Their combined speed will be 1m/s
Given the following parameters;
- The velocity of the first hockey player is 1m/s.
- The velocity of the second hockey player is 2m/s.
If the player collides and stick together, the combined speed is expressed as;
- Combined speed = differences in velocity
- Combined speed = 2m/s - 1m/s
- Combined speed = 1m/s.
Hence their combined speed will be 1m/s
Learn more on collision here: brainly.com/question/7538238
X -> Y + 2Z
So there are 2 different particles. 1 mol of X produces
1 mol of Y and 2 moles of Z.
Kps = [Y] [Z]^2
We will call “s” (solubility) the molarity of X
So the molarity of Y+ is also “s” (same number)
And the molarity of Z is “2s” (twice as much)
Kps = s*(2s)^2 = s*4s^2=4s^3
If s is multiplied by 2:
Kps = 4*(2s)^3=4*2^3*s^3=4*8*s^3
So Kps is multiplied by 8.
Answer:
Vd = 2.42 ×10⁻⁴ m/s
Explanation:
Given: A = 3.00×10⁻⁶ m², I = 7.00 A, ρ = 2.70 g/cm³
To find Drift Velocity Vd=?
Sol
the formula is Vd = I/nqA (n is the number of charge per unit volume)
n = No. of electron in a mole ( Avogadro's No.) / Volume
Volume = Molar mass / density ( molar mass of Al =27 g)
V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³
n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)
n= 6.02 × 10 ²⁸
Now
Vd = (7A) / ( 6.02 × 10 ²⁸ × 1.6 × 10⁻¹⁹ C × 3.00×10⁻⁶ m²)
Vd = 2.42 ×10⁻⁴ m/s
Answer:
Explanation:
The forces compare together as a result of the fact that the force exerted by that of the ball and the force exerted by that of the wall both have the same magnitude.
