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2 years ago

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

1 answer:

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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Give an example in which a small force exerts a large torque. give another example in which a large force exerts a small torque.

goldenfox [79]

Since the torque involves the product of force times lever arm, a small force can exert a greater torque than a larger force if the small force has a large enough lever arm.

With a large force exerts a small torque is a gate, hinged in its vertical line (axis). When pushed from a point near to the hinge, a very large amount is needed to open the gate.

3 0

3 years ago

Read 2 more answers

The greater the of an object the more force is needed to cause acceleration

d1i1m1o1n [39]

the greater the mass of an object the more force is needed to cause acceleration

8 0

3 years ago

Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its

kkurt [141]

Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

$\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2}$ ,

Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .

Initial flux: $\Phi(0)= 0$ .

Final flux: $\Phi(0.7) = \rm 1.1136\times 10^{-4}\; Wb$ .

Average EMF, which is the same as the average rate of change in flux:

$\displaystyle \frac{\Phi(0.7) - \Phi(0)}{0.7} \approx\rm 1.62\times 10^{-4}\; V$ .

8 0

3 years ago

Can someone help with my physics homework? please

Murrr4er [49]

Answer:

a) 19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options: (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) 24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

Explanation:

a) If force applied in the cart is uniform, that is, constant in magnitude and direction and is parallel to distance travelled by the car, the work done on the cart is defined by the following equation:

$W = F\cdot \Delta s$ (1)

Where:

$F$ - Force applied by the cart, measured in newtons.

$\Delta s$ - Distance travelled by the car, measured in meters.

$W$ - Work done on the cart, measured in joules.

If we know that $F = 888\,N$ and $\Delta s = 22\,m$ , then the work done on the cart is:

$W =(888\,N)\cdot (22\,m)$

$W = 19536\,J$

19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

d) If we know that $F = 1111\,N$ and $\Delta s = 22\,m$ , then the work on the cart is:

$W = (1111\,N)\cdot (22\,m)$

$W = 24442\,N$

24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The gas consumption is directly proportional to the square of velocity and mass of the cart and, hence, to the work done on the cart. In consequence, we conclude that the bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

3 0

3 years ago

A proton of mass m is at rest when it is suddenly struck head-on by an alpha particle (which consistsof 2 protons and 2 neutrons

antiseptic1488 [7]

Answer:

3/5 v

Explanation:

The computation of speed will the alpha particle have after the collision is shown below:-

In a perfectly elastic the kinetic energy and collision the momentum are considered.

The velocity of the particles defines the below equation:

$VA_f=(\frac{m_A-m_B}{m_A+m_B})VA_i+(\frac{2m_B}{m_A+m_B})VB_i$