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Karo-lina-s [1.5K]
2 years ago
10

A crane is used to lower weights into a lake for an underwater construction project. Determine the tension in the cable of the c

rane due to a concrete sphere with a radius of 2 m when the sphere is lowered into the water. The concrete has a specific gravity of 2.8.
Physics
1 answer:
atroni [7]2 years ago
5 0

Answer:

Tension in the cable = 591292.8 N or 591.3 kN

Explanation:

Tension in the cable = Weight of object in air - upthrust/bouyancy due to the liquid

Weight of object in air = mass * acceleration due to gravity, g

Since mass is not given, it is obtained from the formula, mass = density * volume

Volume of the sphere = 4/3πr³ where r, radius = 2 m; π = 22/7

volume = 4/3 * 22/7 * (2m)³

volume of sphere = 33.52 m³

Density of concrete sphere = specific gravity * density of water

where, specific gravity = 2.8, density of water = 1000 kg/m³

density of concrete =  2.8 * 1000 kg/m³

density of concrete = 2800 kg/m³

acceleration due to gravity, g = 9.8 m/s²

Thus, weight of concrete = 2800 kg/m³ * 33.52 m³ * 9.8 m/s² = 919788.8 N

Upthrust = density of water * volume of sphere * g

Upthrust = 1000 kg/m³ * 33.52 m³ * 9.8 m/s² = 328496 N

The tension in the cable is the calculated as below;

Tension in the cable = Weight of object in air - upthrust/bouyancy due to the liquid

Tension in cable = 919788.8 N - 328.496 N = 591292.8 N

Therefore, tension in the cable is 591292.8 N or 591.3 kN

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Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t
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10.2 m

Explanation:

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where

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D is the distance of the screen from the slit

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In this problem we have:

\lambda=683 nm = 683\cdot 10^{-9} m is the wavelength of the light

d=1.1 mm = 0.0011 m is the width of the slit

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Solving for D, we find the distance of the screen from the slit:

D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m

6 0
3 years ago
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally h
zysi [14]

Complete question:

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)

Answer:

The peak emf generated by the coil is 15.721 kV

Explanation:

Given;

Radius of coil, r = 0.250 m

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time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s

magnetic field strength, B = 0.425 T

Induced peak emf = NABω

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A is the area of the coil

A = πr²

ω is angular velocity

ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s =  60 rev/s

Induced peak emf = NABω

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                               = 15721.16 V

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Therefore, the peak emf generated by the coil is 15.721 kV

5 0
3 years ago
A particle moves along a straight line. Its position at any instant is given by x = 32t− 38t^3/3 where x is in metre and t in se
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Answer:

The acceleration of the object is -69.78 m/s²

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postion of the particle:

x = 32t - 38\frac{t^3}{3} \\\\

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a = \frac{dv}{dt} = -76t\\\\recall , \ t = 0.918 \ s\\\\a = -76(0.918)\\\\a = -69.78 \ m/s^2

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