A 20 kg truck drives in a circle of radius 4 m at 10m/s. What is the
1 answer:
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Answer:
Energy required = 3169.34 Joules.
Explanation:
The quantity of energy (Q) required can be determined by;
Q = mcΔθ
Where: m is the mass, c is the specific heat and Δθ is the change in temperature.
But, m = 96.7 kg, c = 0.874 J/(kg),
=
and
=
.
So that,
Q = mc( -
)
= 96.7 x 0.874 x ( -
)
= 96.7 x 0.874 x 37.5
= 3169.3425
Q = 3169.34
= 3.2 KJ
The amount of energy required is 3169.34 Joules.
Answer:
The acceleration is a = 2.75 [m/s^2]
Explanation:
In order to solve this problem we must use kinematics equations.
where:
Vf = final velocity = 13 [m/s]
Vi = initial velocity = 2 [m/s]
a = acceleration [m/s^2]
t = time = 4 [s]
Now replacing:
13 = 2 + (4*a)
(13 - 2) = 4*a
a = 2.75 [m/s^2]
Answer:
Explanation:
The rod will act as pendulum for small oscillation .
Time period of oscillation
angular frequency ω = 2π / T
=
b )
ω = 20( given )
velocity = ω r = ω l
Let the maximum angular displacement in terms of degree be θ .
1/2 m v ² = mgl ( 1 - cosθ ) ,
[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]
.5 ( ω l )² = gl( 1 - cos θ )
.5 ω² l = g ( 1 - cosθ )
1 - cosθ = .5 ω² l /g
cosθ = 1 - .5 ω² l /g
θ can be calculated , if value of l is given .