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Elena-2011 [213]
3 years ago
6

A radio station tower was built in two sections. From a point 87 feet from the base of the tower, the angle of elevation of the

top of the first section is 25 degrees, and the angle of the top of the second section is 40 degrees. to the nearest foot, what is the height of the top section of the tower?
Physics
1 answer:
Fiesta28 [93]3 years ago
3 0

Answer:

tan theta = H / B        where H is height of triangle and B the length of base

H1 = 87 * tan 25 =  40.6 feet

H2 = 87 * tan 40 = 73 feet

H2 - H1 = 32.4 feet

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Use the circuit diagram to decide if the lightbulb will light. Justify your answer.
podryga [215]

Answer:

The lightbulb will NOT light.

Explanation:

You put me in a difficult position.  I can't help it, but the "sample answer" is by far the best way to explain this, briefly and correctly.  There's no other choice but to copy it.

This is a short circuit. The branch without the bulb has almost no resistance, so all the current will flow through that branch instead of flowing through the bulb.

<em>If</em> the lower switch were <u>opened</u>, THEN we would have a series circuit.  Current would no longer have any other choice but to flow through the bulb, and the bulb would light.

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3 years ago
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rusak2 [61]

Answer:

9.6

Explanation:

to convert km to miles multiply by 1.609

7 0
4 years ago
physics A river flows at a speed vr = 5.37 km/hr with respect to the shoreline. A boat needs to go perpendicular to the shorelin
vova2212 [387]

Answer: Vb is the vector  (-5.37m/s,  8.59 m/s), with a module 10.13m/s

then the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9

Explanation:

We can use the notation (x, y) where the river flows in the x-axis and the pier is on the y-axis.

We have Vr = (5.37m/s, 0m/s)

Now, if the boat wants to move only along the y-axis (perpendicularly to the shore).

The velocity of the boat Vb will be:

Vb = (-c*sin(32). c*cos(32))

Then we should have that:

5.37 m/s - c*sin(32) = 0

c = (5.37/sin(32))m/s = 10.13 m/s

the velocity in the y-axis is:

10.13m/s*cos(32) = 8.59 m/s

So Vb = (-5.37m/s,  8.59 m/s)

the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9 where i used Vb as the module of the boat's velocity.

7 0
3 years ago
Help me with this question, please
Lunna [17]

Answer:

First one is voltage

second is resistance

third is electric current

hope this helped!!

5 0
3 years ago
1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t
tiny-mole [99]

Explanation:

Charges,q_1=8\ \mu C=8\times 10^{-6}\ C

q_2=-5\ \mu C=-5\times 10^{-6}\ C

The distance between charges, r = 10 cm = 0.1 m

We need to find the magnitude and direction of the electric force. It is given by :

F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\\\F=36\ N

So, the required force between charges is 36 N and it is towards positive charge i.e. +8 μC.

6 0
3 years ago
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