(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
1 answer:
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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Answer:
a)
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:
b) The acceleration of the car can be calculated as follows:
We used the horizontal component of the force because the cart is moving in that direction.
Hence, the acceleration of the car is 1.5 m/s².
I hope it helps you!
Answer:
Fₓ = 0, = 0 and
<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 = 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
= 0
<em> = - 3.115 10⁻¹⁵ N</em>
Is this about using triangular prisms in spectrometer experiments ? See enclosed
