Answer:
The magnitude of the electric force on a protein with this charge is 
Explanation:
Given that,
Electric field = 1500 N/C
Charge = 30 e
We need to calculate the magnitude of the electric force on a protein with this charge
Using formula of electrostatic force
Where, F = force
E = electric field
q = charge
Put the value into the formula
Hence, The magnitude of the electric force on a protein with this charge is
Answer:
Xc= 17.267 Ω, Z= 415.5 Ω, I= 0.537 A
Explanation:
Em = 223 V
f= 300 Hz, R = 222 Ω, L = 147 mH, C = 23.1 μF
a)
Capacitive reactance = Xc=?
Xc= 
Xc=1/2pi *399*23.1*10^-6
Xc= 17.267 Ω
b).
Z=
Xl= 2π * f * L
Xl= 2π * 399 * 147 * 
Xl= 368.5 Ω
Z= = 
Z= 415.5 Ω
c).
Current:
I= V / Z= Em / Z
I= 223/415.5
I= 0.537 A
Answer:
Explanation:
according to third equation of motion
2as=vf²-vi²
vf²=2as+vi²
vf=√2as+vi²
vf=√2as+vi
vf=√2*2*4+3
vf=√16+3
vf=4+3=7
so final velocity is 7 m/s
Answer:20/47 meter per second
Explanation:
Mass of arrow(ma)=0.25kg
Velocity of arrow(va)=12m/s
Mass of target(mt)=6.8kg
Velocity of target(vt)=0 since target is at rest
Conservation of linear momentum says that :
maxva+mtxvt=(ma+mt)V
V=(maxva+mtxvt)/(ma+mt)
V=(0.25x12+6.8x0)/(0.25+6.8)
V=3/(7.05)
V=20/47 meter per second
I wouldn't be 1000 but I have a feeling your best bet will be B
