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ss7ja [257]
2 years ago
6

A stoplight with weight 100 N is suspended at the midpoint of a cable strung between two posts 200 m apart. The attach points fo

r the cable on the posts are each 15 m above the ground level. The cable is tightened until its tension is 1,000 N. How far is the stoplight attach point above ground
Physics
1 answer:
Tasya [4]2 years ago
3 0

There are 3 forces acting on the stoplight:

• its weight <em>W</em>, with magnitude <em>W</em> = 100 N, pointing directly downward

• two tension forces <em>T</em>₁ and <em>T</em>₂ with equal magnitude <em>T</em>₁ = <em>T</em>₂ = <em>T</em> = 1000 N, both making an angle of <em>θ</em> with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ <em>F</em> = <em>T</em>₁ sin(<em>θ</em>) + <em>T</em>₂ sin(180° - <em>θ</em>) - <em>W</em> = 0

We have sin(180° - <em>θ</em>) = sin(<em>θ</em>) for all <em>θ</em>, so the above reduces to

2<em>T</em> sin(<em>θ</em>) = <em>W</em>

2 (1000 N) sin(<em>θ</em>) = 100 N

sin(<em>θ</em>) = 0.05

<em>θ</em> ≈ 2.87°

If <em>y</em> is the vertical distance between the stoplight and the ground, then

tan(<em>θ</em>) = (15 m - <em>y</em>) / (100 m)

Solve for <em>y</em> :

tan(2.87°) = (15 m - <em>y</em>) / (100 m)

<em>y</em> = 15 m - (100 m) tan(2.87°)

<em>y</em> ≈ 9.99 m

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What substances were found in the innermost regions (within about the inner 0.3 AUAU) of the solar system before planets began t
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Hydrogen and helium compounds.

Explanation:

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3 years ago
An object is placed 4.0 cm to the left of a convex lens with a focal length of +8.0 cm . Where is the image of the object?
Serjik [45]

The image of the object is 8cm to the left of the lens (D)

<h3></h3>

What is the image of an object?

The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.

It is calculated thus:

1÷v = 1÷f - 1÷u

<h3>How to calculate the image of an object</h3>

From the formula

1÷v = 1÷f - 1÷u

<h3>Where </h3>

V = image distance fromthe object

U = object

f = focal length

Substitute the values

1÷v = 1÷8 - 1÷ 4

1÷v = - 1÷8

Make v the subject of formula

v = -8cm

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Europa, a satellite of Jupiter, is believed to have a liquid ocean of water (with a possibility of life) beneath its icy surface
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Answer:

so maximum velocity for walk on the surface of europa is  0.950999 m/s

Explanation:

Given data

legs of length r =  0.68 m

diameter = 3100 km

mass = 4.8×10^22 kg

to find out

maximum velocity for walk on the surface of europa

solution

first we calculate radius that is

radius = d/2 = 3100 /2 = 1550 km

radius = 1550 × 10³ m

so we calculate no maximum velocity that is

max velocity = √(gr)    ...............1

here r is length of leg

we know g = GM/r²   from universal gravitational law

so G we know 6.67 × 10^{-11} N-m²/kg²

g = 6.67 × 10^{-11} ( 4.8×10^22 ) / ( 1550 × 10³ )

g = 1.33 m/s²

now

we put all value in equation 1

max velocity = √(1.33 × 0.68)

max velocity = 0.950999 m/s

so maximum velocity for walk on the surface of europa is  0.950999 m/s

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