Answer:
The store energy in the inductor is 0.088 J
Explanation:
Given that,
Inductor = 100 mH
Resistance = 6.0 Ω
Voltage = 12 V
Internal resistance = 3.0 Ω
We need to calculate the current
Using ohm's law


Put the value into the formula


We need to calculate the store energy in the inductor



Hence, The store energy in the inductor is 0.088 J
Answer:
2. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/gºC? c=0.90J/g. 9 (2 sigs.)
Explanation:
Answer:
v = 98.75 km/h
Explanation:
Given,
The distance driver travels towards the east, d₁ = 135 km
The time period of the travel, t₁ = 1.5 h
The halting time, tₓ = 46 minutes
The distance driver travels towards the east, d₂ = 215 km
The time period of the travel, t₁ = 2 h
The average speed of the vehicle before stopping
v₁ = d₁/t₁
= 135/1.5
= 90 km/h
The average speed of vehicle after stopping
v₂ = d₂/t₂
= 215/2
= 107.5 km/h
The total average velocity of the driver
v = (v₁ +v₂) /2
= (90 + 107.5)/2
= 98.75 km/h
Hence, the average velocity of the driver, v = 98.75 km/h
Answer:
The oxidation number of a monatomic (composed of one atom) ion is the same as the charge of the ion. For example, the oxidation numbers of K+, Se2−, and Au3+ are +1, -2, and +3, respectively. The oxidation number of oxygen in most compounds is −2.
Explanation:
Answer:
a) 0.64 b) 2.17m/s^2 c) 8.668joules
Explanation:
The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,
Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move
Frictional force = mgsin20o + 5N = 6.71+5N = 11.71
The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44
Coefficient of static friction = 11.71/18.44= 0.64
Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)
b) coefficient of kinetic friction = frictional force/ normal force
Fr = 0.4* mgcos 20o = 7.375N
F due to motion = ma = total force - frictional force
Ma = 11.71 - 7.375 = 4.335
a= 4.335/2(mass of the block) = 2.17m/s^2
C) work done = net force *distance = 4.335*2= 8.67Joules