B) atmosphere pressure, i believe
Answer:
a) 0.1832 A
b) 11.91 Volts
c) 2.18 Watt , 0.0168 Watt
Explanation:
(a)
R = external resistor connected to the terminals of the battery = 65 Ω
E = Emf of the battery = 12.0 Volts
r = internal resistance of the battery = 0.5 Ω
i = current flowing in the circuit
Using ohm's law
E = i (R + r)
12 = i (65 + 0.5)
i = 0.1832 A
(b)
Terminal voltage is given as
= i R
= (0.1832) (65)
= 11.91 Volts
(c)
Power dissipated in the resister R is given as
= i²R
= (0.1832)²(65)
= 2.18 Watt
Power dissipated in the internal resistance is given as
= i²r
= (0.1832)²(0.5)
= 0.0168 Watt
First method
initial distance = 16m
final distance= 43 m
total distance covered= final -initial
=43m -16m
=27m
Second method
Si= 16m
Sf =43 m
t= 12 s
first we will find V
V = (Sf-Si)/ t
V =( 43- 16)/ 12
V = 27/12 ⇒ V= 9/4
V= distance / time
distance= V×time
distance = (9/4) ×12
distance =27
Answer:
Acids break down fabrics and can cause burns if the acids are strong.
Explanation:
A protective apron or lab coat is important when working with acids because acids break down fabrics and can cause burns if the acids are strong.
- An acid is a substance that interacts with water to produce excess hydroxonium ions in an aqueous solution.
- A strong acid ionizes completely in solution.
- When they come in contact with a fabric, they break them down violently.
- So, if they come in contact with the skin, it causes a violent break down of body tissues.
- The apron acts a protective layer.
Answer:
a) L = 3.29 10⁻⁴ H, b)U = 5.33 10⁻² J
Explanation:
a) The inductance is a solenoid this given carrier
L =
The magnetic field inside the solenoid is
B = μ₀
hence the magnetic flux
Ф_B = B. A = μ₀
we substitute in the expression of inductance
L = N² μ₀ A /l
let's find the area of each turn
A = π r²
A = π 0.02²
A = 1.2566 10⁻³ m²
let's calculate
L = 250² 4π 10⁻⁷ 1.2566 10⁻² / 0.3
L = 3.29 10⁻⁴ H
b) The stored energy is
U = ½ L i²
let's calculate
U = ½ 3.29 10⁻⁴ 18²
U = 5.33 10⁻² J