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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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Likurg_2 [28]

Answer:

Yes

Explanation:

Antarctica is usually always cold.

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What do is mean by environment friendly behaviour?​
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Answer:
Being environmentally friendly means having a lifestyle that is better for the environment
5 0
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Middle School Physics 5+3 pts
Murljashka [212]
A) The ball on the small ball is far smaller than the force on the basketball.

B) The total momentum before and after the collision remains constant.

C) We know momentum is conserved so we do:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.1 x 5 + 0.6 x 0 = 0.1 x -4 + 0.6 x v₂
v₂ = 1.5 m/s
5 0
3 years ago
A geosynchronous satellite orbits Earth at a distance of 42,250 km from the center of Earth and has a period of 1 day. What is t
yan [13]

Answer:

The centripetal acceleration of the satellite is a=0.22\ m/s^2.

Explanation:

Given that,

The distance covered by a geosynchronous satellite, d = 42250 km

The time taken by the satellite to covered distance, t = 1 day = 24 hours

Since, 24 hours = 86400 seconds

Let v is the speed of the satellite. It is given by the total distance divided by total time taken such that :

v=\dfrac{d}{t}

v=\dfrac{2\pi d}{t}

v=\dfrac{2\pi\times 42250 \times 10^3}{86400 }

v = 3072.5 m/s

The centripetal acceleration of the satellite is given by :

a=\dfrac{v^2}{d}

a=\dfrac{(3072.5)^2}{42250 \times 10^3}

a=0.22\ m/s^2

So, the centripetal acceleration of the satellite is a=0.22\ m/s^2. Hence, this is the required solution.

7 0
3 years ago
Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to
stiks02 [169]

Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:

"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."

Answer:

\mu_{sB}=0.126

\mu_{sC}=0.168

Explanation:

In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:

Sum of torques:

\sum \tau_{A}=0

N(3m)-W(1.5m)=0

When solving for N we get:

N=\frac{W(1.5m)}{3m}

N=\frac{(1962N)(1.5m)}{3m}

N=981N

Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:

First, the forces in y.

\sum F_{y}=0

-F_{By}+N_{c}=0

F_{By}=N_{c}

Next, the forces in x.

\sum F_{x}=0

-f_{sB}-f_{sC}+P_{x}=0

We can find the x-component of force P like this:

P_{x}=360N(\frac{4}{5})=288N

and finally the torques about C.

\sum \tau_{C}=0

f_{sB}(1.75m)-P_{x}(0.75m)=0

f_{sB}=\frac{288N(0.75m)}{1.75m}

f_{sB}=123.43N

With the static friction force in point B we can find the coefficient of static friction in B:

\mu_{sB}=\frac{f_{sB}}{N}

\mu_{sB}=\frac{123.43N}{981N}

\mu_{sB}=0.126

And now we can find the friction force in C.

f_{sC}=P_{x}-f_{xB}

f_{sC}=288N-123.43N=164.57N

f_{sC}=N_{c}\mu_{sC}

and now we can use this to find static friction coefficient in point C.

\mu_{sC}=\frac{f_{sC}}{N}

\mu_{sC}=\frac{164.57N}{981N}

\mu_{sB}=0.168

3 0
3 years ago
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