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hram777 [196]
4 years ago
15

(1 point) A horizontal clothesline is tied between 2 poles, 20 meters apart. When a mass of 5 kilograms is tied to the middle of

the clothesline, it sags a distance of 2 meters. What is the magnitude of the tension on the ends of the clothesline

Physics
1 answer:
sweet-ann [11.9K]4 years ago
4 0

Answer:

Ft = 50.25 N

Explanation:

As you can see in the attached diagram, the angle θ is given by

tanθ = opp/adj

The opposite is the sag of distance 2 m and the adjacent is the half of the horizontal clothesline that is 20 m.

tanθ = 2/10

θ = tan⁻¹(2/10)

θ = 11.30°

There are two forces along the y axis, one is the F = mg acting downwards and the other is the Ftsinθ component of tension force in the string.

The sum of forces along y-axis is

5Ftsin(11.30) - mg = 0

Ft = mg/5sin(11.30)

Ft = 5*9.8/5sin(11.30)

Ft = 9.8/0.195

Ft = 50.25 N

Therefore, the magnitude of the tension on the ends of the clothesline is 50.25 N.

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The atomic number is equivalent to an atom's
Helen [10]

Answer:

Amount of Protons

Explanation:

If this is what you are asking for, the atomic number is equal to the amount of protons, while the atomic mass is equal to the amount of protons *plus* neutrons.

3 0
3 years ago
. Emily pushes a 38.8 kg grocery cart of groceries by exerting a 76.0 N force on the handle inclined at 40.0 degrees below the h
Lana71 [14]

Answer:

a) F_{x} = 58.2 N

F_{y} = 48.9 N

b) a = 1.5 m/s²

Explanation:

a) The horizontal and vertical components of Emily's force can be found knowing the angle and the exerted force.

Since the handle is inclined at 40.0° below the horizontal we have:

F_{x} = |F|*cos(\theta) = 76.0 N*cos(40) = 58.2 N

F_{y} = |F|*sin(\theta) = 76.0 N*sin(40) = 48.9 N

b) The acceleration of the car can be calculated as follows:

F_{x} = ma        

We used the horizontal component of the force because the cart is moving in that direction.

a = \frac{F_{x}}{m} = \frac{58.2 N}{38.8 kg} = 1.5 m/s^{2}

Hence, the acceleration of the car is 1.5 m/s².    

I hope it helps you!

3 0
3 years ago
Two crates, of mass m1 = 71 kg and m2 = 130 kg , are in contact and at rest on a horizontal surface. Force F = 610 N is exerted
Anvisha [2.4K]

Answer: Respective answers are 1.27 m/s^2 and 90.17 N

Explanation:

Explanation requires basic knowledge of forces in physics. The solutions are attached in this response. To summarize:

- Newton's Second Law -> F = ma

- Force of friction = Fn (normal force) x µk (coefficient of kinetic friction)

- Because the surface is leveled, Fn = Fg, Fg = mg

- The crates exert the same amount of force on each other because of the rule of action // reaction (Newton's Third Law)

Download pdf
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>
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3 0
4 years ago
An electron moves with velocity v⃗ =(5.8i−6.7j)×104m/s in a magnetic field B⃗ =(−0.81i+0.60j)T.
Minchanka [31]

Answer:

Fₓ = 0,  F_{y} = 0  and  F_{z}<em> = - 3.115 10⁻¹⁵   N</em>

Explanation:

The magnetic force given by the expression

       F = q v xB

the bold are vectors,  the easiest analytical way to determine this force in solving the determinant

   F = q \left[\begin{array}{ccc}i&j&k\\5.8&-6.7&0\\-0.81&0.6&0\end{array}\right]  10^{4}

   F = 1.6 10⁻¹⁵ [ i( 0-0) + j (0-0) + k^( 5.8 0.60 - 0.81 67) ]

   F =i^0 + j^0   - k^  3.115 10⁻¹⁵   N

   

Fₓ = 0

F_{y} = 0

F_{z}<em> = - 3.115 10⁻¹⁵   N</em>

6 0
4 years ago
Explain why it is not advisable to use small values of i in performing triangular glass prism experiment
kifflom [539]
Is this about using triangular prisms in spectrometer experiments ? See enclosed

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