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svlad2 [7]
4 years ago
13

A mouse spots an owl flying 10.1 feet over a nearby bush. The mouse is 30 feet away from the bush, and it sees the owl at a cert

ain angle of elevation. Which trigonometric equation can be used to solve for x, the angle of elevation from the mouse to the owl?

Physics
1 answer:
Anna11 [10]4 years ago
6 0

Answer:

Explanation:

Given

fly is at height of h=10.1\ ft

mouse is at a distance of 30 ft from bush

if \theta  is the angle of elevation then

using trigonometric relation

\tan \theta =\frac{h}{x}

\tan \theta =\frac{10.1}{30}

\theta =tan^{-1}(\frac{10.1}{30})

\theta =18.57^{\circ}

           

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An electric light is plugged into a 120-V outlet. If the current in the bulb is 0.50 A, how much electrical energy does the bulb
ioda

Answer:

= 54,000 Joules or 54 kJ

Explanation:

Electrical energy is given by the formula;

E = VIt; where V is the potential difference in volts, I is the current and t is the time in seconds.

Therefore;

Electrical energy = 120 V × 0.50 A × 15 ×60 seconds

                            = 54,000 Joules

Thus; the electrical energy is 54,000 joules or 54 kJ

7 0
3 years ago
true or false Both the large loose rocks and the small loose rocks used to be part of earth's solid rock layer
salantis [7]
Hello Micu212006 


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8 0
3 years ago
You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac
dusya [7]

Answer:

20.62361 rad/s

489.81804 J

Explanation:

I_i = Initial moment of inertia = 9.3 kgm²

I_f = Final moment of inertia = 5.1 kgm²

\omega_i = Initial angular speed = 1.8 rev/s

\omega_f = Final angular speed

As the angular momentum of the system is conserved

I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

6 0
3 years ago
Spring compressed 10cm by 100N force and held in place with Pin. Pin is pulled and block is pushed Up the incline. Uk(coefficien
otez555 [7]

The compression of 10 cm by a 100 N force on the plane that has a

coefficient of friction of 0.39 give the following values.

  • The velocity of the block after the Spring extends 7 cm is approximately 1.73 m/s
  • The height at which the block stops rising is approximately 1.1415 m
  • The length of the incline is approximately 1.536 m

<h3>How can the velocity and height of the block be calculated?</h3>

Mass of the block, m = 3 kg

Spring \ constant, K = \dfrac{100 \, N}{0.1 \, m}  = \mathbf{ 1000\, N/m}

Coefficient of kinetic friction, \mu_k = 0.39

Therefore, we have;

Friction force = \mathbf{\mu_k}·m·g·cos(θ)

Which gives;

Friction force = 0.39 × 3 × 9.81 × cos(48°) ≈ 7.68

Work done by the motion of the block, <em>W</em> ≈ 7.68 × d

The work done = The kinetic energy of the block, which gives;

\mathbf{\dfrac{1}{2} \times k \cdot x^2 }= 7.68 \cdot d

The initial kinetic energy in the spring is found as follows;

K.E. = 0.5 × 1000 N/m × (0.1 m)² = 5 J

The initial velocity of the block is therefore;

5 = 0.5·m·v²

v₁ = √(2 × 5 ÷ 3) ≈ 1.83

Work done by the motion of the block, <em>W</em> ≈ 7.68 N × 0.07 m ≈ 0.5376 J

Chane in kinetic energy, ΔK.E. = Work done

ΔK.E. = 0.5 × 3 × (v₁² - v₂²)

Which gives;

ΔK.E. = 0.5 × 3 × (1.83² - v₂²) = 0.5376

Which gives;

  • The velocity of the block after the Spring extends 7 cm, v₂ ≈ <u>1.73 m/s</u>

The height at which the block will stop moving, <em>h</em>, is given as follows;

At \ the \ maximum \ height, \ h, \ we \ have ; \  \dfrac{1}{2} \times 1000 \times 0.1^2 = 7.68 \times x

Which gives;

Length \ of \ the \ incline \ at \ maximum \ height, \ x_{max} =\dfrac{  7.68 }{ \dfrac{1}{2} \times 1000 \times 0.1^2  } \approx 1.536

The distance up the inclined, the block rises, at maximum height is therefore;

x_{max} ≈ 1.536 m

Therefore;

h = 1.536 × sin(48°) ≈ 1.1415

  • The height at which the block stops rising, h ≈ <u>1.1415 m</u>

From the above solution for the height, the length of the incline is he

distance along the incline at maximum height which is therefore;

  • Length of the incline, x_{max} = 1.536 m

Learn more about conservation of energy here:

brainly.com/question/7538238

5 0
2 years ago
The change in pitch of a train's horn as it passes while you are standing still can be explained by
alexgriva [62]

The change in pitch of a train's horn as it passes while you are
standing still can be described by the Doppler effect, but that
doesn't explain it.

8 0
3 years ago
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