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Tcecarenko [31]
3 years ago
11

A mason stretches a string between two points 70 ft apart on the same level with a tension of 20 lb at each end. If the string w

eighs 0.18 lb, determine the sag h at the middle of the string.
Physics
1 answer:
sergejj [24]3 years ago
4 0

Answer:

Explanation:

Let tension be T in each string . Let angle with horizontal be θ in the middle

The sum of vertical components of tension of two string will balance the weight

2Tsinθ = mg

2 x 20 sinθ = .18

sinθ = .18 / 40 = .0045

θ = .25783 degree

If sag be y

y / 70 = tan .25783

y = 70 x tan.25783

= .3150 m

= 31.50 cm .

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The joule and the kilowatt-hour are both units of energy. 15 kw · h is equivalent to how many joules? answer in units of j.
choli [55]

The solution for the problem is:

1 Watt = 1 Joule per second 
1 Watt*second = 1 Joule 

a Kilowatt is 1,000 Watts 
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15 W*h = 15,000 Watt*hour = 15,000 Watt * 3,600 seconds = 54,000,000 Watt*second 

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3 years ago
What is the mass moment of inertia of a 20kg sphere with a radius of 0.2m about a point on the sphere's perimeter
Kobotan [32]

Answer:

I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).

The moment  of inertia about the center of a sphere is 2 / 5 M R^2.

By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is  I = 2/5 M R^2 + M R^2 = 7/5 M R^2

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If the moon phase is seen as a waxing crescent moon in london, what phase of the moon would be seen in new york if it's viewed a
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3 years ago
Two identical twins hold on to a rope, one at each end, on a smooth, frictionless ice surface. They skate in a circle about the
777dan777 [17]

Answer:

Part a)

L = 2683.2 kg m^2/s

Part b)

v' = 8.60 m/s

Part c)

W = 4326.7 J

Explanation:

Part a)

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so here we will use

L = mv r + mvr

L = 2(78 \times 4.30 \times 4)

L = 2683.2 kg m^2/s

Part b)

Since angular momentum is conserved here as there is no external torque

so we will have

2(m v r) = 2( m v' \frac{r}{2})

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v' = 8.60 m/s

Part c)

Work done by both of them = change in kinetic energy

so we have

W = 2(\frac{1}{2}mv'^2 - \frac{1}{2}mv^2)

W = m(v'^2 - v^2)

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W = 4326.7 J

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3 years ago
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