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Mice21 [21]
2 years ago
11

5. is the earth’s magnetic field parallel to the ground at all locations? if not, where is it parallel to the surface? is its st

rength the same at all locations? if not, where is it greatest?
Physics
1 answer:
Setler79 [48]2 years ago
3 0

The Earth's magnetic field lines appear to emanate perpendicularly from the poles and go round along the Earth from the  South Pole to the North Pole. Therefore, the Earth's magnetic field is generally not parallel to the ground at every location. It is parallel to the earth at the equator near the earth's surface.

The Earth's magnetic field, also known as the geomagnetic field, is a magnetic field that extends from the Earth's interior into space, where it interacts with the solar wind, a stream of charged particles emanating from the Sun.

The magnetic field  varies in strength over the earth's surface. It is strongest at the poles and weakest at the equator.

To know more about Earth's magnetic field,

brainly.com/question/13019369

#SPJ4

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When is the angular momentum of a system constant?
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I think it’s a but I’m not 100% sure
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3 years ago
A projectile is launched at ground level with an initial speed of 50.0m/s at an angle of 30° above the horizontal. It strikes a
tino4ka555 [31]

Answer:

x = 129.9 m

y = 30.9 m

Explanation:

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

v_{i} = 50 m/s

Angle = 30°

Time = t = 3 s

horizontal component of velocity = v_{i_{x}} = v_{i}cos30°

v_{i_{x}} = 50cos30°

v_{i_{x}} = 43.3 m/s

Vertical component of velocity = v_{i_{y}} = v_{i}Sin30°

v_{i_{y}} = 50Sin30°

v_{i_{y}} = 25 m/s

This is a projectile motion, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

But the vertical component of velocity varies with time and there is an acceleration along vertical direction which is equal to gravitational acceleration g.

Horizontal distance = x =  v_{i_{x}}t

x =  43.3*3

x = 129.9 m

Vertical Distance = y = v_{i_{y}}t -0.5gt²

y = 25*3 - 0.5*9.8*3²

y = 75 - 44.1

y = 30.9 m

3 0
3 years ago
The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizo
Norma-Jean [14]

Answer:

0.528m

Explanation:

a)58.7 cm = 0.587 m

Let g = 9.8m/s2. When the frog jumps from ground to the highest point its kinetic energy is converted to potential energy:

E_p = E_k

mgh = mv^2/2

where m is the frog mass and h is the vertical distance traveled, v is the frog velocity at take-off

v^2 = 2gh = 2*9.8*0.587 = 11.5

v = \sqrt{11.5} = 3.4 m/s

b) Vertical and horizontal components of the velocity are

v_v = vsin(\alpha) = 3.4sin(58^0) = 2.877 m/s

v_h = vcos(\alpha) = 3.4cos(58^0) = 1.8 m/s

The time it takes for the vertical speed to reach 0 (highest point) under gravitational acceleration g = -9.8m/s2 is

\Delta t = \Delta v / g = \frac{0 - 2.877}{-9.8} = 0.293s

This is also the time it takes to travel horizontally, we can multiply this with the horizontal speed to get the horizontal distance it travels

s_h = v_ht = 1.8*0.293 = 0.528 m

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3 years ago
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