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alekssr [168]
3 years ago
11

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

angle of elevation of the kite (the angle between the ground where the boy is standing and the kite string)?

Physics
1 answer:
vagabundo [1.1K]3 years ago
3 0
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)

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IrinaVladis [17]

Answer:

A) T1 = 269.63 K

T2 = 192.59 K

B) W = -320 KJ

Explanation:

We are given;

Initial volume: V1 = 7 m³

Final Volume; V2 = 5 m³

Constant Pressure; P = 160 KPa

Mass; m = 2 kg

To find the initial and final temperatures, we will use the ideal gas formula;

T = PV/mR

Where R is gas constant of helium = R = 2.0769 kPa.m/kg

Thus;

Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K

Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K

B) world one is given by the formula;

W = P(V2 - V1)

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A box that weighs 5.00×10^2 N is sliding down a ramp at a constant speed. The angle the ramp makes with the horizontal is 25°. W
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0.466 (3 sig. fig.)

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A gas bottle contains 0.650 mol of gas at 730 mm Hg pressure. If the final pressure is 1.15 atm, how many moles of gas were adde
AlekseyPX

Answer:

0.779 mol

Explanation:

Since the gas is in a bottle, the volume of the gas is constant. Assuming the temperature remains constant as well, then the gas pressure is proportional to the number of moles:

p \propto n

so we can write

\frac{p_1}{n_1}=\frac{p_2}{n_2}

where

p1 = 730 mm Hg = 0.96 atm is the initial pressure

n1 = 0.650 mol is the initial number of moles

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n2 is the final number of moles

Solving for n2,

n_2 = n_1 \frac{p_2}{p_1}=(0.650 mol)\frac{1.15 atm}{0.96 atm}=0.779 mol

6 0
3 years ago
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