Answer:
c) 8.3 x 10^24 molecules
Explanation:
First of all, we convert the volume of the glass of water from mL to :
The relationship between mass and volume is:
where
is the density of water
M is the total mass of the glass of water
is the volume of the water in the glass
Solving for M,
Now we know that the mass of a single molecule of water is
The total mass of the water in the glass can be written as
where
N is the number of molecules in the glass of water
Solving for N, we find:
The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:
<h3>What are perfectly elastic collision?</h3>
Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.
In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.
Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.
Since some of the Kinetic energy is converted to potential energy of the body;
Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.
Learn more about elastic collisions at: brainly.com/question/7694106
Answer:
1. Unit for work is measured in Joules.
Formula for work is power * time taken
2 The unit for force is Newton
A work done in lifting a jar of water
3. The unit for measuring power is Joules.
The formula for measuring power is work done/time taken
We use v=IR and assuming the resistance doesn’t change we can also say that the voltage and current (I) are directly proportional which means the voltage also decreases by 1/2
Answer:
The charge on the third object is − 21.7nC
Explanation:
From Gauss's Law
Φ = Q/ε₀
where;
Φ is the total electric flux through the shell = − 533 N⋅m²/C
Q is the total charge Q in the shell = ?
ε₀ is the permittivity of free space = 8.85 x 10⁻¹²
From this equation; Φ = Q/ε₀
Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²
Q = −4.7 X 10⁻⁹ C = -4.7nC
Q = q₁ + q₂ + q₃
− 4.7nC = − 14.0 nC + 31.0 nC + q₃
− 4.7nC − 17nC = q₃
− 21.7nC = q₃
Therefore, the charge on the third object is − 21.7nC