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kap26 [50]
3 years ago
12

A child is pushing a merry-go-round. The angle through which the merry-go-round has turned varies with time according to θ(t)=γt

+βt3θ(t)=γt+βt3, where γ=γ= 0.450 rad/srad/s and β=β= 1.10×10−2 rad/s3rad/s3. Part A Calculate the angular velocity of the merry-go-round as a function of time. Express your answer in terms of the variables ββ, γγ, and tt. ωz(t)ωz(t) = nothing rad/srad/s SubmitRequest Answer Part B What is the initial value of the angular velocity? ωzωz = nothing rad/srad/s SubmitRequest Answer Part C Calculate the instantaneous value of the angular velocity ωzωz at t=t= 5.25 ss . ωzωz = nothing rad/srad/s SubmitRequest Answer Part D Calculate the average angular velocity ωav−zωav−z for the time interval t=0t=0 to t=t= 5.25 ss .
Physics
1 answer:
Lana71 [14]3 years ago
7 0

Answer:

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Explanation:

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In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera
Talja [164]

Answer:

η = 40 %  

Explanation:

Given that

Qa ,Heat addition= 1000  J

Qr,Heat rejection= 600 J

Work done ,W= 400 J

We know that ,efficiency of a engine given as

\eta=\dfrac{W(net)}{Q(heat\ addition)}

Now by putting the values in the above equation ,then we get

\eta=\dfrac{400}{1000}

η = 0.4

The efficiency in percentage is given as

η = 0.4  x 100 %

η = 40 %

Therefore the answer will be 40%.

4 0
3 years ago
A ship sets sail from Rotterdam, The Netherlands, heading due north at 8.00 m/s relative to the water. The local ocean current i
Nuetrik [128]

Answer:

The velocity of the ship relative to the earth V = 9.05 \frac{m}{s}

Explanation:

The local ocean current is  = 1.52 m/s

Direction \theta = 40°

Velocity component in X - direction V_{x} = 1.52 \cos 40°

V_{x} = 1.164 \frac{m}{s}

Velocity component in Y - direction V_{y} = 8 + 1.52 \sin 40°

V_{y} = 8.97 \frac{m}{s}

The velocity of the ship relative to the earth

V = \sqrt{V_{x}^{2} + V_{y} ^{2}  }

Put the values of V_{x} and V_{y} we get,

⇒ V = \sqrt{1.164^{2} + 8.97 ^{2}  }

⇒ V = 9.05 \frac{m}{s}

This is the velocity of the ship relative to the earth.

7 0
3 years ago
An electron moves at a speed of 1.0 x 104 m/s in a circular path of radius 2 cm inside a solenoid. The magnetic field of the sol
iogann1982 [59]

Answer:

(a) B = 2.85 × 10^{-6} Tesla

(b) I =  I = 0.285 A

Explanation:

a. The strength of magnetic field, B, in a solenoid is determined by;

r = \frac{mv}{qB}

⇒ B = \frac{mv}{qr}

Where: r is the radius, m is the mass of the electron, v is its velocity, q is the charge on the electron and B is the magnetic field

B = \frac{9.11*10^{-31*1.0*10^{4} } }{1.6*10^{-19}*0.02 }

  = \frac{9.11*10^{-27} }{3.2*10^{-21} }

B = 2.85 × 10^{-6} Tesla

b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;

B = μ I N/L

⇒    I = B ÷ μN/L

where B is the magnetic field,  μ is the permeability of free space = 4.0 ×10^{-7}Tm/A, N/L is the number of turns per length.

I = B ÷ μN/L

 = \frac{2.85*10^{-6} }{4*10^{-7} *25}

I = 0.285 A

5 0
3 years ago
A certain spring stretches 3 cm when a load of 15 n is suspended from it. how much will the spring stretch if 30 n is suspended
Alik [6]
Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
k= \frac{F}{\Delta x}
where F is the force applied, and \Delta x is the stretch of the spring with respect to its equilibrium position. Using the data, we find
k= \frac{15 N}{3.0 cm}=5.0 N/cm

Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
\Delta x =  \frac{F}{k}= \frac{30 N}{5.0 N/cm}=6 cm
4 0
3 years ago
Read 2 more answers
In a summer storm, the wind is blowing at a velocity of 8 m/s north. Suddenly in 3 seconds, the winds velocity is 23 m/s north.
cricket20 [7]

Answer:a=v-u/t

=23-8/3

=5m/s hope you got your answer

Explanation:

8 0
3 years ago
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