Answer:
can you put on a clearer image this one is hard to see
Answer:
B - Person A's chair rolls backwards.
Explanation:
This can be seen simply through the Conservation of Momentum. When the ball is thrown to be, the ball's momentum(+m) is transfererd to B, so B's chair moves forward(think about it - it makes sense, doesn't it?), meaning A's chair moves backwards.
So, B.
EXTRA TIP: Try drawing a diagram to help you!
Answer:
110.9 m/s²
Explanation:
Given:
Distance of the tack from the rotational axis (r) = 37.7 cm
Constant rate of rotation (N) = 2.73 revolutions per second
Now, we know that,
1 revolution = 
radians
So, 2.73 revolutions = 
Therefore, the angular velocity of the tack is, 
Now, radial acceleration of the tack is given as:
Plug in the given values and solve for 
. This gives,
![a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]](https://tex.z-dn.net/?f=a_r%3D%2817.153%5C%20rad%2Fs%29%5E2%5Ctimes%2037.7%5C%20cm%5C%5Ca_r%3D294.225%5Ctimes%2037.7%5C%20cm%2Fs%5E2%5C%5Ca_r%3D11092.28%5C%20cm%2Fs%5E2%5C%5Ca_r%3D110.9%5C%20m%2Fs%5E2%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5B1%5C%20cm%20%3D%200.01%5C%20m%5D)
Therefore, the radial acceleration of the tack is 110.9 m/s².
Answer:
magnitude of the induced emf in the coil is 0.0153 V
Explanation:
Given data
no of turns = 20
area = 0.0015 m²
magnitude B1 = 4.91 T/s
magnitude B2 = 5.42 T/s
to find out
the magnitude of the induced emf in the coil
solution
we know here
emf = -n A d∅ /dt
so here n = 20 and
A = 0.0015
and d∅ = B2 - B1 = 5.42 - 4.91
d∅ = 0.51 T and dt at 1 sec
so put all value
emf = -n A d∅ /dt
emf = -20 (0.0015) 0.51 / 1
emf = - 0.0153
so magnitude of the induced emf in the coil is 0.0153 V
Answer:
The number of excess electrons on each sphere is 759
Explanation:
Given that,
distance , d = 20 cm
= 0.20 m
let the number of electrons is n
Electric force (F) = k × (n × e)² /d²
3.33 × 
= 9 × 
× (n × 1.602 × 
)² /0.2²
solving for n
n = 759
