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igor_vitrenko [27]
3 years ago
7

Which of the following is NOT an insulator?

Physics
2 answers:
dlinn [17]3 years ago
7 0
Try googling it I guess
sergejj [24]3 years ago
4 0
I don’t know how air would be an insulator so I’m guessing that one isn’t
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How many of each component are shown in the diagram? Check all that apply.
g100num [7]

Can you show the diagram please

3 0
3 years ago
A man 6.00 ft tall approaches a street light 15.0 ft above the ground at the rate of 4.00 ​ft/s. How fast is the end of the​ man
klio [65]

Answer:

\frac{dx}{dt} = 10 ft/s

Explanation:

As per given figure let say the tip of the shadow is at distance "x" from the base of the lamp

so here we have

\frac{x}{15} = \frac{x - y}{6}

so we have

6x = 15 x - 15 y

15 y = 9 x

now we have

5\frac{dy}{dt} = 2\frac{dx}{dt}

5(4) = 2\frac{dx}{dt}

\frac{dx}{dt} = 10 ft/s

7 0
3 years ago
A 14,700 N car is traveling at 25 m/s. The brakes are applied suddenly, and the car slides to a stop. The average braking force
Vlada [557]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

vo = 25 m/sec 
<span>vf = 0 m/sec </span>
<span>Fμ = 7100 N (Force due to friction) </span>
<span>Fg = 14700 N </span>
<span>With the force due to gravity, you can find the mass of the car: </span>
<span>F = ma </span>
<span>14700 N = m (9.8 m/sec²) </span>
<span>m = 1500 kg </span>
<span>Now, we can use the equation again to find the deacceleration due to friction: </span>
<span>F = ma </span>
<span>7100 N = (1500 kg) a </span>
<span>a = 4.73333333333 m/sec² </span>
<span>And now, we can use a velocity formula to find the distance traveled: </span>
<span>vf² = vo² + 2a∆d </span>
<span>0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d </span>
<span>0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d </span>
<span>-625 m²/sec² = (-9.466666666667 m/sec²) ∆d </span>
<span>∆d = 66.0211267605634 m </span>
<span>∆d = 66.02 m</span>
7 0
4 years ago
If 200 ml of tea at 90 °C is poured into a 400 g glass cup initially at 25 °C, what would be the final temperature of the mixtur
sweet-ann [11.9K]

The final temperature of the mixture is 43.62 °C

To solve the question above, we apply the law of calorimetry

The law of calorimetry: which states that if there is no lost of heat the surrounding, heat lost is equal to heat gained, or it can be stated as heat absorbed by a cold body is equal to heat released by a hot body, provided there is no lost of heat to the surrounding.

The law above is expressed mathematically as

Cm(t₁-t₃) = C'm'(t₃-t₂)............. Equation 1

Using equation 1 to solve the question,

Let: C = specific heat capacity of glass cup, m = mass of glass cup, C' = specific heat capacity of tea, m' = mass of tea, t₁ = initial temperature of tea, t₂ = initial temperature of glass cup, t₃ = final temperature of the mixture.

From the question,

Given: m = 400 g = 0.4 kg, C = 840 J/kg°C, m' = 200g (tea is a liquid made of water and the volume of water in ml is thesame a its mass in gram) = 0.2 kg, C' = 4186 J/kg.°C, t₁ = 90°C, t₂ = 25°C

Substitute these values into equation 1 and solve for t₃

0.4(840)(90-t₃) = 0.2(4186)(t₃-25)

336(90-t₃) = 837.2(t₃-25)

30240-336t₃ = 837.2t₃-20930

collect like terms

837.2t₃+336t₃ = 30240+20930

  1173.2t₃ = 51170

t₃ = 51170/1173.2

t₃ = 43.62 °C

Hence, the final temperature of the mixture is 43.62 °C

7 0
3 years ago
Read 2 more answers
What is the electric field strength just outside the flat surface of the conductor?
inna [77]
We can find the answer step-by-step:

1) The electric charges on a conductor must lie entirely on its surface. This is because the charges have same sign, so the force acting between each other is repulsive therefore the charges must be as far apart as possible, i.e. on the surface of the conductor.

2) We consider a cylinder perpendicular to the surface of the conductor, that crosses the surface with its section. We then apply Gauss law, which states that the flux of the electric field through this cylinder is equal to the total charge inside it divided the electrical permittivity:
\Phi =  \frac{Q}{\epsilon_0}

3) The electric field outside the surface is perpendicular to the surface itself (otherwise there would be a component of the electric force parallel to the surface, which would move the charge, violating the condition of equilibrium). The electric field inside the conductor is instead zero, because otherwise charges would move violating again equilibrium condition. Therefore, the only flux is the one crossing the section A of the cylinder outside the surface: 
\Phi = E A

4) The total charge contained in the cylinder is the product between the section, A, and the charge density \sigma on the surface of the conductor:
Q=\sigma A

5) Substituting the flux and the charge density inside Gauss law, we can find the electric field just outside the surface of the conductor:
EA= \frac{\sigma A}{\epsilon_0}
therefore
E= \frac{\sigma}{\epsilon_0}
4 0
4 years ago
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