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ankoles [38]
2 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]2 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]2 years ago
4 0

Answer:

15 ft

Explanation:

took the test got it right

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A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
2 years ago
A uniform magnetic field makes an angle of 30o with the z axis. If the magnetic flux through a 1.0 m2 portion of the xy plane is
Irina18 [472]

Answer:

(b) 10 Wb

Explanation:

Given;

angle of inclination of magnetic field, θ = 30°

initial area of the plane, A₁ = 1 m²

initial magnetic flux through the plane, Φ₁ =  5.0 Wb

Magnetic flux is given as;

Φ = BACosθ

where;

B is the strength of magnetic field

A is the area of the plane

θ is the angle of inclination

Φ₁ = BA₁Cosθ

5 = B(1 x cos30)

B = 5/(cos30)

B = 5.7735 T

Now calculate the magnetic flux through a 2.0 m² portion of the same plane

Φ₂ = BA₂Cosθ

Φ₂ = 5.7735 x 2 x cos30

Φ₂ = 10 Wb

Therefore, the magnetic flux through a 2.0 m² portion of the same plane is is 10 Wb.

Option "b"

3 0
2 years ago
A plane took 7 hours to travel 2020 km. For the first 4 hours,
kicyunya [14]

Answer:

-6.6 km/h

Explanation:

In 7hr plane travelled 2020km;

For the first 4hr the average speed was 310km/h;

d=st, s=d/t;

Distance covered in first 4h is d = 310km/h×4h = 1240km;

See the image attached for further solution

8 0
2 years ago
Read 2 more answers
In an isolated system, the total heat given off by warmer substances equals the total heat energy gained by cooler substances. N
galina1969 [7]

Answer:

The temperature of the cooler substance was close to the room temperature. Therefore, the system experienced less change

Explanation:

Generally, in a closed system containing two bodies at different temperatures, there is a flow of heat energy from the body at a higher temperature to the body at a lower temperature. The effect is more significant when there is a large temperature difference between the bodies. However, if the temperature difference is small or insignificant, the change will be less.

3 0
2 years ago
A 15-g bullet moving at 300 m/s passes through a 2.0 cm thick sheet of foam plastic and emerges with a speed of 90 m/s. Let's as
Shkiper50 [21]

Answer:

Explanation:

a) Change in momentum, Δp = mΔv = m(v - u) = (15 * 10-3) * (90 - 300) = -3.15 kg-m2

b) Acceleration of the bullet, a = (v2 - u2) / 2s = (902 - 3002) / (2 * 0.02) = -2047500 m/s2

So, the bullet is in contact with the plastic for the time,  \bigtriangleupt = \frac{(v - u)}{a} =\frac{(90 - 300)}{(-2047500)} = 1.03 \times 10^{-4} s

c) Average force, F_{avg} =\frac{\bigtriangleup p}{\bigtriangleup t} =\frac{(-3.15)}{(1.03 \times 10^{-4})} = 30.6 kN

8 0
2 years ago
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