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SCORPION-xisa [38]
3 years ago
8

A string attached to a kite was maintained at an angle of 65.0° with the ground. If 120 m of string was reeled in to return the

kite back to the ground, what was the vertical height of the kite? (Assume the string was straight and did not sag)
Physics
1 answer:
Anni [7]3 years ago
4 0

Answer:

The hypotenuse is measured at 120 meters of string, and you need to solve for the leg of the triangle that is horizontal. The degree is 40, so use trigonometry to figure it out.

Cosin (40) is equal to around .766

Adjacent/Hypotenuse

x/120 = cos40

Answer: 91.92533.

If you use 3 significant figures it should be 91.9 meters.

Explanation:

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The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of g
Morgarella [4.7K]

Here is the full question:

The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:  

k=\sqrt{\frac{I}{M} }

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.

Answer:

a) 0.85 m

b) 0.98 m

c) 0.76 m

Explanation:

Given that: the radius of gyration  k=\sqrt{\frac{I}{M} }

So, moment of rotational inertia (I) of a cylinder about it axis = \frac{MR^2}{2}

k=\sqrt{\frac{\frac{MR^2}{2}}{M} }

k=\sqrt{{\frac{MR^2}{2}}* \frac{1}{M} }

k=\sqrt{{\frac{R^2}{2}}

k={\frac{R}{\sqrt{2}}

k={\frac{1.20m}{\sqrt{2}}

k = 0.8455 m

k ≅ 0.85 m

For the spherical shell of radius

(I) = \frac{2}{3}MR^2

k = \sqrt{\frac{\frac{2}{3}MR^2}{M}  }

k = \sqrt{\frac{2}{3} R^2}

k = \sqrt{\frac{2}{3} }*R

k = \sqrt{\frac{2}{3}}  *1.20

k = 0.9797 m

k ≅ 0.98 m

For the solid sphere of  radius

(I) = \frac{2}{5}MR^2

k = \sqrt{\frac{\frac{2}{5}MR^2}{M}  }

k = \sqrt{\frac{2}{5} R^2}

k = \sqrt{\frac{2}{5} }*R

k = \sqrt{\frac{2}{5}}  *1.20

k = 0.7560

k ≅ 0.76 m

6 0
4 years ago
A cart is given an initial velocity of 5.0 m/s and
larisa86 [58]

Answer:

cart displacement is 66 m

Explanation:

given data

velocity = 5 m/s

acceleration = 2 m/s²

time = 6 s

to find out

What is the

magnitude of cart displacement

solution

we will apply here equation of motion to find displacement that is

s = ut + 0.5×at²    .............1

here s id displacement and u is velocity and a is acceleration and time is t here

put all value in equation 1

s = ut + 0.5×at²

s = 5(6) + 0.5×(2)×6²

s = 66

so cart displacement is 66 m

5 0
4 years ago
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