Answer
when there are ten they don't grow so well but when there is less than 10 they tend to grow
The field strength needed to produce a 24.0 V peak emf is 0.73T.
To find the answer, we need to know about the expression of emf.
What's the expression of peak emf produced in a rotating rectangular loops?
- The peak emf produced in a rotating loops= N×B×A×w
- N= no. of turns of the loop, B= magnetic field, A= area of loop and w= angular frequency
- So, B = emf/(N×A×w)
<h3>What's the magnetic field applied to the loop, when rectangular coil with 300 turns of dimensions 5.00 cm by 5.22 cm rotates at 400 rpm produce a 24.0 V peak emf?</h3>
- N= 300, A= 5cm × 5.22cm = 0.05m × 0.0522m = 0.00261 m²
- Emf= 24V, w= 2π×400 rpm= 2π×(400rps/60) = 42 rad/s
- Now, B= 24/(300×0.00261×42)
B= 24/(300×0.00261×42) = 0.73T
Thus, we can conclude that the magnetic field is 0.73T.
Learn more about the electromagnetic force here:
brainly.com/question/13745767
#SPJ4
Answer:
v₃ = 3.33 [m/s]
Explanation:
This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.
In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.

where:
m₁ = mass of the car = 1000 [kg]
v₁ = velocity of the car = 10 [m/s]
m₂ = mass of the truck = 2000 [kg]
v₂ = velocity of the truck = 0 (stationary)
v₃ = velocity of the two vehicles after the collision [m/s].
Now replacing:
![(1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]](https://tex.z-dn.net/?f=%281000%2A10%29%2B%282000%2A0%29%3D%281000%2B2000%29%2Av_%7B3%7D%5C%5Cv_%7B3%7D%3D3.33%5Bm%2Fs%5D)
Answer:

Explanation:
From the question we are told that:
Height of window 
Height of window off the ground 
Time to fall and drop
Generally the Newton's equation motion is mathematically given by

Where



Generally the Newton's equation motion is mathematically given by

Where





Therefore the ball’s initial speed
