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bekas [8.4K]
3 years ago
6

A heavy ball with a weight of 110 N is hung from the ceiling of a lecture hall on a 4.9-m-long rope. The ball is pulled to one s

ide and released to swing as a pendulum, reaching a speed of 5.0 m/s as it passes through the lowest point.
Required:
What is the tension in the rope at that point?
Engineering
1 answer:
katovenus [111]3 years ago
5 0

Answer:T = 167.3 N

Explanation:

Given that the

Weight mg = 110 N

The mass m of the ball will be

m = 110/9.8 = 11.22 kg

As the direction of the ball’s velocity is changing, the force responsible for this is centripetal force F. And

F = mV^2/r

Where

V = 5.0 m/s

r = L = 4.9 m

m = 11.22

Substitute all these parameters into the formula

F = (11.22 × 5^2)/4.9

F = 280.6/4.9

F = 57.27 N

Tension T = F + mg

Substitute F and mg into the formula

T = 57.27 + 110

T = 167.3 N

Therefore, the tension in the rope at that point is 167.3 N

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Answer:

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Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

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After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

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Thus,

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So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

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