The number of radians (d) that the wheel have turned may be calculated by the equation,
d = (v1)(t) + 0.5 x a x t²
where v1 is the initial velocity, a is the acceleration, and t is time. Substituting the known values
d = (18 rad/s)(3s) + (0.5) x (-2 rad/s²) x (3s)²
The value of d is 63 radians.
Answer:
1/9 E0
Explanation:
The computation is shown below:
As we know that
where,
E = Electric field strength
k = Coulomb's constant
Q = charge on the sphere
r = distance from the center of the sphere
It is given that
The radius of the larger sphere is three times larger than that of the smaller sphere i.e
hence, the last option is correct
Answer:
Explanation:
Given data
For Part (a) Speed
The speed v is given by
For Part (b) minimum coefficient of friction
To determine the friction of coefficient we know that friction force f is given by:
The first centripetal force Fc₁ is given by:
The second centripetal force Fc₂ is given by:
The additional friction force is given by:
Answer:
Explanation:
Given that,
Efficiency of Carnot engine is 47%
η =47%=0.47
The wasted heat is at temp 60°F
TL=60°F
Rate of heat wasted is 800Btu/min
Therefore, rate of heat loss QL is
QL' = 800×60 =48000
The power output is determined from rate of heat obtained from the source and rate of wasted heat.
Therefore,
W' = QH' - QL'
Note QH' = QL' / (1-η)
W' = QL' / (1-η) - QL'
W'=QL' η / (1-η)
W'= 48000×0.47/(1-0.47)
W'=42566.0377 BTU
1 btu per hour (btu/h) = 0.00039 horsepower (hp)
Then, 42566.0377×0.00039
W'=16.6hp
Which is approximately 17hp
b. Temperature at source
Using ratio of wanted heat to temp
Then,
TH / TL = QH' / QL'
TH = TL ( QH' / QL')
Since, QH' = QL' / (1-η)
Then, TH= TL( QL' /QL' (1-η))
TH=TL/(1-η)
TL=60°F, let convert to rankine
°R=°F+459.67
TL=60+459.67
TL=519.67R
TH=519.67/(1-0.47)
TH=980.51R
Which is approximately 1000R
The impulse is equal to the final moment of the wall where the moments and part of the ball from the player