Answer:
Force,
Explanation:
There are two positive charges, 2 C and 5 C, separated by a distance of 20 cm in a vacuum. What is the value of the modulus of the electrostatic force between them?
We have,
Charge on object 1, q₁ = 2 C
Charge on object 2, q₂ = 5 C
Distance between charges, r = 20 cm = 0.02 m
The electric force between two charges is given by the formula as follows :
Plugging all the values we get :
So, the electric force between two charges is .
Answer:
The same numbers of shell
Explanation:
Elements in the same periods have the same numbers of shells of electrons. This implies that across the period, electrons are filled into the same shells until it filled. So, As you move across the period, elements have the same number of shells
It must be noted that during an elastic collision both the momentum and kinetic energy are conserved. For the kinetic energy, it can be solved through the equation,
KE = 0.5mv²
Equating the kinetic energies before and after collision,
0.5(60)(8 m/s)² = (0.5)(60)(x²) + (0.5)(0.45)(35 m/s)²
The value of x from the equation is approximately 7.40 m/s
A person could push, pull or lift an object
Answer:
the sound intensity at the position of the microphone is 7.4 × 10⁻⁴ W/m²
Explanation:
Given the data in the question;
Sound power P = 26.0 W
Area of microphone A = 1.00 cm²
Radius r = 53.0 m
sound intensity at the position of the microphone = ?
Now, intensity at the position of the microphone can be determined using the following expression;
= P / 4πr²
We substitute
= 26.0 / ( 4 × π × (53.0 )² )
= 26.0 / ( 4 × π × 2809 )
= 26.0 / 35298.935
= 26.0 / ( 4 × π × (53.0 )² )
= 0.000736566
= 7.4 × 10⁻⁴ W/m²
Therefore, the sound intensity at the position of the microphone is 7.4 × 10⁻⁴ W/m²