Answer:
Vdc=10V
Explanation:
in a closed loop consisting of a super charged capacitor and an inductor, the super charge capacitor acts as a supply when the loop is closed, at t=0, the emf stored in the capacitor is 10V (q/c); and at that same time Vl= voltage across the inductor or loop too would be 10V,
if the loop remains closed for a longer period, the inductor would absorb energy from the capacitor till it dissipates all charges with itself.
Answer:
The speed will be "18km/s". A further explanation is given below.
Explanation:
According to the question, the values are:
Wavelength,



As we know,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
⇒ 
or,
⇒ 
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.