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ankoles [38]
3 years ago
10

Emir is standing in a treehouse and looking down at a swingset in the yard next door. The angle of depression from Emir's eyelin

e to the swingset is 33.69°, and Emir is 10 feet from the ground. How many feet is the base of the tree from the swingset? Round your answer to the nearest foot.
Physics
2 answers:
lara [203]3 years ago
4 0

Answer:

15 ft

Explanation:

We use tangent of angles. Let Ф = angle of depression of Emir's line of sight = angle of elevation of base of swingset from Emir = 33.69°. Let x = height of tree = 10 ft and y = distance of base of tree from swingset.

So tanФ = x/y and y = x/tanФ

y = x/tanФ

= 10 ft/tan33.69°

= 15 ft

nlexa [21]3 years ago
4 0

Answer:

15 ft

Explanation:

took the test got it right

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Answer:

B

Explanation:

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3 years ago
(a) Calculate the buoyant force on a 2.00-L helium balloon.
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Answer:

0.0239364 N

0.0057879 N

Explanation:

\rho = Density of the gas

g = Acceleration due to gravity = 9.81 m/s²

V = Volume

Mass of rubber = 1.5 g

Buoyant force is given by

F_b=\rho gV\\\Rightarrow F_b=1.22\times 9.81\times 2\times 10^{-3}\\\Rightarrow F_b=0.0239364\ N

The buoyant force is 0.0239364 N

Net vertical force is given by

F_n=F_b-W_{He}-W_{r}\\\Rightarrow F_n=0.0239364-0.175\times 2\times 10^{-3}\times 9.81-1.5\times 10^{-3}\times 9.81\\\Rightarrow F_n=0.0057879\ N

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garik1379 [7]
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4 0
3 years ago
A 600 N force acts on an object with a mass of 50 kg. What is the resulting acceleration of the object?
DochEvi [55]

Answer:

<h3>The answer is 12 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{600}{50}  =  \frac{60}{5}  \\

We have the final answer as

<h3>12 m/s²</h3>

Hope this helps you

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3 years ago
You move a 2.5 kg book from a shelf that is 1.2 m above the ground to a shelf that is 2.6 m above the ground. What is the change
Sophie [7]
The change in potential energy of an object is given by
U=mg \Delta h
where
m is the mass of the object
g is the gravitational acceleration
\delta h is the increase in altitude of the object

In our problem, m=2.5 kg is the mass of the book, g=9.81 m/s^2 and 
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U=mg\Delta h=(2.5 kg)(9.81 m/s^2)(1.4 m)=34.3 J
8 0
4 years ago
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