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3 years ago

Why are magnetic fields evidence of sea floor spreading

Physics

1 answer:

xenn [34]3 years ago

8 0

Eruptions of molten material, magnetic stripes in the rock of the ocean floor, and the ages of the rocks themselves.

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An electron moving with a velocity v⃗ = 5.0 × 107 m/s i^ enters a region of space where perpendicular electric and a magnetic fi

stiv31 [10]

Answer:

Magnetic field, $B=2\times 10^{-4}\ T$

Explanation:

Given that,

Velocity of electron, $v=5\times 10^7\ m/s$

It enters a region of space where perpendicular electric and a magnetic fields are present.

Magnitude of electric field, $E=10^4\ V/m$

We need to find the magnetic field will allow the electron to go through the region without being deflected.

Magnetic force on the electron, $F_m=qvB\ sin\theta$ .......(1)

Electric force on the electron, F = q E........(2)

From equation (1) and (2) we get:

$qvB\ sin\theta=qE$

$B=\dfrac{E}{v}$

$B=\dfrac{10^4\ V/m}{5\times 10^7\ m/s}$

B = 0.0002 T

$B=2\times 10^{-4}\ T$

Hence, this is the required solution.

3 0

3 years ago

Y’all I have best describes the purpose of a bar graph

Sholpan [36]

so you can see all the different categories at once. both as a whole and on an individual scale.

3 0

3 years ago

Determine the maximum theoretical speed that may be achieved over a distance of 66 m by a car starting from rest, knowing that t

My name is Ann [436]

Answer:

v=32.49 m/s

Explanation:

Given that

Distance ,d= 66 m

Initial speed of the car ,u = 0 m/s

Coefficient of friction ,μ = 0.8

Lets take the total mass of the car = m

The acceleration of the car is given as

a = μ g ( g= 10 m/s² )

Now by putting the values in the above equation we get

a= 0.8 x 10 m/s²

a= 8 m/s²

We know that ,final speed is given as

v²= u ²+ 2 a d

Now putting the value

v²=0² + 2 x 8 x 66

v²= 1056

v=32.49 m/s

3 0

3 years ago

What will happen when these two waves interact?

oee [108]

B would be your answer

4 0

3 years ago

The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the drag force?

timurjin [86]

Answer: The drag force goes up by a factor of 4

Explanation:

The Drag Force equation is:

$F_{D}=\frac{1}{2}C_{D}\rho A_{D}V^{2}$ (1)

Where:

$F_{D}$ is the Drag Force

$C_{D}$ is the Drag coefficient, which depends on the material

$\rho$ is the density of the fluid where the bicycle is moving (air in this case)

$A_{D}$ is the transversal area of the body or object

$V$ the bicycle's velocity

Now, if we assume $C_{D}$ , $\rho$ and $A_{D}$ do not change, we can rewrite (1) as:

$F_{D}=C.V^{2}$ (2)

Where $C$ groups all these coefficients.

So, if we have a new velocity $V_{n}$ , which is the double of the former velocity:

$V_{n}=2V$ (3)

Equation (2) is written as:

$F_{D}=C.V_{n}^{2}=C.(2V)^{2}$

$F_{D}=4CV^{2}$ (4)

Comparing (2) and (4) we can conclude the Drag force is four times greater when the speed is doubled.

7 0

3 years ago