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3 years ago

The angle between magnetic north and the north to which a compass needle points is known as magnetic declination.

1 answer:

7 0

From among the choices provided, the better choice is the upper-case letter '<em>T </em>'. That symbol can conveniently be used to represent the words "true" or "truth", which is exactly the reason that it is the better choice for a response, since the complicated statement at the beginning of the question is completely true in its every detail, nuance, jot and tittle.

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You pull a suitcase along the floor by exerting 43N at an angle. The force of friction is 27 N and the suitcase moves at a const

Mumz [18]

The angle of the force is $51.1^{\circ}$

Explanation:

To solve this problem, we can apply Newton's second law along the horizontal direction of motion of the suitcase:

$\sum F_x = ma_x$

where

$\sum F_x$ is the net force along the x-axis

m is the mass of the suitcase

$a_x$ is the acceleration along the x-axis

The suitcase is moving at constant speed, so the acceleration is zero:

$a_x=0$

Therefore the net force must also be zero:

$\sum F_x = 0$ (1)

We have two forces acting along the horizontal direction:

- The component of the push (forward) in the horizontal direction, $F cos \theta$ , with

F = 43 N

$\theta$ = angle of the force with the horizontal

- The force of friction, $F_f = 27 N$ , backward

So the net force can be written as

$\sum F_x = F cos \theta - F_f$ (2)

Combining (1) and (2),

$F cos \theta - F_f = 0$

And so we can find the angle:

$\theta = cos^{-1}(\frac{F_f}{F})=cos^{-1}(\frac{27}{43})=51.1^{\circ}$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

8 0

2 years ago

on 28 may _______ at Chaghi, Balochistan; Dr A.Q Khan commanded to perform six successful nuclear explsions.

KATRIN_1 [288]

I think the answer would be 1998

8 0

2 years ago

What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2 A

Nostrana [21]

Complete Question

A very long, straight solenoid with a cross-sectional area of 2.39cm2 is wound with 85.7 turns of wire per centimeter. Starting at t= 0, the current in the solenoid is increasing according to i(t)=( 0.162A/s2) t2. A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A ?

Answer:

The value is $\epsilon = 1.83 *10^{-5} \ V$

Explanation:

From the question we are told that

The cross-sectional area is $A = 2.39 \ cm^2 = \frac{2.39}{10000} = 0.000239 \ m^2$

The number of turns is $N = 85.7 \ turns/cm = 8570 \ turns / m$

The initial time is t = 0s

The current on the solenoid is $I(t) = (0.162 \ A/s^2) t^2$

The number of turns of the secondary winding is $n = 5 \ turns$

Generally At I = 3.2 A

$3.2 = (0.162 )t^2$

=> $t^2 = 19.8$

=> $t = 4.4 \ s$

Generally induced emf is mathematically represented as

$\epsilon = A * \mu_o * n * N \frac{d(I)}{dt}$

$\epsilon = 0.000239 * 4\pi * 10^{-7} * 8570 * 5 * (0.162) * 2t$

$\epsilon = 0.000239 * 4\pi * 10^{-7} * 8570 * 5 * (0.162) * 2 * 4.4$

$\epsilon = 1.83 *10^{-5} \ V$

3 0

2 years ago

When you boil water in a pot on the stove, the metal pot heats up faster than the water because the pot

nika2105 [10]

Is a conductor of the heat. Hope this helps!

4 0

2 years ago

Read 2 more answers

You compress a spring by a distance of 0.2 m. The spring has a spring constant of 37 N/m. When you release the spring, it snaps

Serggg [28]

At that point it is no longer trying to uncompress nor is it trying to stretch. This is the same thing as a pendulum at the bottom of its swing, no longer falling but not yet rising against gravity. Thus the kinetic energy there is the same as the potential energy when it is compressed. The energy of compression is
$\frac{1}{2}k x^{2}$
This gives E=0.5(37)(0.2)²=0.74J
This is the same as the kinetic energy when it is at natural length

7 0

3 years ago

Read 2 more answers