Answer:
There will be 1800 W power consumption in heater
Explanation:
We have given current flowing in the heater I = 15 A
Voltage on which heater is operating V = 120 volt
We have to find the power consumption in the heater
We know that power consumption is given by P = VI
So power consumption in heater = 120 × 15 = 1800 W
So there will be 1800 W power consumption in heater
Answer:
= 4.3 × 10 ⁻¹⁴ m
Explanation:
The alpha particle will be deflected when its kinetic energy is equal to the potential energy
Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C
Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷C
Kinetic energy of the alpha particle = 5.28 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV)
= 8.459 × 10⁻¹³
k electrostatic force constant = 9 × 10⁹ N.m²/c²
Kinetic energy = potential energy = k q₁q₂ / r where r is the closest distance the alpha particle got to the gold nucleus
r = ( 9 × 10⁹ N.m²/c² × 3.2 × 10⁻¹⁹ C × 1.264 × 10⁻¹⁷C) / 8.459 × 10⁻¹³
= 4.3 × 10 ⁻¹⁴ m
Answer:
The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.
Explanation:
To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
where,
are the initial pressure and temperature of the gas.
are the final pressure and temperature of the gas.
We are given:
Putting values in above equation, we get:
The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.
If you remember the formula for potential energy,
then this question is a piece-o-cake.
<em>Potential energy = (mass) x (<u>acceleration of gravity</u>) x (height) .</em>
-- The object's mass is the same everywhere.
-- You said that the height is the same both times.
-- How about the acceleration of gravity ?
Compared to gravity on Earth, it's only 16.5 percent as much on the Moon.
So naturally, from the formula, you'd expect the Potential Energy to be less
on the Moon.
