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andrew11 [14]
3 years ago
15

A rectangular loop of wire carries current I in the clockwise direction. The loop is in a uniform magnetic field B that is paral

lel to the plane of the loop, in the direction toward the bottom of the page. The length of the rectangle is b and the width is a. What is the net force on the loop by the magnetic field
Physics
1 answer:
Eva8 [605]3 years ago
3 0

Answer:

Explanation:

Area of the loop = a b

current = I

magnetic moment of the loop M  = area x current

= ab I

Torque on the loop = MB sinθ

here θ = 90

Torque = MB

= abIB

In this case net force on the loop will be zero because here torque is created by two equal and opposite force acting on two opposite sides of the loop so net force will be zero .

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In a little league baseball game, the 145 g ball enters the strike zone with a speed of 14.0m/s . the batter hits the ball, and
Novosadov [1.4K]

Answer:

5365 N

Explanation:

v = Final velocity = 23 m/s

u = Initial velocity = -14 m/s (opposite direction)

m = Mass of ball = 145 g

t = Time taken = 1 ms

Impulse is given by

J=m(v-u)

Impulse is also given by

J=Ft

Ft=m(v-u)\\\Rightarrow F=\dfrac{m(v-u)}{t}\\\Rightarrow F=\dfrac{0.145\times (23-(-14))}{1\times 10^{-3}}\\\Rightarrow F=5365\ N

The magnitude of the average force exerted by the bat on the ball is 5365 N

8 0
3 years ago
You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. Yo
Vitek1552 [10]

Answer:

λ = 5.734 x 10⁻⁷ m = 573.4 nm

Explanation:

The formula of the Young's Double Slit experiment is given as follows:

\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta x d}{L}

where,

λ = wavelength = ?

L = distance between screen and slits = 8.61 m

d = slit spacing = 1.09 mm = 0.00109 m

Δx = distance between consecutive bright fringes = \frac{4.53\ cm}{10} = 0.00453 m

Therefore,

\lambda = \frac{(0.00453\ m)(0.00109\ m)}{8.61\ m}

<u>λ = 5.734 x 10⁻⁷ m = 573.4 nm</u>

6 0
3 years ago
An eagle is flying horizontally at a speed of 3.10 m/s when the fish in her talons wiggles loose and falls into the lake 6.10 m
Alinara [238K]

Answer:

10.93m/s with the assumption that the water in the lake is still (the water has a speed of zero)

Explanation:

The velocity of the fish relative to the water when it hits the water surface is equal to the resultant velocity between the fish and the water when it hits it.

The fish drops on the water surface vertically with a vertical velocity v. Nothing was said about the velocity of the water, hence we can safely assume that the velocity if the water in the lake is zero, meaning that it is still. Therefore the relative velocity becomes equal to the velocity v with which the fish strikes the water surface.

We use the first equation of motion for a free-falling body to obtain v as follows;

v = u + gt....................(1)

where g is acceleration due to gravity taken as 9.8m/s/s

It should also be noted that the horizontal and vertical components of the motion are independent of each other, hence we take u = 0 as the fish falls vertically.

To obtain t, we use the second equation of motion as stated;

h=ut+gt^2/2.................(2)

Given; h = 6.10m.

since u = 0 for the vertical motion;  equation (2) can be written as follows;

h=\frac{1}{2}gt^2............(3)

substituting;

6.1=\frac{1}{2}*9.8*t^2\\6.1=4.9t^2\\hence\\t^2=6.1/4.9\\t^2=1.24\\t=\sqrt{1.24}=1.12s

Putting this value of t in equation (1) we obtain the following;

v = 0 + 9.8*1.12

v = 10.93m/s

5 0
3 years ago
What is a sentence for allele
slavikrds [6]
A different form of gene is known as an allele.
5 0
3 years ago
Read 2 more answers
A stone is thrown horizontally from 2.4m above the ground at 35m/s. The wall is 14m away and 1m high.At what height the stone wi
KIM [24]

The stone reaches the wall at a height of <u>1.62 m</u>.

The stone lands at a point <u>24.5 m</u> from the point of projection.

The stone is projected horizontally with a velocity u at a height <em>h</em> from the ground. The wall is located at a distance <em>x</em> from the point of projection. The stone takes a time <em>t</em> to reach the wall and in the same time the stone falls a vertical distance <em>y</em>.

The horizontal distance <em>x</em> is traveled with a constant velocity <em>u</em>.

x=ut

Calculate the time taken <em>t</em>.

t=\frac{x}{u} \\ =\frac{14m}{35 m/s} \\ =0.40s

The stone's initial vertical velocity is zero. It falls through a distance <em>y</em> in the time <em>t</em> under the action of acceleration due to gravity <em>g</em>.

y=\frac{1}{2} gt^2\\ \frac{1}{2} (9.81m/s^2)(0.40s)^2\\ =0.784m

The height  <em>h₁ </em>of the stone above the ground when it reaches the wall  is given by,

h_1=h-y\\ =(2.4m)-(0.784m)\\ =1.616m=1.62m

When the stone reaches the wall, its height from the ground is <u>1.62m.</u>

The stone thus crosses over the wall, since the height of the wall is 1 m. It reaches the ground at a distance <em>R</em> from the point of projection. If the time taken by the stone to reach the ground is <em>t₁, </em>then,

h=\frac{1}{2} gt_1^2

Calculate the time taken by the stone to reach the ground.

t_1=\sqrt{\frac{2h}{g} } \\=\sqrt{\frac{2(2.4m)}{9.81m/s^2} } \\ =0.699 s

The horizontal distance traveled by the stone is given by,

R=ut_1 \\ =(35m/s)(0.699s)\\ =24.5m

The stone lands at point 24.5 m from the point of projection and 10.5 m from the wall.

3 0
3 years ago
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