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Pani-rosa [81]
1 year ago
8

To support a tree damaged in a storm, a 12-foot wire is secured from the ground to the tree at a point 10 feet off the ground. T

he tree meets the ground at a right angle. At approximately what angle does the wire meet the ground?
A)33.6°
B)39.8°
C)50.2°
D)56.4°
Physics
1 answer:
podryga [215]1 year ago
6 0

Answer:

The wire meet the ground at an angle of 56.4 degrees

Explanation:

It is given that,

To support a tree damaged in a storm, a 12-foot wire is secured from the ground to the tree at a point 10 feet off the ground.

The hypotenuse is, H = 12 foot

The perpendicular distance is, P = 10 feet

The angle between the tree and the ground is 90 degrees

Using Pythagoras theorem as :

sin\theta=\dfrac{P}{H}

sin\theta=\dfrac{10}{12}

\theta=56.4^{\circ}

So, the wire meet the ground at an angle of 56.4 degrees. Hence, the correct option is (d).                                                    

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Answer:

0.0059

Explanation:

According to the question the seismic activity density is given by

\text{Seismic activity density}=\frac{\text{Number of Earthquakes over a given time span}}{\text{The land area affected}}

Here,

Number of Earthquakes over a given time span = 424

The land area affected = 71300 mi²

So,

\text{Seismic activity density}=\frac{424}{71300}\\\Rightarrow \text{Seismic activity density}=0.0059

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1 year ago
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Explanation:

Given

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a=\frac{\mathrm{d} v}{\mathrm{d} t}

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1 year ago
A 1-m-long monopole car radio antenna operates in the AM frequency of 1.5 MHz. How muchcurrent is required to transmit 4 W of po
Zanzabum

Answer:

The current needed to transmit Power of 4 W is 28.47 A

Solution:

As per the question:

Length of the antenna, L_{a} = 1 m

Frequency, \vartheta = 1.5 MHz = 1.5\times 10^{6} Hz

Power transmitted, P_{t} = 4 W

Now,

For a monopole antenna:

\lambda_{a} = \frac{c}{\vartheta}

where

\lambda_{a} = wavelength transmitted by the antenna

c = speed of light in vacuum

\lambda_{a} = \frac{3\times 10^{8}}{1.5\times 10^{6}} = 200 m

Now,

Since, the value of \lambda_{a} >> L_{a} thus the monopole is a Hertian monopole.

The resistance is calculated as:

R = \frac{1}{2}(\frac{dL_{a}}{\lambda_{a}})^{2}\times 80\pi^{2}

R = \frac{1}{2}(\frac{1}{200)^{2}\times 80\pi^{2} = 9.869\times 10^{- 3} = 9.869 m\Omega

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2 years ago
The robot arm is elevating and extending simultaneously. At a given instant, θ = 30°, ˙ θ = 10 deg / s = constant θ˙=10 deg/s=co
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Explanation:

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