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Aneli [31]
3 years ago
15

A 10 m long clothesline is strung so that it is perfectly horizontal. when a shirt is hung in the exact center the line sags to

create an angle that is 1 degree below the horizontal on each end. how far below the original height of the line is the center of the clothesline
Physics
1 answer:
Leto [7]3 years ago
3 0

Answer:

x = 0.0873 m

Explanation:

given,

length of clothesline = 10 m

the line sags to create an angle that is 1 degree below the horizontal on each end.

As mass is hang at the center the angle made let the deflection be 'x'

as the shirt is hang at the center distance will be equal to 5 m.

now

tan \theta = \dfrac{d_e}{d}

tan \theta = \dfrac{x}{5}

tan 1^0 = \dfrac{W}{5}

W = 5 \times 0.0175

x = 0.0873 m

Hence, the cloth line will be x = 0.0873 m from the original length of the clothesline.

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To calculate the change in kinetic energy, you must know the force as a function of _______. The work done by the force causes t
QveST [7]

Answer:

(c) position

Explanation:

From the work-energy theorem, the workdone by a force on a body causes a change in kinetic energy of the body.

But, remember that the work done (W) by a force (F) on a body is the product of the force and the distance d, moved by the body caused by the force. i.e

W = F x d

This distance is a measure of the position of the body at a given instance.

Therefore, the work done is given by the force as a function of distance (or position).

3 0
3 years ago
You have two contentious friends, Chris and Pat, and you’ve quickly discovered that they need you to resolve arguments they have
pshichka [43]

Answer:

v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points,

Explanation:

To resolve the debate, it must be shown that the two have part of the reason, the space or distance between the two points divided by time is the average speed between the points.

             v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points, in the only case that it is so is when there is no acceleration.

Therefore neither of them is right.

3 0
3 years ago
A rocket consists of a shuttle and a fuel tank. The rocket flies through the air with a momentum of 180000kgm/s at a velocity of
babunello [35]

Answer:

Answer is 1000kg

Hope it help

6 0
2 years ago
A 1000-turn solenoid is 50 cm long and has a radius of 2.0 cm. It carries a current of 18.0 A. What is the magnetic field inside
vitfil [10]

Answer:

The value  is  B =  0.0452 \  T

Explanation:

From the question we are told that

   The number of turns is  N  =  1000

    The length is  L =  50 cm =  0.50 m  

    The radius is  r =  2.0 cm  =  0.02 m

     The current is I  =  18.0 A

   

Generally the magnetic field is mathematically represented as

         B = \mu_o  * \frac{N }{L}  *  I

Here \mu_o is the permeability of free space with value  

     \mu_o  =  4\pi * 10^{-7} N/A^2

So

     B =  4\pi * 10^{-7}  *   \frac{1000}{0.50} *  18.0

=>   B =  0.0452 \  T

3 0
3 years ago
Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
BARSIC [14]

Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
3 years ago
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