What causes a compass needle to point to geographic north?

A rectangular current loop is located near a long straight wire that carries a current of 12A. The current in the loop is 5A. De

masya89 [10]

Explanation:

The net magnetic force acting on the loop is the superposition of the forces on each side of the loop. The forces on the two sides perpendicular to the long straight wire will cancel (right hand rule). The magnitude of the magnetic force between two wires with currents I_1 = 12 A and I_2 = 5 A of length L = 0.5 m ( assuming not given in question).

$F = \frac{\mu_o I_1 I_2 L}{2 \pi d}$

so the forces on the near and far wires are

$F_n = 2.7\times 10^{-4} N$

$F_f = 1.2\times 10^{-4} N$

The force F_n is an attractive one, while F_f is a repulsive one (right hand rule). The magnitude of the force is therefore

F = F_n - F_f

$F=1.5\times10^{-4}$

8 0

3 years ago

Which is the BEST explanation for why the speed of sound changes for solids, liquids, and gases?

garik1379 [7]

the reason why sound travels faster in solid is because the molecules in solid is more close to each other and they allow the waves of the sound to travel quickly or easily. and same reason why the sound travels fast in liquid that gas, it's because the molecules in liquid is closer to each other than the molecules in gas but they're not as close as solid hence why solid is the fastest at traveling sound

5 0

3 years ago

Can anyone solve this?

love history [14]

Answer:

F = 39.2 N

Explanation:

Since, the object is in uniform motion. Therefore, the frictional force on object will be:

Frictional Force = μk N = μk mg

where,

μk = coefficient of kinetic friction = 0.2

m = mass of crate = 10 kg

g = 9.8 m/s²

Therefore,

Frictional Force = (0.2)(10 kg)(9.8 m/s²)

Frictional Force = 19.6 N

The horizontal component of force must be equal to this frictional force to continue the uniform motion:

F Sin 30° = 19.6 N

F = 19.6 N/Sin 30°

3 0

3 years ago

A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for

nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

And the pull force is:

$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds of the block, measured in meters per second.

$\Sigma F$ - Horizontal net force, measured in newtons.

$\Delta t$ - Impact time, measured in seconds.

Now we clear the final speed in (2):

$v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $m = 14\,kg$ , $\Sigma F = 98\,N$ and $\Delta t = 1.40\,s$ , then final speed of the ice block is:

$v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}$

$v_{f} = 9.8\,\frac{m}{s}$

The final speed of the block of ice is 9.8 meters per second.

6 0

3 years ago

a lorry travels 3600m on a test track accelerating constantly at 3m/s squared from standstill. what is the final velocity (3 sig

suter [353]

The final velocity of the truck is found as 146.969 m/s.

Explanation:

As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.

Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.

$2as =v^{2} - u^{2} \\\\v^{2} = 2as + u^{2}$

Then, the final velocity will be

$v^{2} = 2 * 3 * 3600 + 0 = 21600\\ \\v=\sqrt{21600}=146.969 m/s$

Thus, the final velocity of the truck is found as 146.969 m/s.

8 0

3 years ago