I believe it she should use the first aid kit next
Given Information:
KEa = 9520 eV
KEb = 7060 eV
Electric potential = Va = -55 V
Electric potential = Vb = +27 V
Required Information:
Charge of the particle = q = ?
Answer:
Charge of the particle = +4.8x10⁻¹⁸ C
Explanation:
From the law of conservation of energy, we have
ΔKE = -qΔV
KEb - KEa = -q(Vb - Va)
-q = KEb - KEa/Vb - Va
-q = 7060 - 9520/27 - (-55)
-q = 7060 - 9520/27 + 55
-q = -2460/82
minus sign cancels out
q = 2460/82
Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹
q = 2460(1.60x10⁻¹⁹)/82
q = +4.8x10⁻¹⁸ C
Answer:
limited liability, limitation in expansion, risk bearing, problem of continuity,
Answer: 5 seconds
Explanation:
Given the following :
Inductance (L) = 40 henry
Resistance = 8 ohms
The circuit given above is a Resistor - Inductor (RL) circuit network. The time constant of an RL circuit is the ratio of the circuit Inductance (L) and Resistance (R). Time constant is measured in seconds.
THAT IS;
Time constant = L / R
THEREFORE ;
Time constant = 40 / 8
Time constant = 5 seconds
