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yuradex [85]
3 years ago
7

The resistance of an electric heater is 50 Ω when connected to 120 V. How much energy does it use during 15 min of operation?

Physics
1 answer:
STALIN [3.7K]3 years ago
7 0

Answer:

Explanation:

I = V/R = 120 V/ 50 Ω = 2.4 A

P = VI = 120(2.4) = 288 W = 288 J/s

288 J/s (15 min(60s / min)) = 259,200 J

or the electric company would charge for

288 W / (1000 W/kW)•(15/60) hr = 0.072 kW•hr

At $0.20 / kW•hr, that would be under 1½ cents

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Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.
SSSSS [86.1K]

Answer:

a = 9.94 m/s²

Explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

\rho(r) = ar^2 - br^3

r is the radius of the spherical shell

dr is the thickness

volume of shell

dV = 4 \pi r^2 dr

mass of shell

dM = \rho(r)dV

\rho = \rho_0 - br

now,

dM = (\rho_0 - br)(4 \pi r^2)dr

integrating both side

M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr

M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})

M = \pi R^3(\dfrac{\rho_0}{3}+\rho)

we know,

a = \dfrac{GM}{R^2}

a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}

a =\pi RG(\dfrac{\rho_0}{3}+\rho)

a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)

a = 9.94 m/s²

7 0
3 years ago
Kinematics practice problems Answers: 4. A race car is traveling at +76 m/s when is slows down at -9 m/s2 for 4
arlik [135]

The new velocity after 4 s is 40 m/s

The height of the spaceship above the ground after 5 seconds is 1,127.5 m

The given parameters for the first question;

  • initial velocity of the car, u = 76 m/s
  • acceleration of the car, a = - 9 m/s²
  • time of motion, t = 4 s

The new velocity after 4 s is calculated as;

v = u + at

v = 76 + (-9)(4)

v = 76 - 36

v = 40 m/s

(5)

The given parameters;

  • height above the ground, h = 500 m
  • velocity of spaceship, u = 150 m/s
  • time of motion, t = 5

The height of the spaceship above the ground after 5 seconds is calculated as;

h_y = h_0 + ut - \frac{1}{2}gt^2\\\\h_y = 500 + (150\times 5) -  (0.5\times 9.8 \times 5^2)\\\\h_y = 1,127.5 \ m

Learn more here: brainly.com/question/24527971

5 0
2 years ago
Consider a vortex filament of strength Γ in the shape of a closed circular loop of radius R. Consider also a straight line throu
Sedbober [7]

Answer:

\vec{V} = \frac{\Gamma}{2R}\vec{A}

Explanation:

We define our values according to the text,

R= Radius

\vec{V} =Velocity

\Gamma =Strenght of the vortex filament

From this and in a vectorial way we express an elemental lenght of this filmaent as \vec{dl}. So,

\vec{dl}x\vec{r} = R*dl*\vec{A}

Where \vec{A} imply a vector acting perpendicular to both vectors.

Applying Biot-Savart law, we have,

\vec{V} =\frac{\Gamma}{4\pi}\int\frac{\vec{dl}x\vec{r}}{r^3}

Substituting the preoviusly equation obtained,

\vec{V} = \frac{\Gamma}{4\pi}\int\frac{R*dl*\vec{A}}{R^3}

\vec{V} = \frac{\Gamma}{4\pi R^2}\int^{2\pi R}_0 dl*\vec{A}

\vec{V} = \frac{\Gamma(2\pi R \vec{A})}{4\pi R^2}

So we can express the velocity induced is,

\vec{V} = \frac{\Gamma}{2R}\vec{A}

6 0
3 years ago
A bowling ball encounters a 0.760 m vertical rise on the way back to the ball rack. Ignore frictional losses and assume the mass
Gre4nikov [31]

To solve this exercise we need the concept of Kinetic Energy and its respective change: Initial and final kinetic energy.

Let's start considering that the angular velocity is given by,

\omega = \frac{v}{R}

Where,

V = linear speed

R = the radius

In the case of the initial kinetic energy:

KE_i=\frac{1}{2} mv^2 + \frac{1}{2}I \omega^2

Where I is the moment of inertia previously defined.

KE_i = \frac{1}{2}(m)3.5^2 + \frac{1}{2}* (\frac{2}{5} m R^2) (\frac{3.5}{R})^2

In the case of the final kinetic energy, we have to,

KE_f= mgh+ \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2

KE_f = m * 9.81 * 0.76 + \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2

For conservation of Energy we have, that

KE_f = KE_i, then (canceling the mass and the radius)

\frac{1}{2} 3.5^2 + \frac{1}{2}(\frac{2}{5})(3.5)^2= 9.81 * 0.76 + \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2

8.575= 7.4556+ \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2

1.1194= \frac{1}{2}( v^2 + (\frac{2}{5}) (v)^2)

2.2388= (\frac{7}{5}) (v)^2

v=1.26m/s

7 0
3 years ago
Why does paper attract both positive and negative charges? Explain.
Sedbober [7]
The fibers have lost electrons giving them a positive charge. The rubber gained electrons giving it a negative charge. The positively charged fibers are now attracted to the negatively charged balloon. The negatively charged balloon attracts the paper.
5 0
2 years ago
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