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Mkey [24]
3 years ago
7

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tension in the rope and the angle the rope makes with the horizontal

Physics
1 answer:
tatuchka [14]3 years ago
7 0
The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N
the force of the wind F, acting horizontally, with intensity
F=12 N
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
T \cos \alpha -F=0
T \sin \alpha -W=
By dividing the second equation by the first one, we get
\tan \alpha =  \frac{W}{F}= \frac{24.5 N}{12 N}=2.04
From which we find
\alpha = 63.8 ^{\circ}
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N




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Answer:

The kinetic energy lost in the collision is 48 J

Explanation:

Given;

mass of the first ball, m₁ = 2.0 kg

mass of the second ball, m₂ = 6.0 kg

initial speed of the first ball, u₁ = 12 m/s

initial speed of the second ball, u₂ = 4 m/s

let v be the final velocity of the two balls after the inelastic collision

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2 x 12 + 6 x 4 = v(2 + 6)

48 =  8v

48 / 8 = v

v = 6 m/s

The initial kinetic energy of the balls is calculated as;

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(2)(12²) + ¹/₂(6)(4)²

K.E₁ = 144 + 48

K.E₁ = 192 J

The final kinetic of the balls is calculated as;

K.E₂ = ¹/₂(m₁ + m₂)(v²)

K.E₂ = ¹/₂(2 + 6)(6²)

K.E₂ = ¹/₂(8)(6²)

K.E₂ = 144 J

The lost in kinetic energy of the balls is K.E₂ - K.E₁ = 144 J - 192 J = -48 J

Therefore, the kinetic energy lost in the collision is 48 J

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3 years ago
An 85.0 kg fisherman jumps from a dock at a speed of 4.30 m/s onto their 135.0 kg boat. If the boat was at rest to begin but mov
jeka94

Answer:

Final speed of boat + man is 1.66 m/s

Explanation:

As we know that there is no friction on the system or there is no external force on this system

So here we can use momentum conservation here

mv = (m + M)v_f

so we have

m = 85 kg

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now we have

85 \times 4.30 = (85 + 135) v

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2 years ago
You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acce
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The time needed for the 7th car to pass is 13.2 s

Explanation:

The motion of the train is a uniformly accelerated motion, therefore we can use suvat equations.

We start by analzying the motion of the first car, by using the equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered by the first car in a time t, which corresponds to the length of one car

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The equation can be rewritten as

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The same equation can be written considering the first 7 cars:

7L = ut+\frac{1}{2}at'^2

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7L is the distance covered by the 7 cars

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We still have

u = 0

And the acceleration is constant so it is

a=0.08L

Substituting into the equation, we can find t':

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In attachment the graph of the distance covered versus the time taken: since the motion is uniformly accelerated, the relationship between the two variables is quadratical.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

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#LearnwithBrainly

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Answer:

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