Please help with this, oh and bungee gum contains both the properties of rubber AND gum you see?
1 answer:
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Answer:
e = 10 V
Explanation:
given,
number of the coaxial loops = 10
Cross sectional area = 0.5 m²
magnitude of magnetic field =
B = 3 T + (2 T/s)*t.
B = ( 3+ 2 t ) T
induced potential difference = ?
At time = 2 s
we know,
induced emf
∅ = B . A
e = -10 V
magnitude of induced emf
|e| = |-10 V|
e = 10 V
the induced potential difference in the loop = e = 10 V
<u>Answers:</u>
3. The diagrams showing the forces acting on the golf ball are in the figure attached. Let’s have a detailed look:
a) Here the ball is under 2 forces:
F1 which is called The Normal force and is perpendicular to the surface of the tee where the ball rests
-F1, related to the gravity force, to the weight of the ball, and has the same magnitude, but the opposite direction (That’s why it has a negative sign).
In this case the sum of the forces is 0
b) Here we have again forces F1 and –F1, but in this very moment the club strikes the ball and we have:
F2, the force of the strike
c) While the ball is in its flight, it is under the following forces:
F1, the force of the lift through the air
-F2, the gravity force
-F3, the force of air resistance, also called drag
F4, the tangencial force of the ball flight
4. Here are the sizes and directions of the resultant forces:
i)Two forces of the same magnitude or size are applied to this block, but in opposite directions (in the x-axis). This is expressed as:
F=-10N+10N=0 The resulting force is zero
ii) Two forces of different size and opposite directions in the y-axis are applied to this block. The sum of the forces is:
F=30N-40N=-10N This means the resulting force is 10N applied downwards
iii) In this case the only force applied to the block is -5N applied downwards
iv) Here there are four forces applied to the block.
In the y-axis we have to forces of the same size but opposite directions:
F1=10N-10N=0 This means the applied force in the y-axis is zero
In the x-axis we have two forces of different size and opposite directions:
F2=-15N+10N=-5N This means the resulting force is applied to the –x-side
