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Elena-2011 [213]
2 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]2 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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Alfred Wegener

Explanation:

Alfred Wegener is a german meteorologist who proposed the theory that the continents drifted, and he presented it to the German Geological Society on January 1912.

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What type of energy transfer occurs through light or electromagnetic waves?
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Read 2 more answers
A 24 cm radius aluminum ball is immersed in water. Calculate the thrust you suffer and the force. Knowing that the density of al
Illusion [34]

Answer:

W =1562.53 N

Explanation:

It is given that,

Radius of the aluminium ball, r = 24 cm = 0.24 m

The density of Aluminium, d=2698.4\ kg/m^3

We need to find the thrust and the force. The mass of the liquid displaced is given by :

m=dV

V is volume

Weight of the displaced liquid

W = mg

W=dVg

So,

W=dg\times \dfrac{4}{3}\pi r^3\\\\W=2698.4\times 10\times \dfrac{4}{3}\times \pi \times (0.24)^3\\\\W=1562.53\ N

So, the thrust and the force is 1562.53 N.

7 0
3 years ago
Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12
PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

F_e = k\frac{q_1q_2}{r^2}

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2}  \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}

(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}

Solving the quadratic equation:

q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C

for this values q_1 wil be:

q_1_o =  -1.325 *10^{-6}C | 4.17*10^{-6}C

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0
3 years ago
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