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sdas [7]
3 years ago
12

A painter is placing a ladder to reach the third story window, which is 39 feet above the ground and makes an angle with the gro

und of 70°. How far out from the building does the base of the ladder need to be positioned? Round your answer to the nearest tenth. The base of the ladder needs to be positioned 14.2 feet out from the building.
Physics
2 answers:
tino4ka555 [31]3 years ago
3 0

Answer:

14.2 ft

Explanation:

The ladder form a right angle triangle with the horizontal angle of 70°

the angle faces the wall which is opposite to the ladder, the length of the ladder is the hypotenuse and the distance of the base from the wall is the adjacent

using trigonometrical ratio

tan 70° = opp ( height of the wall) / adjacent, x ( the distance of the base from the wall) = 39 ft / x

x = 39 ft / tan 70° = 14.195 approx 14.2 ft

tangare [24]3 years ago
3 0

Answer:

41.5ft out from the building

Explanation:

The length of the ladder will serve as the hypotenuse of the right angled triangle formed by the set up. The height of the ladder above the ground is 39 feet to the top of the building where the ladder is placed.

Since the angle the ladder make to the base of the ladder is 70°, the opposite side of the set up will serve as the height.

According to SOH CAH TOA

Sin theta = Opposite/Hypotenuse

Sin 70° = 39ft/hypotenuse

Hypotenuse = 39/sin70°

Hypotenuse = 41.5feet

This means that the foot of the ladder must be placed 41.5feet out from the building.

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An inductor is connected to an AC power supply having a maximum output voltage of 3.00 V at a frequency of 280 Hz. What inductan
Sergio [31]

Answer:

L = 0.48 H

Explanation:

let L be the inductance, Irms be the rms current, Vrms be the rms voltage and Vmax  be the maximum voltage and XL be the be the reactance of the inductor.

Vrms = Vmax/(√2)

         = (3.00)/(√2)

         = 2.121 V

then:

XL = Vrms/I  

     = (2.121)/(2.50×10^-3)

     = 848.528 V/A

that is L = XL/(2×π×f)

              = (848.528)/(2×π×(280))

              = 0.482 H

Therefore, the inductance needed to kepp the rms current less than 2.50mA is 0.482 H.

6 0
3 years ago
g Drop the object again and carefully observe its motion after it hits the ground (it should bounce). (Consider only the first b
Anestetic [448]

Answer:

a) quantity to be measured is the height to which the body rises

b) weighing the body , rule or fixed tape measure

c)   Em₁ = m g h

d) deformation of the body or it is transformed into heat during the crash

Explanation:

In this exercise of falling and rebounding a body, we must know the speed of the body when it reaches the ground, which can be calculated using the conservation of energy, since the height where it was released is known.

a) What quantities must you know to calculate the energy after the bounce?

The quantity to be measured is the height to which the body rises, we assume negligible air resistance.

So let's use the conservation of energy

starting point. Soil

          Em₀ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₀ = Em_f

         Em₀ = m g h₀

b) to have the measurements, we begin by weighing the body and calculating its mass, the height was measured with a rule or fixed tape measure and seeing how far the body rises.

c) We use conservation of energy

starting point. Soil

          Em₁ = K = ½ m v²

final point. Higher

          Em_f = U = mg h

         Em₁ = Em_f

         Em₁ = m g h

d) to determine if the energy is conserved, the arrival energy and the output energy must be compared.

There are two possibilities.

* that have been equal therefore energy is conserved

* that have been different (most likely) therefore the energy of the rebound is less than the initial energy, it cannot be stored in the possible deformation of the body or it is transformed into heat during the crash

7 0
3 years ago
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Answer:

a) v_p=9.35m/s

Explanation:

From the question we are told that:

Open cart of mass   M_o=50.0 kg

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