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Ber [7]
2 years ago
5

Where would you find convergent and divergent plate boundaries relative to

Physics
1 answer:
Sergio [31]2 years ago
8 0

Answer:

When two tectonic plates meet, we get a “plate boundary.” There are three major types of plate boundaries, each associated with the formation of a variety of geologic features.

Explanation:

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A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find
pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0=4\pi \cdot 10^{-7} H/m is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

I=1.41 A (current in the wire)

B=5.61\mu T=5.61\cdot 10^{-6} T (strength of magnetic field)

Solving  for r, we find the distance  from the wire:

r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m

4 0
3 years ago
A proton is moving in a circular orbit of radius 12 cm in a uniform 0.31-T magnetic field perpendicular to the velocity of the p
PIT_PIT [208]

Answer:

Explanation:

A proton of charge

q=+1.609×10^-19C

Orbit a radius of 12cm

r=0.12m

Magnetic Field of 0.31T

Angle between velocity and field is 90°

a. Because the magnetic force F supplies the centripetal force Fc.

The magnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by

F = qvB sin θ

And the centripetal force is given as

Fc=mv²/r

Where m is mass of proton

m=1.673×10^-27kg

Then, F=Fc

qvB sin θ=mv²/r

qBSin90=mv/r

rqB=mv

Then, v=rqB/m

v=0.12×1.609×10^-19×0.31/1.673×10^-23

v=3577692.78m/s

v=3.58×10^6m/s

b. Since,

F=qVBSin90

F=1.609×10^-19×3.58×10^6×0.31

F=1.785×10^-13 N.

6 0
3 years ago
If we use 1 millimeter to represent 1 light-year, how large in diameter is the Milky Way Galaxy?
storchak [24]

Answer:

d.100 meters

Explanation:

The diameter of the Milky Way Galaxy is approximately 100,000 light years.

Here we are using 1 millimiter (1 mm) to represent 1 light-year (1 ly). So, we can set the following proportion:

1 mm : 1 ly = x : 100,000 ly

and by finding x, we find the diameter of the Milky Way Galaxy in the scale used:

x=\frac{(1mm )(100,000 ly)}{1 ly}=100,000 mm = 100 m

so the correct answer is

d. 100 meters

4 0
3 years ago
Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?
frosja888 [35]

Answer:

The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

Explanation:

From the question given above, the following data were obtained:

Height to which the target is located = 50 m

Initial velocity (u) = 20 m/s

To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:

Initial velocity (u) = 20 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 20² – (2 × 10 × h)

0 = 400 – 20h

Collect like terms

0 – 400 = – 20h

– 400 = – 20h

Divide both side by – 20

h = – 400 / – 20

h = 20 m

Thus, the the maximum height to which the cannon ball attained is 20 m.

From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

3 0
3 years ago
Answer 4 and I’ll give brainlist
Sedbober [7]

Answer:

Answer C

Explanation:

The closer two masses are, the greater the gravity force between them.

This is shown in the formula

Fg = GMm/R²

Smaller distance R means greater force Fg

4 0
3 years ago
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