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3 years ago

13

A street lamp weighs 150N. It is supported by two wires that form an angle of 120° with each other. The tensions in each wire ar

e equal. What is the tension in each wire?

1 answer:

____ [38]3 years ago

4 0

Answer:

60

so you take 120÷2 wires

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

3 years ago

Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.8 x 105 kg on springs that hav

Vikki [24]

Answer:

The force constant is $k =1.316 *10^{7} \ N/m$

The energy stored in the spring is $E = 1.68 *10^{7} \ J$

Explanation:

From the question we are told that

The mass of the object is $M = 4.8*10^{5} \ kg$

The period is $T = 1.2 \ s$

The period of the spring oscillation is mathematically represented as

$T =2 \pi \sqrt{ \frac{M}{k}}$

where k is the force constant

So making k the subject

$k = \frac{4 \pi ^2 M }{T^2}$

substituting values

$k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}$

$k =1.316 *10^{7} \ N/m$

The energy stored in the spring is mathematically represented as

$E = \frac{1}{2} k x^2$

Where x is the spring displacement which is given as

$x = 1.6 \ m$

substituting values

$E = \frac{1}{2} (1.316 *10^{7}) (1.6)^2$

$E = 1.68 *10^{7} \ J$

7 0

3 years ago

Which of the following is not a use for a weather radar?

Elodia [21]

Weather radar is used to locate precipitation<span>, calculate its motion, and estimate its type rain, snow, </span>hail etc

so, d.measuring temperature <span>is not a use for a weather radar

</span>effect of global warming-<span>heavier rainfall and flooding</span>

5 0

3 years ago

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How many minutes of daylight do we gain after winter solstice

serg [7]

Answer:

Depends on which hemisphere you are belong to and how much distance you are away from Ecuador.

Explanation:

Minutes of daylight is equal on everywhere only on the equinox days (21 March and 23 September). On other days it depends on the place that you are belong to. On winter solstice, places on Ecuador have 12 hours daylight. North side of Ecuador have less, south side of Ecuador have more hour of daylight.

4 0

3 years ago

A second order measuring system has a known natural frequency of 2000 Hz and damping ratio of 0.8. Would it be satisfactory to u

Anna007 [38]

Answer:

The expected dynamic error is 0.019

The phase shift is -23.10°C

Explanation:

The explanation is shown on the first uploaded image

7 0

3 years ago