PLEASE HELP WORTH 30 POINTS!!!

The area vector of a square loop of 5 turns of a conductor each with side length of 0.2 m carrying a current of 2 A is antiparal

lord [1]

Answer:

the magnitude and the direction of the total magnetic field is 0.4 Am² antiparallel to the area vector

Explanation:

Given that:

The area vector of a square loop has 5 numbers of turns i.e n = 5

each with side length = 0.2 m

Current I = 2 A

uniform magnetic field = 50.0 T

Now; the magnitude of the total magnetic field B is calculated as :

B = IA

where;

I = current

A = area ( n × l²)

B = I ( n × l²)

B = 2 × 5 × 0.2²

B = 0.4 Am²

The direction of the magnetic moment is antiparallel to the area vector;

Hence ; the magnitude and the direction of the total magnetic field is 0.4 Am² antiparallel to the area vector

3 0

3 years ago

A particle moves in a straight line and has acceleration given by a(t) = 12t + 10. Its initial velocity is v(0) = −5 cm/s and it

soldi70 [24.7K]

Answer:

The position function is $s_{t}=2t^3+5t^2-5t+9$ .

Explanation:

Given that,

Acceleration $a =12t+10$

Initial velocity $v_{0} = -5\ cm/s$

Initial displacement $s_{0}=9\ cm$

We know that,

The acceleration is the rate of change of velocity of the particle.

$a = \dfrac{dv}{dt}$

The velocity is the rate of change of position of the particle

$v=\dfrac{dx}{dt}$

We need to calculate the the position

The acceleration is

$a_{t} = 12t+10$

$\dfrac{dv}{dt} = 12t+10$

$a_{t}=dv=(12t+10)dt$

On integration both side

$\int{dv}=\int{(12t+10)}dt$

$v_{t}=6t^2+10t+C$

At t = 0

$v_{0}=0+0+C$

$C=-5$

Now, On integration again both side

$v_{t}=\int{ds_{t}}=\int{(6t^2+10t-5)}dt$

$s_{t}=2t^{3}+5t^2-5t+C$

At t = 0

$s_{0}=0+0+0+C$

$C=9$

$s_{t}=2t^3+5t^2-5t+9$

Hence, The position function is $s_{t}=2t^3+5t^2-5t+9$ .

7 0

3 years ago

A 330 kg piano slides 3.6 m down a 28o incline and is kept from accelerating by a man who is pushing back on it parallel to the

e-lub [12.9K]

Answer:

a. 652.68N

b. -2349.65J

c. -3116.12J

d. 5465.77J

e. Zero

Explanation:

a. According to equilibrium of forces, the force of gravity is equal to the sum of the frictional force and force exerted by the man in the opposite direction (since they're both resistant forces).

Fg = Fm + Fr

Fm = Fg - Fr

Fm = mgsin(28°) - umgcos(28°)

u = coefficient of frictional force.

Fm = 330*9.8*sin28 - 0.4*330*9.8*cos28

Fm = 1518.27 - 865.59

Fm = 652.68N

b. Work done by man is:

Wm = -Fm * d

Wm = -652.68 * 3.6

Wm = -2349.65J

c. Work done by friction force:

W(Fr) = -Fr * d

W(Fr) = -865.59 * 3.6

W(Fr) = -3116.12J

d. Work done by gravity:

Wg = Fg * d

Wg = 1518.27 * 3. 6

Wg = 5465.77J

e. Net work done on the piano is:

Work done by friction + work done by gravity + work done by man

= -3116.12 + 5464.77 + (-2349.65)

= 0J

7 0

3 years ago

Since this object is traveling at terminal velocity, what must be the value of Fd? (HINT: how to the lengths of both arrows comp

Juli2301 [7.4K]

The magnitude of the air drag when the object is traveling at terminal velocity is C. 850 Newtons

Explanation:

There are only two forces acting on the object here:

- The force of gravity, of magnitude $F_g = 850 N$ , acting downward

- The air drag, $F_d$ , acting upward

Therefore, the equation of motion for the object is

$F_g - F_d = ma$

where m is the mass of the object and a its acceleration.

The object in this problem is traveling at terminal velocity: this means that the acceleration is zero, so

a = 0

Therefore the equation becomes

$F_d = F_g$

which means that the magnitude of the air resistance is equal to the magnitude of the force of gravity:

$F_d = 850 N$

Learn more about acceleration and Newton laws of motion here:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

8 0

4 years ago

A 4,000 kg boat floats with one-third of its volume submerged. if two more people get into the boat, each of whom weighs 690 n,

AveGali [126]

The weightiness of the added water displaced is equivalent to the joined weight of the two extra people who come to be into the boat:

m water g = 2 x 690 N

= 1,380 N

The mass of the water displace is then

m water g = 1,380 N

= 1,380 N / 9.8 m/s^2

= 141 kg

Compute the calculation for density for the volume of water displace and practice this outcome for the mass of the water displace to get the answer:

p water = mass of water / volume of water

volume of water = mass of water / p water

= 141 kg / 1000 kg /m^3 eliminate kilogram

= 0.14 m^3 the additional volume of water that is displaced

4 0

4 years ago