Explanation:
a) Power = work / time = force × distance / time
P = Fd/t
P = (85 kg × 9.8 m/s²) (4.6 m) / (12 s)
P ≈ 319 W
b) P = Fd/t
0.70 (319 W) = (m × 9.8 m/s²) (4.6 m) / (9.6 s)
m = 47.6 kg
Let the cold water go up x degrees.
Let the hot water go down 100 - x degrees.
The formula for heat exchange is m*c*delta t
Givens
Ice
deltat = x
m = 0.50 kg
c = 4.18
Hot water
deltat = 100 - x
m = 1.5 kg
c = 4.18
Formula
The heat up = heat down
0.50 * c * x = 1.5 * c * (100 - x) Divide both sides by c
Solution
0.50 *x = 1.5*(100 - x) Remove the brackets.
0.5x = 150 - 1.5x Add 1.5x to both sides.
0.5x + 1.5x = 150 - 1.5x + 1.5x Combine like terms
2x = 150 Divide by 2
x = 75
Answer
A
Answer:
a) the magnitude of the force is
F= Q(
) and where k = 1/4πε₀
F = Qqs/4πε₀r³
b) the magnitude of the torque on the dipole
τ = Qqs/4πε₀r²
Explanation:
from coulomb's law
E = 
where k = 1/4πε₀
the expression of the electric field due to dipole at a distance r is
E(r) = 
, where p = q × s
E(r) = 
where r>>s
a) find the magnitude of force due to the dipole
F=QE
F= Q()
where k = 1/4πε₀
F = Qqs/4πε₀r³
b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces
τ = F sinθ × s
θ = 90°
note: sin90° = 1
τ = F × r
recall F = Qqs/4πε₀r³
∴ τ = (Qqs/4πε₀r³) × r
τ = Qqs/4πε₀r²
Answer:
Coefficient of friction.
Explanation:
The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :
N is normal force.
= coefficient of friction
Maybe you can split up the questions. I will try to answer your first question.
1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)
2. Momentum: p = m₁v₁ + m₂v₂
m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B
Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0
Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0
3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.
4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7
5.You figure out.
