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goblinko [34]
4 years ago
8

A horizontal steel bar of uniform mass density and mass 5kg is supported by two vertical steel cables atteached to the ends of t

he bar. When a 75kg mass hangs from the center of the bar the tension in each cable is?
Physics
1 answer:
Alex4 years ago
3 0

Answer:

400 N

Explanation:

Since the steel bar is uniform, its center of mass must be at its geometric center as well, which is at the same spot as the 75kg mass hanging. So the total mass is 75 + 5 = 80 kg.

Suppose g = 10m/s2, we can then calculate the weight of the system:

W = Fg = 80 * 10 = 800 N

As the weight is at the center with 2 ropes supporting on 2 ends, weight should be evenly distributed on each rope. So each rope is subjected to a tension force of

T = 800 / 2 = 400 N

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A 2,300-kg truck is traveling down a highway at 32 m/s. What is the kinetic energy of the truck?
kondor19780726 [428]

m = mass of the truck = 23 00 kg

v = speed of the truck down the highway = 32 m/s

K = kinetic energy of the truck = ?

kinetic energy of the truck down the highway is given as

K = (0.5) m v²

inserting the values

K = (0.5) (2300) (32)²

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6 0
3 years ago
Arrow_forward
garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

k = 250 N/m

(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

T = 0.281 s

(c) EE = ½ kx²

EE = ½ (250 N/m) (0.0500 m)²

EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

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3 years ago
A group of students conduct an experiment with a block of wood sliding down an incline. They find that the final energy of the b
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1) 51 m

2) Some energy was transformed to other forms. (This question)

3) 3.24 J

4) 45 J

5) 1020 J

Explanation:

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An open beaker of pure water has a water potential of.
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Answer: Having Pure Water Is Zero.

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