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2 years ago

Consider a loop of wire placed in a uniform magnetic field. Which factors affect the magnetic flux Φm through the loop?

Physics

1 answer:

dimaraw [331]2 years ago

4 0

Answer:

* The value of the magnetic field changes either in time or space

* The waxed area changes, the bow is fitting in size

* The angle between the field and the area changes

Explanation:

Magnetic flux is the scalar product of the magnetic field over the area

Ф = ∫ B. dA

where B is the magnetic field and A is the area

Let's look at stationary, for which factors affect flow

* The value of the magnetic field changes either in time or space

* The waxed area changes, the bow is fitting in size

* The angle between the field and the area changes

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A car of mass 1100kg moves at 24 m/s. What is the braking force needed to bring the car to a halt in 2.0 seconds? N

LenaWriter [7]

13200N

Explanation:

Given parameters:

Mass = 1100kg

Velocity = 24m/s

time = 2s

unknown:

Braking force = ?

Solution:

The braking force is the force needed to stop the car from moving.

Force = ma = $\frac{mv}{t}$

m is the mass of the car

v is the velocity

t is the time taken

Force = $\frac{1100 x 24}{2}$ = 13200N

Learn more:

Force brainly.com/question/4033012

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1 year ago

A jogger runs 10.0 blocks do east, 5.0 blocks due South, and another two. Zero blocks do east. Assume all blocks are equal size,

Annette [7]

Jogger moves in three displacements

d1 = 10 blocks East

d2 = 5 blocks South

d3 = 2 blocks East

now we can say

total displacement towards East direction will be

$d_x = 10 + 2= 12 blocks$

Total displacement towards South

$d_y = 5 block$

now to find the net displacement we can use vector addition

$d = \sqrt{d_x^2 + d_y^2}$

$d = \sqrt{12^2 + 5^2}$

$d = 13 blocks$

<em>so magnitude of net displacement will be equal to 13 blocks</em>

6 0

1 year ago

A professor's office door is 0.99 m wide, 2.2 m high, 4.2 cm thick; has a mass of 27 kg, and pivots on frictionless hinges. A "d

ANEK [815]

Answer:

$I=8.8209\ kg.m^2$

$\alpha=0.6348\ rad.s^{-2}$

Explanation:

Given:

width of door, $w=0.99\ m$

height of the door, $h=2.2\ m$

thickness of the door, $t=4.2\ cm$

mass of the door, $m=27\ kg$

torque on the door, $\tau=5.6\ N.m$

<em>∵Since the thickness of the door is very less as compared to its other dimensions, therefore we treat it as a rectangular sheet.</em>

For a rectangular sheet we have the mass moment of inertia inertia as:

$I=\frac{1}{3} m.w^2$

$I=\frac{1}{3}\times 27\times 0.99^2$

$I=8.8209\ kg.m^2$

We have a relation between mass moment of inertia, torque and angular acceleration as:

$\alpha=\frac{\tau}{I}$

$\alpha=\frac{5.6}{8.8209}$

$\alpha=0.6348\ rad.s^{-2}$

6 0

1 year ago

What is the value of work done on an object when a 0.1x102-newton force moves it 30 meters and the angle between the force and t

disa [49]

Answer:306joules

Explanation:

3 0

2 years ago

A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit

makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

2.87 = -1.442 + 1.8 u

u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0

2 years ago