Answer:
The velocity is
Henrietta is at distance from the under the window
Explanation:
From the question we are told that
The speed of Henrietta is
The height of the window from the ground is
Generally the time taken for the lunch to reach the ground assuming it fell directly under the window is
=>
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Generally the time taken for the lunch to reach Henrietta is mathematically represented as
Here is the time duration that elapsed after Henrietta has passed below the window the value is given as 4 s
Now
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Generally the distance covered by Henrietta before catching her lunch is
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Generally the speed with which Bruce threw her lunch is mathematically represented as
Answer:
Explanation:
When the central shaft rotates , the seat along with passenger also rotates . Their rotation requires a centripetal force of mw²R where m is mass of the passenger and w is the angular velocity and R is radius of the circle in which the passenger rotates.
This force is provided by a component of T , the tension in the rope from which the passenger hangs . If θ be the angle the rope makes with horizontal ,
T cos θ will provide the centripetal force . So
Tcosθ = mw²R
Tsinθ component will balance the weight .
Tsinθ = mg
Dividing the two equation
Tanθ =
Hence for a given w , θ depends upon g or weight .
Answer:
109656.25 Nm
Explanation:
= Final angular velocity = 1.5 rad/s
= Initial angular velocity = 0
= Angular acceleration
t = Time taken = 6 s
m = Mass of disk = 29000 kg
r = Radius = 5.5 m
Torque is given by
The torque specifications must be 109656.25 Nm
Answer
Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)
Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.
ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)
Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC
30.7pC/εo = 3.47 V∙m <----- C)
ic(0.5ns) = 29.7ma <----- D)