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Nadusha1986 [10]
2 years ago
5

Will give correct answer brainliest

Physics
1 answer:
Sav [38]2 years ago
4 0

Answer:

.....

Explanation:

i. sorry? i really hope this helps!

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A boy flying a kite is standing 30 ft from a point directly under the kite. if the string to the kite is 50 ft long, what is the
vagabundo [1.1K]
Draw a diagram to illustrate the problem as shown in the figure below.
h =  height of the kite above ground.

By definition, the angle of elevation is
cos \theta = \frac{30}{50} =0.6
Therefore
\theta = cos^{-1} 0.6 = 53.1 \, deg.

Answer: 53° (nearest integer)

3 0
3 years ago
A ramp is used to load furniture onto a moving truck. The person does 1240 J of work pushing
Len [333]

Answer:

The efficiency of the ramp is, Eff = 6.63 %

Explanation:

Given,

The work done by the person pushing the furniture up the ramp is, W₁ = 1240 J

The work done by the ramp is, W₀ = 822 J

The efficiency of the ramp is given by the formula,

                                 <em> Eff = ( W₀ / W₁ ) x 100%</em>

                                        = ( 822 / 12400 ) x 100%

                                        = 6.63 %

Hence, the efficiency of the ramp is, Eff = 6.63 %

6 0
3 years ago
A batter swings and hits a pitched baseball far over the left field wall. at the moment the baseball contacts the bat, which obj
Lyrx [107]
The baseball bat not the baseball
7 0
3 years ago
Which of the following is a subsurface event takes place during the rock cycle
alukav5142 [94]

Answer:

The answer is A. Cementing...

Explanation:

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4 0
3 years ago
Read 2 more answers
A 2.30-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. a solid sph
Marina86 [1]

Solution:


initial sphere mvr = final sphere mvr + Iω 
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m² 
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω 
where: ω = 2.87 rad/s 

So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)² 
E = 12.64 J becomes PE = mgh, so 
12.64 J = 2.3 kg * 9.8m/s² * h 
h = 0.29 m 

h = L(1 - cosΘ) → where here L is the distance to the CM 
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ 
Θ = arccos((1-0.29)/1) = 44.77 º 

8 0
3 years ago
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