Answer:
2 m/s^2, west
Explanation:
Vf=final velcoity
Vi=initial velocity
t=timw
=
= - 2 m/s^2
The - changes direction and makes it opposite
2 m/s, west
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:
0.65 kg*m/s and 0.165 kg*m/s
Explanation:
Step one:
given data
mass m= 0.5kg
initial velolcity u=1.3m/s
final velocity v= 0.97m/s
Required
The change in momentum
Step two:
We know that the expression for impulse is given as
Ft= mv
Ft= 0.5*1.3
Ft= 0.65 kg*m/s
The expression for the change in momentum is given as
P= mΔv
substitute
Pt= 0.5*(1.3-0.97)
Pt= 0.5*0.33
Pt=0.165 kg*m/s
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hope this helps!