Answer:
the mass of the bullet is 10.5 g
Explanation:
Given;
initial velocity, u₁ = 280 m/s
final velocity of the bullet, v₁ = 70 m/s
final velocity of the block, v₂ = 0.2 m/s
mass of the block, m₂ = 11 kg
initial velocity of the block, u₂ = 0
let the mass of the bullet = m₁
Apply the principle of conservation of linear momentum for elastic collision to calculate the mass of the bullet.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
280m₁ + 11(0) = 70m₁ + 11 x 0.2
280m₁ = 70m₁ + 2.2
280m₁ - 70m₁ = 2.2
210m₁ = 2.2
m₁ = 2.2/210
m₁ = 0.0105 kg
m₁ = 10.5 g
Therefore, the mass of the bullet is 10.5 g
Answer:
a = 8 m/s^2, Ffriction = 10 N, μk = 0.205
Explanation:
a. Force = Mass*Acceleration,
(since you didn't add the units..."5 block"....for the mass, I will assume it to be in kg, per SI units)
40 N = 5 kg*acceleration,
a = 40/5 = 8 m/s^2
b. As you know newtons second law (F=m*a) is actually in the form Fnet = m*a. Which means that if the friction force comes into play, it would be Fapplied - Ffriction = m*a.
Fapplied - Ffriction = m*a,
40 - Ffriction = 5*6,
40 - Ffriction = 30,
Ffriction = 40 - 30 = 10 N
c. The coefficient of kinetic friction is calculated by the formula "Ffriction = μk*Fnormal".
10 = μk*Fnormal (Fnormal = m*g = 5*9.8)
10 = μk*49,
μk=10/49 ≈ 0.205
We need to use Wien's Law
Wavelength = 0.0028976 [m.K] / T
This establishes a relation between the wavelength and temperature of a black body (any body that absorbs radiation, such as the stars)
T = 0.0028976 [m.K]/290 E-9[m] = 9991.724 K
