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Whitepunk [10]
3 years ago
11

The magnetic north pole changes location but the geographic north pole does not true or false

Physics
1 answer:
a_sh-v [17]3 years ago
3 0
The North Magnetic Pole is the point on the surface of Earth's Northern Hemisphere at which the planet's magnetic field points vertically downwards (in other words, if a magnetic compass needle is allowed to rotate about a horizontal axis, it will point straight down). There is only one location where this occurs, near (but distinct from) the Geographic North Pole and the Geomagnetic North Pole. So yes true
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A 1.50x103-kilogram car is traveling east at 30 meters per second.
frez [133]

Answer:

39000\ \text{kg m/s}

West

Explanation:

m = Mass of car = 1.3\times 10^{3}\ \text{kg}

t = Time = 9 seconds

u = Initial velocity = 30 m/s

v = Final velocity = 0

Impulse is given by

J=m(v-u)\\\Rightarrow J=1.3\times 10^3(0-30)\\\Rightarrow J=-39000\ \text{kg m/s}

The magnitude of the total impulse applied to the car to bring it to rest is 39000\ \text{kg m/s}.

The direction is towards west as the sign is negative.

7 0
2 years ago
3. Provide two examples of static electric charge.
Rzqust [24]

Answer: 1.  walking across a carpet and touching a metal door handle            2. pulling your hat off and having your hair stand on end.

Explanation

:)

5 0
3 years ago
A point charge is placed 100 m from a 6 uC charge generating an electric field. What is the
kvv77 [185]

The strength of the electric field is 5 N/C

Explanation:

The magnitude of the electric field produced by a single-point charge is given by:

E=\frac{kQ}{r^2}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

Q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

Q=6 \mu C = 6\cdot 10^{-6}C is the charge producing the field

r = 100 m is the distance from the charge at  which we want to calculate the field

Substituting into the equation, we find the s trength of the electric field:

E=\frac{(8.99\cdot 10^9)(6\cdot 10^{-6})}{(100)^2}=5.4 N/C \sim 5 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

3 0
3 years ago
Two wires each carry 10.0 A of current (in opposite directions) and are 2.50 mm apart. What is the magnetic field 37.0 cm away a
lyudmila [28]

Answer:

see answer below

Explanation:

Before we do any kind of calculation, we need to convert the proper units of the exercise. All the units of distance must be in meters, so, let's change distance of the wire, and the magnetic field to meters:

Separation between the wires are 2.5 mm:

2.5 mm * (1 m / 1000 mm) = 0.0025 m

The distance of P from the bottom of the wires is 37 cm:

37 cm * (1 m/100 cm) = 0.37 m

The distance of P from the top of the wires is just the sum of the two distances:

R = 0.37 + 0.0025 = 0.3725 m

Now that we have the distance, we can determine the magnetic field, using the following expression:

B = B(bottom) - B(top)   or just B₂ - B₁

And B = μ₀ I / 2πR

Replacing in the above expression we have:

B = μ₀ I / 2π ( 1/R₂ - 1/R₁)

Now we can determine the magnetic field:

B = (4πx10⁻⁷ * 10 / 2π) (1/0.37 - 1/0.3725)

<h2>B = 3.63x10⁻⁸ T</h2><h2></h2>

Which means that the magnetic field is out of the page.

Hope this helps

4 0
3 years ago
Can someone pls help ASAP:(
diamong [38]
I’m not sure but I think it’s
△ m=5 and △= -3 and so

Answer: 5/△-3 m/s

So sorry if it’s wrong
6 0
3 years ago
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