Response:
The man's speed is 0.144 m/s
Explanation:
This exemplifies conservation of momentum.
The momentum of the ball prior to being caught must equal the momentum of the man-ball system after catching the ball.
Mass of the ball = 0.65 kg
Mass of the man = 54 kg
Speed of the ball = 12.1 m/s
The momentum of the ball before impact can be calculated as mass multiplied by velocity
= 0.65 x 12.1 = 7.865 kg-m/s
After catching the ball, the momentum of the combined system is
(0.65 + 54)Vf = 54.65Vf
Where Vf denotes their final shared velocity.
Setting the initial momentum equal to the final momentum,
7.865 = 54.65Vf
Vf = 7.865/54.65 = 0.144 m/s
Answer:
A) 5.1*10^10m B) 5.4*10^6m
Explanation:
Utilizing the formula for surface radiation P (energy per second in Watts) = emissivity constant * surface area * Stefan-Boltzmann constant * Temperature in Kelvin^4 *
2.7*10^31 = 1* 5.67*10^-8*A*11000^4
Rearranging to solve for A = 2.7*10^31 / (5.67*10^-8*1.46*10^16) = 0.3261*10^23m^2
Assuming the shape is spherical, the surface area is = 4πR^2 (radius of Rigel)
R = √(0.3261*10^23 / 4*π) = 5.1 * 10^10m
B) repeating the same calculation
2.1 *10^23 = 1*A*5.67*10^-8*10000^4 where A is the surface area of Procyon
Rearranging gives A = 2.1*10^23/(5.67*10^-8*10^16)
A = 0.37*10^15
Assuming the star is spherical;
A = 4πR^2 where R is Procyon's radius
R = √(0.37*10^15/4π) = 5.4*10^6m
Response:
(b) 10 Wb
Clarification:
Given;
angle of the magnetic field, θ = 30°
initial area of the plane, A₁ = 1 m²
initial magnetic flux through the plane, Φ₁ = 5.0 Wb
The equation for magnetic flux is;
Φ = BACosθ
where;
B denotes the magnetic field strength
A represents the area of the plane
θ is the inclination angle
Φ₁ = BA₁Cosθ
5 = B(1 x cos30)
B = 5/(cos30)
B = 5.7735 T
Next, calculate the magnetic flux through a 2.0 m² section of the same plane:
Φ₂ = BA₂Cosθ
Φ₂ = 5.7735 x 2 x cos30
Φ₂ = 10 Wb
<pHence, the magnetic flux through a 2.0 m² area of the same plane is
10 Wb.Option "b"
