Answer:
The work done by the gravel to stop the truck is 520.44 kJ
Explanation:
<u>Step 1</u>: Data given
Mass of the truck = 3047.8 kg
The ramp has an angle of 9.5 °
Velocity of the truck = 20.68 m/s
distance = 26.6 meters
<u>Step 2:</u> Calculate initial kinetic energy
sin 9.5° = 0.165
h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m
Ek = 1/2m*Vo² = 1/2*3047.8*20.68² = 651714.7 Joule = 651.7 kJ = initial kinetic energy
<u>Step 3: </u>Calculate potential energy
Epot = U = m*g*h = 3047.8*9.81*4.39 = 131256.25 Joule = 131.26 kJ
<u>Step 4:</u> What work is done by the truck on the gravel?
Frictional energy Ef = 651.7 kJ - 131.26 kJ = 520.44 kJ
Answer:
The statement "If a positively charged rod is brought close to a positively charged object, the two objects will repel
" applies to electric charges.
Explanation:
There are only two types of electric charges. Both having own magnitude but different charge.
1. Positive charge
2. Negative charge
Like charges repel each other and opposite charges always attract each other.
When a positively charged rod is brought close to a positively charged object, the rod and the object will repel.
The correct answer is: Angular velocity =
rad/s
Explanation:
The angular velocity is given as:
ω =
--- (1)
Where T = 165 * (365 days) * (24 hours/day) * (60 minutes/hour) * (60 seconds/minute) = 5203440000 s
Plug in the value in (1):
ω =
rad/s
From an energy balance, we can use this formula to solve for the angular speed of the chimney
ω^2 = 3g / h sin θ
Substituting the given values:
ω^2 = 3 (9.81) / 53.2 sin 34.1
ω^2 = 0.987 /s
The formula for radial acceleration is:
a = rω^2
So,
a = 53.2 (0.987) = 52.494 /s^2
The linear velocity is:
v^2 = ar
v^2 = 52.949 (53.2) = 2816.887
The tangential acceleration is:
a = r v^2
a = 53.2 (2816.887)
a = 149858.378 m/s^2
If the tangential acceleration is equal to g:
g = r^2 3g / sin θ
Solving for θ
θ = 67°
Answer:
<h2>The angular velocity just after collision is given as</h2><h2>
</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>
Explanation:
As per given figure we know that there is no external torque about hinge point on the system of given mass
So here we will have
now we can say
so we will have
Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass
So we can use angular momentum conservation about the hinge point
