The work done by the battery is equal to the charge transferred during the process times the potential difference between the two terminals of the battery:
where q is the charge and
is the potential difference.
In our problem, the work done is W=39 J while the potential difference of the battery is
, so we can find the charge transferred by the battery:
I would rather be hit by the deflated ball because it wouldn't hurt as bad because it wouldn't have a lot of weight to hurt me in anyway
Answer:
The mass of the ice block is equal to 70.15 kg
Explanation:
The data for this exercise are as follows:
F=90 N
insignificant friction force
x=13 m
t=4.5 s
m=?
applying the equation of rectilinear motion we have:
x = xo + vot + at^2/2
where xo = initial distance =0
vo=initial velocity = 0
a is the acceleration
therefore the equation is:
x = at^2/2
Clearing a:
a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2
we use Newton's second law to calculate the mass of the ice block:
F=ma
m=F/a = 90/1.283=70.15 kg
Answer: 71.93 *10^3 N/C
Explanation: In order to calculate the electric field from long wire we have to use the Gaussian law, this is:
∫E*dr=Q inside/εo Q inside is given by: λ*L then,
E*2*π*r*L=λ*L/εo
E= λ/(2*π*εo*r)= 4* 10^-6/(2*3.1415*8.85*10^-12*2 )= 71.93 * 10^3 N/C
Answer b protons and electrons