states that when there is an interaction between charged objects there is a non-contact force which acts over some dist
Dmitry [639]
That's Coulomb's law, but it's much better than that.  If you know
the charges on the objects and the distance between them, then
Coulomb actually tells you the strength and direction of the force.
3 0
3 years ago
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the resistivity of a given wire of cross session area 0.7mm² is 49×10⁴.calcuate the resistance of a 2m lenght of wire​
sp2606 [1]

R = 1.4GΩ.

The relation between the resistance and the resistivity is given by the equation R = ρL/A, where ρ is the resistivity of a given material, L is the length and A is the cross-sectional area of the material.

To calculate the resistance of a wire of L = 2m, ρ = 49x10⁴Ω.m and A = 0.7mm² = 0.7x10⁻³m² we have to use the equation  R = ρL/A.

R = [(49x10⁴Ω.m)(2m)/0.7x10⁻³m²

R = 98x10⁴Ω.m²/0.7x10⁻³m²

R = 1.4x10⁹Ω = 1.4GΩ

6 0
3 years ago
How much charge must pass by a point in a wire in 1.5 s for the current inb the wire to be 2.0 A?
skelet666 [1.2K]

Answer:

3 Coulombs

Explanation:

Q = Current x time

Q = 2.0 x 1.5

Q = 3 Coulombs

3 0
4 years ago
You're driving at 50 mph and then speed up to 70 mph. how much more "force" does the engine in your car have to produce to maint
Fittoniya [83]
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3 years ago
A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex
irina1246 [14]

a)

F_{E_x}=1.19\cdot 10^{-3}N (+x axis)

F_{B_x}=0

F_{B_y}=0

b)

F_{E_x}=1.19\cdot 10^{-3} N (+x axis)

F_{B_x}=0

F_{B_y}=3.21\cdot 10^{-3}N (+z axis)

c)

F_{E_x}=1.19\cdot 10^{-3} N (+x axis)

F_{B_x}=3.21\cdot 10^{-3} N (+y axis)

F_{B_y}=3.21\cdot 10^{-3}N (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

F=qE

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

q=+4.9\mu C=+4.9\cdot 10^{-6}C is the charge

E_x=+242 N/C is the electric field, along the x-direction

So the electric force (along the x-direction) is:

F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N

towards positive x-direction.

The magnetic force instead is given by

F=qvB sin \theta

where

q is the charge

v is the velocity of the charge

B is the magnetic field

\theta is the angle between the directions of v and B

Here the charge is stationary: this means v=0, therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

F_{E_x}=1.19\cdot 10^{-3} N (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- B_x: this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, \theta=0^{\circ}, so the force due to this field is zero.

- B_y: this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, \theta=90^{\circ}. Therefore, \theta=90^{\circ}, so the force due to this field is:

F_{B_y}=qvB_y

where:

q=+4.9\cdot 10^{-6}C is the charge

v=345 m/s is the velocity

B_y = +1.9 T is the magnetic field

Substituting,

F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

F_{E_x}=1.19\cdot 10^{-3}N (+x axis)

For the field B_x, the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

F_{B_x}=qvB_x

And by substituting,

F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field B_y, the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

F_{B_y}=qvB_y

And by substituting,

F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0
4 years ago
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