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andreyandreev [35.5K]
3 years ago
6

An object, experiencing no friction, keeps moving at a constant speed. What can we say about the net force on the object?

Physics
1 answer:
sleet_krkn [62]3 years ago
3 0

Answer:

Explanation:

Let's look at a mathematical representation of this. The equation for tis is just a souped up version of Newton's 2nd Law:

F - f = ma. It an object is moving at a constant speed, the acceleration of that object is 0. That changes this equation to

F = f which states that the applied Force equals the frictional force, choice a.

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(c) If η = 60% and TC = 40°F, what is TH, in °F?
arlik [135]
2b2t hope that helps
5 0
3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
Question 2 of 10
Mekhanik [1.2K]

the awnser to ur question is D

6 0
3 years ago
9. How much energy does a 100 W light bulb use in half an hour? If the light bulb converts
yulyashka [42]

The energy used by the light bulb in half an hour is 180000 J and the amount of thermal energy generated is 158400 J.

What is Energy?

Energy is the ability or the capacity to do work.

To calculate the energy of the light bulb we use the formula below

Formula:

  • E = Pt.......... Equation 1

Where:

  • E = Energy used by the bulb in a half-hour
  • P = Power of the bulb
  • t = Time

Given:

  • P = 100 W
  • t = 1/2 hour = 30 minutes = (30×60) = 1800 seconds

Substitute these values into equation 1

  • E = (100×1800)
  • E = 180000 J

  • If the light converts 12% of electric energy to light energy, then  88% of the energy is used to generate thermal energy

Therefore,

  • Thermal energy = (180000×88/100) = 158400 J

Hence, the energy used by the light bulb in half an hour is 180000 J and the amount of thermal energy generated is 158400 J.

Learn more about energy here: brainly.com/question/21927255

#SPJ1

6 0
2 years ago
horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes
lbvjy [14]

Answer:

The tension is  T =  103.96N

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

           The distance between the two poles is D =14 m

          The mass tied between the two cloth line is  m = 3Kg

         The distance it sags is d_s = 1m

The objective of this solution is to obtain the magnitude of the tension on the ends of the  clothesline

Now the sum of the forces on the y-axis is zero assuming  that the whole system is at equilibrium

       And this can be mathematically represented as

                             \sum F_y = 0

 To obtain \theta we apply SOHCAHTOH Rule

 So    Tan \theta = \frac{opp}{adj}

          \theta = tan^{-1} [\frac{opp}{adj} ]

            = tan^{-1} [\frac{1}{7}]

          =8.130^o

=>  \  \ \ \ \ \ \ \ 2T sin\theta -mg =0

=>  \  \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}

=>  \  \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }

=>  \  \ \ \ \ \ \ \  T =\frac{29.4}{2sin(8.130)}

=>  \  \ \ \ \ \ \ \  T = 103.96N

             

                 

5 0
3 years ago
Read 2 more answers
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