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adelina 88 [10]
3 years ago
15

An electromagnet produces a magnetic eld of 0.95 T in a cylindrical region of radius 3.0 cm between its poles. A straight wire c

arrying a current of 15.0 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic eld. What magnitude of force is exerted on the wire?
Physics
1 answer:
Daniel [21]3 years ago
6 0

Answer:

0.855 N

Explanation:

Using

F = BILsinФ.................... Equation 1

Where F = magnitude of the force exerted on the wire, B = magnetic Field, I = current, L = Length of the wire in the magnetic Field, Ф = angle between the wire and the magnetic field.

Given: B = 0.95 T, I = 15.0 A, L = 2r, r = radius = 3.0 cm = 0.03 m L = 2×0.03 = 0.06 m, Ф = 90° (perpendicular)

Substitute into equation 1

F = 0.95(15)(0.06)sin90°

F = 0.855 N.

Hence the magnitude of force that is exerted on the wire = 0.855 N

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A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long a
IrinaK [193]

The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

<h3>What's the expression of range of a projectile motion?</h3>
  • Range = U²× sin(2θ)/g
  • U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
  • U=√{Range×g/sin(2θ)}
  • Here, range= 2.20m, = 36.5°
  • U= √{2.20×9.8/sin(73)}

U= √{2.20×9.8/sin(73)} = 22.5m/s

<h3>What's the expression of time of flight in projectile motion?</h3>
  • Time of flight= (2×U×sinθ)/g
  • So, T= (2×22.5×sin36.5°)/9.8

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

Learn more about the range and time period of projectile motion here:

brainly.com/question/24136952

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4 0
2 years ago
15.One girl was involved in a bus crash. She told the police that she was seated in the last line of seats and that when the bus
Alona [7]

Answer: It is not likely.

Explanation:

When the bus is moving forward, all the objects inside of it also are moving forward.

Now, as the objects inside the buss are not fixed to the bus, if the bus suddenly stops the objects inside of it will keep moving forward, because of the conservation of the momentum, defined as the quantity of motion (Similar to when you are in a car and it suddenly stops, you can feel the forward impulse).

Then is not likely that, in a case where the bus stops suddenly, an object inside the bus flies backward in opposite direction to the previous movement of the bus.

5 0
3 years ago
A. 24.89<br> B. 25.89<br> C. 17.74<br> D. 19.73
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Answer: D

Explanation:

Just did it got an 100

5 0
2 years ago
If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a
PolarNik [594]

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s w_o = 12 rad/s

Angular Velocity at time 0.15s w_f = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( w_f - w_o ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

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3 years ago
Two spaceships leave Earth in opposite directions, each with a speed of 0.60c with respect to Earth. (a) What is the SPEED of sp
Studentka2010 [4]

Answer:

The relative speed of 1 relative to 2 is 0.88c

Explanation:

In relativistic mechanics the relative speed between 2 objects moving in different direction is given by

v_{ab}=\frac{v_{a}+v_{b}}{1+\frac{v_{a}v_{b}}{c^{2}}}

Since it is given that

v_{a}=0.6c\\\\v_{b}=0.6c

Applying values in the formula we get

v_{ab}=\frac{0.6c+0.6c}{1+\frac{(0.6c)^{2}}{c^{2}}}\\\\v_{ab}=0.88c

8 0
3 years ago
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