Answer:
a) 16 N
b) 2.13 m/s²
Explanation:
Draw a free body diagram of the tv stand. There are four forces:
Weight force mg pulling down,
Normal force N pushing up,
Friction force Nμ pushing left,
and applied force P pulling right.
Sum of forces in the y direction:
∑F = ma
N − mg = 0
N = mg
The net force in the x direction is:
∑F = P − Nμ
∑F = P − mgμ
∑F = 25 N − (7.5 kg) (10 m/s²) (0.12)
∑F = 16 N
Net force equals mass times acceleration:
∑F = ma
16 N = (7.5 kg) a
a = 2.13 m/s²
Answer:
20 Joules
Explanation:
Work is done whenever a force moves a body through a certain distance in the direction of the force. So, work done is the product of force and distance moved.
Therefore, we have;
Work done = Force x distance
i.e Wd = Fs
Given that: F = 20 N and s = 1 m, then;
Wd = 20 N x 1 m
= 20 Nm
The work done by the father is 20 Joules(Nm).
Answer:
I = 21.13 mA ≈ 21 mA
Explanation:
If
I₁ = 5 mA
L₁ = L₂ = L
V₁ = V₂ = V
ρ₁ = 1.68*10⁻⁸ Ohm-m
ρ₂ = 1.59*10⁻⁸ Ohm-m
D₁ = D
D₂ = 2D
S₁ = 0.25*π*D²
S₂ = 0.25*π*(2*D)² = π*D²
If we apply the equation
R = ρ*L / S
where (using Ohm's Law):
R = V / I
we have
V / I = ρ*L / S
If V and L are the same
V / L = ρ*I / S
then
(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂
If
S₁ = 0.25*π*D² and
S₂ = 0.25*π*(2*D)² = π*D²
we have
ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)
⇒ I₂ = 4*ρ₁*I₁ / ρ₂
⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m
⇒ I₂ = 21.13 mA
Answer:
Speed =0.283m/ s
Direction = 47.86°
Explanation:
Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane
MU1 =MU2cos38 + MV2cos y ...x plane
0 = MU2sin38 - MV2sin y .....y plane
Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2
Substitute into equation above
.46 = .34cos38 + V2cos y ...equ1
.34sin38 = V2sin y...equ2
.19=V2cos Y...x
.21=V2sin Y ...y
From x
V2 =0.19/cost
Sub V2 into y
0.21 = 0.19(Sin y/cos y)
1.1052 = tan y
y = 47.86°
Sub Y in to x plane equ
.19 = V2 cos 47.86°
V2=0.283m/s