Ask question

Ask question

All categories

3 years ago

14

The electric field in a region of space increases from 0 0 to 3450 N/C 3450 N/C in 4.40 s. 4.40 s. What is the magnitude of the

induced magnetic field B B around a circular area with a diameter of 0.460 m 0.460 m oriented perpendicularly to the electric field?

1 answer:

slavikrds [6]3 years ago

7 0

Answer: B = 1380T

Explanation: please find the attached file for the solution

You might be interested in

Un móvil se desplaza con una rapidez inicial de 100 km/h, se le aplican los frenos con una

KIM [24]

Answer:

in English please I am quite puzzled

4 0

3 years ago

A nasa spacecraft measures the rate r of at which atmospheric pressure on mars decreases with altitude. the result at a certain

Lesechka [4]

Answer: $4.21 \times 10^{-10} J/cm^4$

$1 kPa= 10^3 Pa$

$1 km=10^5 cm$

$1kPa/km=0.01 Pa/cm$

$1kPa/km=10^{-8} J/cm^4$

$\Rightarrow r= 0.0421 kPa/km= 0.0421 kPa/km \times \frac{10^{-8} J/cm^4}{1 kPa/km}= 0.0421 \times 10^{-8}J/cm^4=4.21 \times 10^{-10} J/cm^4$

3 0

3 years ago

Read 2 more answers

A stone is dropped into a well. The sound of the splash is heard 6.25 s later. What is the depth of the well? (Take the speed of

Naya [18.7K]

Answer:

depth of well is 163.30 m

Explanation:

Given data

speed of sound = 343 m/s

timer = 6.25 s

to find out

depth of well

solution

let us consider depth d

so equation will be

depth = 1/2 ×g ×t² ..............1

and

depth = velocity of sound × time .................2

here we have given time 6.25 that is sum of 2 time

when stone reach at bottom that time

another is sound reach us after stone strike on bottom

so time 1 + time 2 = 6.25 s

so from equation 1 and 2 we get

1/2 ×g ×t² = velocity of sound × time

1/2 ×9.8 × t1² = 343 × (6.25 - t1 )

t1 = 5.77376 sec

so height = 1/2 ×g ×t²

height = 1/2 ×9.8 × (5.773)²

height = 163.30 m

3 0

3 years ago

Which of the following is a limited resource, which can take longer to be replenished than it takes to be used up?

Zarrin [17]

What is the list of choices?

8 0

2 years ago

Read 2 more answers

[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r

blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b) it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity $V_{f}$ for an object traveling distance h taking the initial vertical component of velocity as $V_{i}$ the kinematics equation is written as

$V_{f}^{2}=V_{i}^{2}+2ah$ where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

$V_{f}^{2}=V_{i}^{2}+2gh$

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

$V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}$ = 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

$V_{f}^{2}=V_{i}^{2}+2gh$

Since we have final velocity of 25 m/s

$V_{i}^{2}=2gh-V_{f}^{2}$

$V_{i}=\sqrt{(V_{f}^{2}-2gh)}$

$V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}$ = 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0

3 years ago