1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Bad White [126]
3 years ago
15

Suppose you pull two opposite sides of the loop away from each other, such that the other two sides get closer, reducing the loo

p area. Suppose as you look down on the loop, the magnetic field also points downward through the loop. Which direction (clockwise or counterclockwise) does the induced current flow, from your perspective?
Physics
1 answer:
HACTEHA [7]3 years ago
6 0

Answer:

Explanation:

Magnetic field is oriented downwards . area of loop is decreasing so downward flux is decreasing . To increase the downward flux , current induced will be clockwise so that flux generated by induced current increases the downwards flux . It is in accordance with Len'z law.

You might be interested in
1. According to paragraph 3 in the text, MOST of the electromagnetic waves from
Fed [463]

Answer:

Most of the EM waves from the sun that reach Earth are infrared waves, visible light, and UV radiation.

Explanation:

I hope this helps! Have a good day!

5 0
3 years ago
Which of the following is a way in which earth material is returned to the surface?
kvv77 [185]

Answer:

Explanation:

soil

4 0
3 years ago
Read 2 more answers
At an air show, a stunt pilot performs a vertical loop-the-loop in a circle of radius 3.63 x 103 m. During this performance the
san4es73 [151]

Answer:

189 m/s

Explanation:

The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.

So, F = W

mv²/r = mg

v² = gr

v = √gr where  v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m

So, v = √gr

v = √(9.8 m/s² × 3.63 × 10³ m)

v = √(35.574 × 10³ m²/s²)

v = √(3.5574 × 10⁴ m²/s²)

v = 1.89 × 10² m/s

v = 189 m/s

5 0
3 years ago
Collar P slides outward at a constant relative speed along rod AB, which rotates counterclockwise with a constant angular veloci
diamong [38]

Answer:

a= 23.65 ft/s²

Explanation:

given

r= 14.34m

ω=3.65rad/s

Ф=Ф₀ + ωt

t = Ф - Ф₀/ω

= (98-0)×\frac{\pi}{180}/3.65

98°= 1.71042 rad

1.7104/3.65

t= 0.47 s

r₁(not given)

assuming r₁ =20 in

r₁ = r₀ + ut(uniform motion)

u = r₁ - r₀/t

r₀ = 14.34 in= 1.195 ft

r₁ = 20 in = 1.67 ft

= (1.667 - 1.195)/0.47

0.472/0.47

u= 1.00ft/s

acceleration at collar p

a=rω²

= 1.67 × 3.65²

a = 22.25ft/s²

acceleration of collar p related to the rod = 0

coriolis acceleration = 2ωu

= 2× 3.65×1 = 7.3 ft/s²

acceleration of collar p

= 22.5j + 0 + 7.3i

√(22.5² + 7.3²)

the magnitude of the acceleration of the collar P just as it reaches B in ft/s²

a= 23.65 ft/s²

4 0
3 years ago
A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal
Sati [7]

(a) The ball has a final velocity vector

\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j

with horizontal and vertical components, respectively,

v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}

v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}

The horizontal component of the ball's velocity is constant throughout its trajectory, so v_{x,i}=v_{x,f}, and the horizontal distance <em>x</em> that it covers after time <em>t</em> is

x=v_{x,i}t=v_{x,f}t

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s

The vertical component of the ball's velocity at time <em>t</em> is

v_{y,f}=v_{y,i}-gt

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}

So, the initial velocity vector is

\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j

which carries an initial speed of

\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}

and direction <em>θ</em> such that

\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height <em>y</em> at time <em>t</em> is

y=v_{y,i}t-\dfrac12gt^2

so that when it lands in the seats at <em>t</em> ≈ 6.38 s, it has a height of

y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}

6 0
3 years ago
Other questions:
  • Spot welding is used to fuse two sheets of metal together at one small spot. Two copper electrodes pinch the sheets together at
    9·1 answer
  • How does a cockcroft walton generator work?
    14·1 answer
  • According to the Second Law of Thermodynamics, a. any process during which the entropy of the universe increases will be product
    5·1 answer
  • What is in between the nucleus and the electrons in an atom?
    13·1 answer
  • all of the following methods are ways to correct a run-on sentence except ,placing a semicolon between the compete sentences.put
    9·1 answer
  • Can someone please help me on thisss
    13·1 answer
  • Two spherical objects have equal masses and
    15·1 answer
  • Which is a “big idea” for matter and change?
    11·1 answer
  • Find the acceleration of the blocks in figure if friction forces are negligible. What is the tension in the cord connecting them
    13·1 answer
  • which relationship best represents the relationship between the magnitude of the centripetal acceleration and the speed of an ob
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!