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Serga [27]
3 years ago
15

Telephone and electrical lines are allowed to sag between poles so that the tension will not be too great when something hits or

sits on the line. Suppose that a line were stretched almost perfectly horizontally betweern two poles that are 30 m apart. If a 0.25-kilogram bird perches on the wire midway between the poles and the wire sags 1.0 cm, what would the tension in the wire be?
Physics
1 answer:
navik [9.2K]3 years ago
7 0

Answer:

Tension in the wire= 1837.5 N

Explanation:

The answer along with solution can be found in the image attached.

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According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about
NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, r=4\ AU=1.496\times 10^{11}\ m

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

T^2=\dfrac{4\pi^2}{GM}r^3

M is the mass of the sun

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3    

T^2=\sqrt{9.96\times 10^{14}}\ s

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0
3 years ago
A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l
artcher [175]

Answer:

at y=6.29 cm the charge of the two distribution will be equal.

Explanation:

Given:

linear charge density on the x-axis, \lambda_1=8\times 10^{-6}\ C

linear charge density of the other charge distribution, \lambda_2=-6\times 10^{-6}\ C

Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.

Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.

<u>we know, the electric field due to linear charge is given as:</u>

E=\frac{\lambda}{2\pi.r.\epsilon_0}

where:

\lambda= linear charge density

r = radial distance from the center of wire

\epsilon_0= permittivity of free space

Therefore,

E_1=E_2

\frac{\lambda_1}{2\pi.x.\epsilon_0}=\frac{\lambda_2}{2\pi.(0.11-x).\epsilon_0}

\frac{\lambda_1}{x} =\frac{\lambda_2}{0.11-x}

\frac{8\times 10^{-6}}{x} =\frac{6\times 10^{-6}}{0.11-x}

x=0.0629\ m

∴at y=6.29 cm the charge of the two distribution will be equal.

9 0
3 years ago
for a moving object distance covered by it is always greater than or equal to the displacement of the object in a given time. ex
AleksandrR [38]
<h3><u>Answer</u></h3>

  • Distance is equal to the Total Distance covered by a body, from the initial till the final point.

  • Displacement is equal to the shortest distance between two points.

  • So we known that Distance can only be equal to or greater than the displacement and can never be shorter than the displacement.

  • This is just common sense how can anything be shorter than the shortest path itself. But it can be equal to the shortest path
<h3>━━━━━━━━━━━━━━</h3>

<h3><u>Know </u><u>More</u></h3>

☯ Distance is a scalar quantity and has only magnitude but no direction.

☯ Displacement is a vector quantity and has both magnitude and direction.

☯ Distance can only have +ve values whereas displacement can be +ve, -ve or even be zero.

6 0
3 years ago
TIMED! URGENT! REALLY APPRECIATE HELP!! TYSM!!!!!!
Diano4ka-milaya [45]
The answer is 27.03 I just multiplied the two numbers
3 0
3 years ago
Read 2 more answers
Finding the spring constant as shown, spring 3, which has an unknown spring constant k3, replaces spring 2. the mass of the weig
nevsk [136]
Replaces spring 2. the mass of the weight and pulley are unchanged: m=5.8 kg and mp=1.7 kg
6 0
4 years ago
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