Heat supplied to the gold will raise the temperature of the gold from 20 degree Celsius to 90 degree Celsius.
Mass of the gold (m) = 0.072 kg
Temperature change (ΔT) = 90 - 20 = 70 degree Celsius
Specific heat capacity of the gold (c) = 136 J/kg C
Heat supplied = m × c × ΔT
Heat supplied = 0.072 × 136 × 70
Heat supplied = 685.44 Joules
Hence, the heat supplied to the gold to raise the temperature from 20 degree Celsius to 90 degree Celsius = 685.44 Joules
Answer:
0.4
Explanation:
Given that a particular inductor is connected to a circuit where it experiences a change in current of 0.8 amps every 0.10 sec. If the inductor has a self-inductance of 2.0 V, what is the inductance
Using the power formula
P = IV
Substitute all the parameters
P = 0.8 × 2
P = 1.6 W
But P = I^2 R
Substitute power and current
1.6 = 0.8^2 R
R = 1.6 / 0.64
R = 2.5 ohms
Inductance = reciprocal of resistance
Inductance = 1 / 2.5
Inductance = 0.4
Answer:
In two significant figure 360K
Explanation:
The temperature difference (ΔT) can be calculated as the boiling temperature minus the freezing temperature in Fahrenheit.
Hence,
ΔT = 212 - 32
ΔT = 180°F
To convert to °F to kelvin, we use the formula below
= (°F - 32) × 5/9 + 273.15
= (180°F - 32) × 5/9 + 273.15
= 355.37K ⇔ 360K
Potential energy = (mass) x (gravity) x (height)
1 joule = (1,000 kg) x (9.8 m/s²) x (height)
Height = 1 joule / (9,800 newtons)
= 1/9800 meter
= 0.000102 meter
= 0.102 millimeter (rounded)
