Answer: counterclockwise
Explanation:
As the area of the loop decreases steadily, the flux through the loop also decreases. By Lenz’s law, any induced current will tend to oppose the decrease. Using Flemming right hand rule (Fleming's right-hand rule which shows the direction of induced current) we know that magnetic field inside the loop due to a counterclockwise current comes out of the plane. Therefore a counterclockwise current will create a stronger magnetic field inside
the loop, tending to increase the flux.
When a wire loop is moved in the direction of the current or a wire loop is being pulled through a uniform magnetic field there would be no induced current ( current loop is said to be zero)
Answer:
Yes, both of them are correct.
During the internal examination, the pathologist drains the intestines, removes any undigested food and feces, and examines the contents of the stomach. This examination could give the pathologist clues of the time of death, and the location of death. The process of digesting and defecation vary from person to person, the entire process is generally considered to take approximately 40 hours in adults
Answer:
"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2."
Explanation:
The full question has not been provided, so I just copied this into the web and found this answer and explanation on quizlet:
"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2.
D sin θ = m λ
if the wavelengths are the same, then if the angle is smaller, the slit width must be larger. The top photo shows a pattern that is more closely spaced. That means the angle is smaller. The slit width must be larger."
This answer/explanation should be correct, as we are looking at bright fringes and the formula being used corresponds to the parameters of the question.
Hope this helps!
As the box compresses the spring, the spring performs
-1/2 (85 N/m) (0.065 m)² ≈ -0.18 J
of work on the box. By the work energy theorem, the total work performed on the box (which is done only by the spring since there's no friction) is equal to the change in the box's kinetic energy. At full compression, the box has zero instantaneous speed, so
<em>W</em> = ∆<em>K</em> ==> -0.18 J = 0 - 1/2 (2.5 kg) <em>v</em> ²
where <em>v</em> is the box's speed when it first comes into contact with the spring. Solve for <em>v</em> :
<em>v</em> ² ≈ 0.14 m²/s² ==> <em>v</em> ≈ 0.38 m/s
