Answer:
The work done on the suitcase is, W = 1691 J
Explanation:
Given data,
The force on the suitcase is, F = 89 N
The distance Russell dragged the suitcase, S = 19 m
The work done on the suitcase by Russell is equal to the work done on the suitcase to overcome the friction
The work done on the suitcase by Russell is given by the formula
W = F · S
Substituting the given values,
W = 89 N x 19 m
W = 1691 J
Hence, the work done on the suitcase is, W = 1691 J
Answer:
775.48 W
Explanation:
given,
diameter of disk = 0.6 cm
length of the disk = 0.4 m
T₁ = 450 K T₂ = 450 K T₃ = 300 K
= 1.33
now,
the value of view factor (F₁₂)corresponding to 1.33
F₁₂ = 0.265
F₁₃ = 1 - 0.265 = 0.735
now,
net rate of radiation heat transfer from the disk to the environment:
= 2 F₁₃ A₁ σ (T₁⁴ - T₃⁴)
= 2 x 0.735 x π x (0.3)² x (5.67 x 10⁻⁸ W/m²) (450⁴ - 300⁴)
= 775.48 W
Net radiation heat transfer from the disks to the environment = 775.48 W
Answer:
Cart A
Explanation:
Momentum can be computed by finding the product of mass and velocity. To solve this, you can use the formula below to find the greatest momentum:
p = mv
where:
p = momentum (kgm/s) m = mass (kg) v = velocity (m/s)
Because carts are moving along with the weights, we need to consider the whole system. This means that you need to add in the masses and the mass of the cart.
<u>Cart A:</u>
m = 200kg + 0 kg = 200 kg
v = 4.8 m/s
p = 200kg x 4.8 m/s = 960 kg-m/s
<u>Cart B:</u>
m = 200kg + 20 kg = 220 kg
v = 4.0 m/s
p = 220kg x 4.0 m/s = 880 kg-m/s
<u>Cart C:</u>
m = 200kg + 40 kg = 240 kg
v = 3.8 m/s
p = 240kg x 3.8 m/s = 912 kg-m/s
<u>Cart D:</u>
m = 200kg + 60 kg = 260 kg
v = 3.5 m/s
p = 260kg x 3.5 m/s = 910 kg-m/s
As you can see, Cart A has the greatest momentum.
