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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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The acceleration of a particle traveling along a straight line isa=14s1/2m/s2, wheresis in meters. Ifv= 0,s= 1 m whent= 0, deter
Yuliya22 [10]

Answer:

0.78m/s

Explanation:

We are given that

Acceleration=a=\frac{1}{4}s^{\frac{1}{2}}m/s^2

v=0, s=1 when t=0

We have to find the particle's velocity at s=2m

We know that

a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=\frac{dv}{ds}v

vdv=ads

\int_{0}^{v} vdv=\int_{1}^{s}0.25s^{\frac{1}{2}}ds

\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2})^{s}_{1}

By using formula:\int x^ndx=\frac{x^{n+1}}{n+1}+C

\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2}}-1

Substitute s=2

\frac{v^2}{2}=\frac{0.50}{3}((2^{1.5})-1)

\frac{v^2}{2}=\frac{0.50}{3}\times 1.83

v^2=2\times 0.305=0.61

v=\sqrt{0.61}=0.78m/s

Hence, the velocity of particle at s=2m=0.78m/s

3 0
3 years ago
Thinking about an analogue clock, calculate the angular displacement in revolutions, radians and degrees for the following: • A
Inessa [10]

Answer:

1.047 rad

Explanation:

A secondhand makes a complete revolution (360 degrees) in 60 seconds, so it displaces 6 degrees for every second of elapsed time (360/60 = 6).

And in 10 seconds, it will make a displacement of 10 degrees (6 * 10 = 60).

Finally converting the result into Radians by multiplication with π/180.

5 0
2 years ago
A frequently quoted rule of thumb in aircraft design is that wings should produce about 1000 N of lift per square meter of wing.
svetlana [45]

Answer:

    v₂ = 63.62 m / s

Explanation:

For this exercise in fluid mechanics we will use Bernoulli's equation

         P₁ + ρ g v₁² +  ρ g y₁ = P₂ +  ρ g v₂² +  ρ g y₂

where the subscript 1 refers to the inside of the wing and the subscript 2 to the top of the wing.

We will assume that the distance between the two parts is small, so y₁ = y₂

        P₁-P₂ =  ρ g (v₂² - v₁²)

pressure is defined by

        P = F / A

we substitute

        ΔF / A =  ρ g (v₂² - v₁²)

         v₂² = \frac{\Delta F}{A \ \rho  \ g} + v_1^2

suppose that the area of ​​the wing is A = 1 m²

we substitute

         v₂² = \frac{1000}{1 \ 1.29 \ 9.8} + 63^2

         v₂² = 79.10 + 3969

         v₂ = √4048.1

         v₂ = 63.62 m / s

7 0
3 years ago
An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary
Talja [164]

Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

Explanation:

Part A)

F_{g} = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

T = Tension force in upward direction

F_{d} = Drag force in upward direction = 1800 N

Force equation for the motion of craft is given as

F_{g} - F_{d} - T = 0

7000 - 1800 - T = 0

T = 5200 N

T = 5.2 x 10³ N

Part B)

F_{g} = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

T = Tension force in upward direction

F_{d} = Drag force in downward direction = 1800 N

Force equation for the motion of craft is given as

T  - F_{g} - F_{d} = 0

T - 7000 - 1800  = 0

T = 8800 N

T = 8.8 x 10³ N

4 0
3 years ago
Water is being boiled in an open kettle that has a 0.52-cm-thick circular aluminum bottom with a radius of 12.0 cm. If the water
tangare [24]

Answer:

T_b=107.3784\ ^{\circ}C

Explanation:

Given:

  • thickness of the base of the kettle, dx=0.52\ cm=5.2\times 10^{-3}\ m
  • radius of the base of the kettle, r=0.12\ m
  • temperature of the top surface of the kettle base, T_t=100^{\circ}C
  • rate of heat transfer through the kettle to boil water, \dot Q=0.409\ kg.min^{-1}
  • We have the latent heat vaporization of water, L=2260\times 10^3\ J.kg^{-1}
  • and thermal conductivity of aluminium, k=240\ W.m^{-1}.K^{-1}

<u>So, the heat rate:</u>

\dot Q=\frac{0.409\times 2260000}{60}

\dot Q=15405.67\ W

<u>From the Fourier's law of conduction we have:</u>

\dot Q=k.A.\frac{dT}{dx}

\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}

where:

A= area of the surface through which conduction occurs

T_b= temperature of the bottom surface

15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}

T_b=107.3784\ ^{\circ}C is the temperature of the bottom of the base surface of the kettle.

6 0
3 years ago
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