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2 years ago

9

Two 25.0N weights are suspended at opposite ends of a rope that passes over a frictionless pulley. What is the tension in the ro

pe?

1 answer:

goldenfox [79]2 years ago

8 0

Answer:

tension in rope = 25.0 N

Explanation:

Two forces act on the suspended weight. The force coming down is the gravitational force and the upward force by the tension in the rope.

Since the suspended weight is not accelerating so that the net force will be zero. Therefore the tension in the rope should be 25 N.

∑F = F - W = 0

so

F = W

so tension in rope = F = T = 25 N

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(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.

faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let call "x₁" distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁ ( force between earth and the object )

F₁ = K * Mt * m₀/ ( x₁)² K is a gravitational constant

F₂ (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂ K*Mt*m₀ / x₁² = K*Ml*m₀ /x₂²

Or simplifying the expression

Mt/ x₁² = Ml/ x₂²

We know that x₁ + x₂ = 384000 Km then

x₁ = 384000 - x₂

Mt/( 384000 - x₂)² = Ml / x₂²

Mt * x₂² = Ml *( 384000 - x₂)²

We need to solve for x₂

Mt * x₂² = Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂² = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂² = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂² + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂ = -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 254880*10∧10 / 1160

x₂ ₁ = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁ = -56.43*10∧5 + 508* 10∧5 / 1160

x₂ ₁ = 451.27*10∧5/1160

x₂ ₁ = 4512.7*10∧4 /1160

x₂ ₁ = 3.89*10∧4 km (distance between the moon and the object)

x₂ ₁ = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁ = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

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3 years ago

А A pool of water of refractive index

babymother [125]

Answer:

Apparent depth = 45 cm

Explanation:

The refractive index of water in a pool, n = 4/3

Real depth, d = 60 cm

We need to find its apparent depth when viewed vertically through air. The ratio of real depth to the apparent depth is equal to the refractive index of the material. Let the apparent depth is d'. So,

$n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{60}{\dfrac{4}{3}}\\\\d'=45\ cm$

So, the apparent depth is 45 cm.

3 0

2 years ago

Atomic math challenge will give brainly and thanks

nevsk [136]

Answer:

1. Hydrogen

Atomic # = 1

Atomic Mass = 1.00794 ( If you round it it's 1.008 )

# of protons = 1

# of neutrons = none

# of electrons = 1

8 0

3 years ago

Read 2 more answers

Arocket launches at an angle of 33.6 degrees from the horizontal at a

babymother [125]

Answer:

Y component = 32.37

Explanation:

Given:

Angle of projection of the rocket is, $\theta=33.6$

Initial velocity of the rocket is, $u=58.5$

A vector at an angle $\theta$ with the horizontal can be resolved into mutually perpendicular components; one along the horizontal direction and the other along the vertical direction.

If a vector 'A' makes angle $\theta$ with the horizontal, then the horizontal and vertical components are given as:

$A_x=A\cos \theta(\textrm{Horizontal or X component})\\A_y=A\sin \theta(\textrm{Vertical or Y component})$

Here, as the velocity is a vector quantity and makes an angle of 33.6 with the horizontal, its Y component is given as:

$u_y=u\sin \theta$

Plug in the given values and solve for $u_y$ . This gives,

$u_y=(58.5)(\sin 33.6)\\u_y=58.5\times 0.55339\\u_y=32.373\approx32.37(\textrm{Rounded to two decimal places})$

Therefore, the Y component of initial velocity is 32.37.

4 0

2 years ago

Does a mirror work because of refraction

Semenov [28]

No. A mirror works because of reflection.

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3 years ago

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