Answer:
yes
Explanation:
I would say yes because it's exerting some type of force as it reacts to the stick to make it jump straight in an upwards direction
(A) We can solve the problem by using Ohm's law, which states:

where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know

and

, therefore we can rearrange Ohm's law to find the current through the device:

(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:

where

is the resistivity of the material of the wire
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.
Answer:
Average force = 3.5 kN
Explanation:
Given:
Mass of Jennifer (m) = 50 kg
Initial velocity = 35 m/s
Time taken to stop body = 0.5 s
Find:
Average force
Computation:
v = u + at
0 = 35 + a(0.5)
Acceleration (a) = - 70 m/s² = 70 m/s²
Average force = ma
Average force = (50(70)
Average force = 3500 N
Average force = 3.5 kN
Answer: 0.2 hours
Explanation: In order to solve this question we have to considerer that a recargeable battery can supply 1800 mA in one hour then we have to determine how long could this battery drive current through a long, thin wire of resistance 34 Ω .
Besides, this battery has a voltage of 12 V
so by using the Ohm law we also know that V=R*I,
Fron this we can obtain:
I= V/R= 12 V/ 34 Ω=0.35 A= 350 mA
then considering that this battery can supply 1800 mA in one hour we have this battery can supply 350 mA in x time in the form:
1hour------- 1800 mA
x hour--------350 mA
time= 350/1800= 0.2 hour
The direction of the electric field would be south.
qE/m = 115
<span> E = 115*m/q </span>
<span> = 115 * 9.1 * 10^(-31) / 1.67*10^(-19) </span>
<span> = 762.87 * 10^(-12) </span>
<span> = 6.27 x 10^-10 N/C
</span>
Hope this answers the question. Have a nice day. Feel free to ask more questions.