Answer:
Capacitance is 0.572×10⁻¹⁰ Farad
Explanation:
Radius = R₁ = 6.25 cm = 6.25×10⁻² m
Radius = R₂ = 15 cm = 15×10⁻² m
Dielectric constant = k = 4.8
Electric constant = ε₀ = 8.854×10⁻¹² F/m
ε/ε₀=k
ε=kε₀
∴ Capacitance is 0.572×10⁻¹⁰ Farad
Put the leash on and have like treats with u and they will follow u that’s how I trained my cat lol
Answer:
The work done on the sled by friction, W = - 4593.75 J
Explanation:
Given data,
The combined mass of sled and the boy, m = 75 kg
The displacement of the boy, S = 25 m
The coefficient of the friction, u = 0.25
The frictional force acting on the boy,
<em>F = u η</em>
Where,
η - is the normal force acting on the boy (mg)
Substituting the values,
F = 0.25 x 75 x 9.8
= 183.75 N
Since the direction of the frictional force is against the direction of motion
F = - 183.75 N
The work done on the sled by friction,
W = F x S
= - 183.75 x 25
= - 4593.75 J
Hence, the work done on the sled by friction, W = - 4593.75 J
Answer:
150 I would believe that it is the correct answer
Answer:
Option D
A type I error is making the mistake of rejecting the null hypothesis when it is actually false.
Explanation:
Error type I is usually represented by alpha symbol and type I error entail making a mistake of rejecting the null hypothesis when it's actually true. Type II error on the other side involves making a mistake of failing to reject null hypothesis when it is actually false. The statement in option D is false because it contradicts the definition of type I error above hence the only false statement in relation to hypothesis testing is option D, A type I error is making the mistake of rejecting the null hypothesis when it is actually false.
