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Lelechka [254]
3 years ago
14

A copper rod is 0.795 m long with a rectangular cross section of dimensions 0.0553 m by 0.0847 m. The ends of the rod are held a

t 52.3°C and 99.5°C, but the rod is otherwise insulated from its environment. Find the heat current (the amount of heat that flows per unit time) through the rod. The thermal conductivity of copper is 385 W/(m⋅K).
Physics
1 answer:
alex41 [277]3 years ago
4 0

Answer:

yoooooooooooooooooooooooooooooooooooooooo

Explanation:

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Calculate the change in length of concrete sidewalk (coefficient of linear expansion for concrete is 12*10^-6/celcius) that is 1
anyanavicka [17]

Answer:

The answer to your question is 5.4 cm

Explanation:

This problem refers to calculate the change in length in one dimension due to a change in temperature.

Data

α = 12 x 10⁻⁶

Lo = 150 meters

ΔT = 30 °C

Formula

                    ΔL/Lo = αΔT

solve for ΔL

                    ΔL = αLoΔT

Substitution

                    ΔL = (12 x 10⁻⁶)(150)(30)

Simplification

                    ΔL = 0054 m = 5.4 cm

7 0
3 years ago
An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N
algol13

Answer:

K.E = 15.57 x 10⁻¹⁷ J

Explanation:

First, we find the acceleration of the electron by using the formula of electric field:

E = F/q

F = Eq

but, from Newton's 2nd Law:

F = ma

Comparing both equations, we get:

ma = Eq

a = Eq/m

where,

E = electric field intensity = 120 N/C

q = charge of electron = 1.6 x 10⁻¹⁹ C

m = Mass of electron = 9.1 x 10⁻³¹ kg

Therefore,

a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)

a = 2.11 x 10¹³ m/s²

Now, we need to find the final velocity of the electron. Using 3rd equation of motion:

2as = Vf² - Vi²

where,

Vf = Final Velocity = ?

Vi = Initial Velocity = 1.4 x 10⁷ m/s

s = distance = 3.5 m

Therefore,

(2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² - (1.4 x 10⁷)²

Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)

Vf = 1.85 x 10⁷ m/s

Now, we find the kinetic energy of electron at the end of the motion:

K.E = (0.5)(m)(Vf)²

K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²

<u>K.E = 15.57 x 10⁻¹⁷ J</u>

4 0
3 years ago
The output voltage of a power supply is normally distributed with mean 5 V and standard deviation 0.02 V. If the lower and upper
-BARSIC- [3]

To solve this problem we will apply the normal distribution, with which we will obtain the probability that the given event will occur. Concepts such as the mean and standard deviation will be present throughout the solution of the problem. Increasing or decreasing the average would change the location or center point of the curve. The change in the standard deviation would lead to the change in the dispersion of the data. As the standard deviation increases, the curve would become flatter.

Let X be the output voltage of power supply

X∼N (5,0.02^2)

A

The lower and upper specifications for voltage are 4.95 V and 5.05 V, respectively

P(4.95

P(4.95

P(4.95

P(4.95

P(4.95

Hence probability that a power supply selected at random will conform to the specifications on voltage is 0.9876

8 0
3 years ago
Approximate the Sun as a uniform sphere of radius 6.96 X 108 m, rotating about its central axis with a period of 25.4 days. Supp
In-s [12.5K]

Answer:

T = 184 seconds

Explanation:

First in order to solve this, we need to know which is the expression to calculate the period. This is an exercise of angular velocity, so:

T = 2π/w

Where w: angular speed (in rad/s)

So, let's calculate first the innitial angular speed:

w = 2π/T

Converting days to seconds:

25.4 days * 24 h/day * 3600 s/h = 2,194,560 s

Then the angular speed:

w = 2π / 2,194,560 = 2.863x10^-6 rad/s

Now, the innitial angular momentum is:

I = (2/5)Mr² replacing data:

I = 2/5* (6.96x10^8)² * M = 1.94x10^17m² * M

so the initial angular momentum would be:

L = Iω = 2.863x10^-6 * 1.94x10^17 M

L = 5.55x10^11 m²/s * M = final angular momentum

Now the  final I = 2/5Mr²

Final I = 2/5 * (6.37x10^6)² * M  = 1.62x10^13m² * M

Then 5.55x10^11m²/s * M = 1.62x10^13m² * M * ω → M cancels

ω = 3.42x10^-2 rad/s

Then the new period

T = 2π/ω = 2*3.14 / 3.42x10^-2

T = 184 seconds

8 0
3 years ago
Two charges separated by 1 m exert 1 N forces on each other. If the charges are pushed to 1/4m separation, the force on each cha
Sloan [31]

Answer:

<em>The force on each charge = 16 N</em>

Explanation:

From coulombs law,

F = 1/4πε₀(q₁q₂)/d²........................ Equation 1

q₁q₂ = F4πε₀d²...................... Equation 2

Where F = force on the two charges, q₁ = charge on the first body, q₂ = charge on the second body, d = distance of separation, 1/4πε₀ = constant of proportionality.

<em>When d = 1 m, F = 1 N,</em>

<em>Constant: 1/4πε₀ = 9×10⁹ Nm²/C²</em>

<em>Substituting these values into equation 2,</em>

<em>q₁q₂ = 1×1²/9×10⁹ </em>

<em>q₁q₂ = 1/9×10⁹  C²</em>

<em>When d = 1/4 m, q₁q₂ = 1/9×10⁹  C² and 1/4πε₀ = 9×10⁹ Nm²/C²</em>

<em>Substituting these values into equation 1</em>

<em>F =  9×10⁹×1/9×10⁹ /(1/4)²</em>

<em>F = 1/(1/16)</em>

<em>F = 16 N</em>

<em>Therefore the force on each charge = 16 N</em>

<em />

4 0
3 years ago
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