A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =
1 answer:
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Answer:
12164.4 Nm
Explanation:
CHECK THE ATTACHMENT
Given values are;
m1= 470 kg
x= 4m
m2= 75kg
Cm = center of mass
g= acceleration due to gravity= 9.82 m/s^2
The distance of centre of mass is x/2
Center of mass(1) = x/2
But x= 4 m
Then substitute, we have,
Center of mass(1) = 4/2 = 2m
We can find the total torque, through the summation of moments that comes from both the man and the beam.
τ = τ(1) + τ(2)
But
τ(1)= ( Center of m1 × m1 × g)= (2× 470× 9.81)
= 9221.4Nm
τ(2)= X * m2 * g = ( 4× 75 × 9.81)= 2943Nm
τ = τ(1) + τ(2)
= 9221.4Nm + 2943Nm
= 12164.4 Nm
Hence, the magnitude of the torque about the point where the beam is bolted into place is 12164.4 Nm
Answer:
Explanation:
Given the following :
Speed (V) = speed of 2.30×10^7 m/s
Acceleration (a) = 1.70×10^13 m/s^2
Using the right hand rule provided by Lorentz law:
B = F / qvSinΘ
Where B = magnitude of the magnetic field
v = speed of the particle
Θ = 90° (perpendicular to the field)
q = charge of the particle
SinΘ = sin90° = 1
Note F = ma
Therefore,
B = ma / qvSinΘ
Mass of proton = 1.67 × 10^-27
Charge = 1.6 × 10^-19 C
B = [(1.67 × 10^-27) × (1.70 × 10^13)] / (1.6 × 10^-19) × (2.30 × 10^7) × 1
B = 2.839 × 10^-14 / 3.68 × 10^-12
B = 0.7715 × 10^-2
B = 7.72 × 10^-3 T
2) Magnetic field will be in the negative y direction according to the right hand thumb rule.
Since Velocity is in the positive z- direction, acceleration in the positive x - direction, then magnetic field must be in the negative y-direction.
