HELP NOW PLEASE ANSWER THE QUESTION THE PICTURE IS ATTACHED BELOW

The plane of a5cm*8cm rectangular loop of wire is parallel to a 0.19T magnetic field the loop carries a current of 6.2 A. What t

Sedaia [141]

Answer:

Torque; τ = 4.712 × 10^(-3) J

Magnetic moment; M = 0.0248 J/T

Explanation:

Torque is gotten from the formula;

τ = BIA

Where;

B is magnetic field

I is current

A is area

We are given;

B = 0.19T

I = 6.2A

Rectangle dimensions = 5cm by 8cm = 0.05m by 0.08m

Thus;

Area; A = 0.05m × 0.08m = 0.004 m²

Thus;

τ = 0.19 × 6.2 × 0.004

τ = 4.712 × 10^(-3) J

Formula for the magnetic moment is given by;

M = IA

M = 6.2 × 0.004

M = 0.0248 J/T

5 0

2 years ago

Sarah, who has a mass of 55 kg, is riding in a car at 20 m/s. She sees a cat crossing the street and slams on the brakes! Her se

avanturin [10]

Answer:

-2200 N

Explanation:

The change in momentum of Sarah is equal to the impulse, which is the product between the force exerted by the seatbelt on Sarah and the time during which the force is applied:

$\Delta p=I\\m \Delta v = F \Delta t$

where

m is the mass

$\Delta v$ is the change in velocity

F is the average force

$\Delta t$ is the duration of the collision

In this problem:, we have:

m = 55 kg is Sarah's mass

$\Delta v = 0-20 = -20 m/s$ is the change in velocity

$\Delta t = 0.5 s$ is the duration of the collision

Solving for F, we find the force exerted by the seatbelt on Sarah:

$F=\frac{m\Delta v}{\Delta t}=\frac{(55)(-20)}{0.5}=-2200 N$

Where the negative sign indicates that the direction of the force is opposite to that of Sarah's initial velocity.

5 0

3 years ago

A 2.1 ✕ 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an avera

Lina20 [59]

Answer:

3.9 m/s

Explanation:

We are given that

Mass of car,m= $2.1\times 10^3 kg$

Initial velocity,u=0

Distance,s=5.9 m

$\theta=19^{\circ}$

Average friction force,f= $4.0\times 10^3 N$

We have to find the speed of the car at the bottom of the driveway.

Net force, $F_{net}=mgsin\theta-f=2.1\times 10^3\times 9.8sin19-4.0\times 10^3$

Where $g=9.8 m/s^2$

Acceleration, $a=\frac{F_{net}}{m}=\frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}$

$v=\sqrt{2as}$

$v=\sqrt{2\times \frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}\times 5.9}$

v=3.9 m/s

7 0

3 years ago

A 1280 kg car is moving 4.92 m/s. a 509 N force then pushes it forward for 28.7 m. what is its final KE?(unit=J)PLEASE HELP

laila [671]

Answer:

The answer to your question is: total energy = 30100.4 J

Explanation:

Kinetic energy (KE) is the energy due to the movement of and object, its units are joules (J)

Data

mass = 1280 kg

speed = 4.92 m/s

Force = 509 N

distance = 28.7 m

Formula

$KE = \frac{1}{2} mv^{2}$

Work = Fd

Process

- Calculate Kinetic energy

- Calculate work

- Add both results

KE = $\frac{1}{2} (1280)(4.92)^{2}$

KE = 15492.1 J

Work = (509)(28.7)

Work = 14608.3 J

Total = 15492.1 + 14608.3

Total energy = 30100.4 J

8 0

2 years ago

Read 2 more answers

A space shuttle takes off from FL and circles Earth several times, finally landing in CA. While the shuttle is in flight, a phot

mixer [17]

Answer:

Both the astronauts and photographer have the same displacement

Explanation:

Displacement is the minimum distance between two point. The initial point of both the astronauts and the photographer was Florida and the final point was California. So, the minimum distance for both of the astronauts and the photographer would be the distance between Florida and California would be the same.

Hence, both the astronauts and photographer will have the same displacement.

3 0

3 years ago