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2 years ago

7

An electric heater has a power rating of 800W. If the total energy used by the electric heater is 480kj, how long was the heater

on for?
A.0.6 seconds
B.10 minutes
C.1.67 seconds
D.60 minutes

1 answer:

maks197457 [2]2 years ago

6 0

So basically I think c is the right answer

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The 2.6-kg collar is released from rest at A and slides down the inclined fixed rod in the vertical plane. The coefficient of ki

neonofarm [45]

Answer:

The attached diagram explains the system,

Sum of Fy = 0

N=9.81

N - mgCos60 = 0

F= ukN= (0.53)(9.81) =

F= 5.12 N

So

F.d= 1/2(mv.v) - mgdsin60

-5.12*0.5 = 0.5*v^2 - 2*(9.81)*(0.5*sin60)

(a) v = 2.436 m/s

For deflection

-F.x = 1/2(mv.v) - mgxsin60 + 1/2 (k*x*x)

by solving for with values of v, m, g, F, k

800x^2 - 11.87 x - 5.938 = 0

by solving the quadratic equation

x = 0.093, -0.079

(b) x = 0.093 m

correct Answer is 0.093m

Explanation:

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3 years ago

To practice problem-solving strategy 22.1: gauss's law. an infinite cylindrical rod has a uniform volume charge density ρ (where

BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

$\int E.dA = \frac{q}{\epcilon_0}$

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

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now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r}{2 \epcilon_0}$

Now if we have a situation that the point lies outside the cylinder

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$q = \int \rho dV$

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now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

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3 years ago

Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19

Alisiya [41]

Answer:

The workdone is $W = 9.28 * 10^{3} J$

Explanation:

From the question we are told that

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The density of the cylinder is $\rho = 0.346 * \rho_w$

Where $\rho_w$ is the density of freshwater which has a constant value

$\rho_w = 1000 kg/m^3$

Now

Let the final height of the device under the water be $= h_f$

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$\rho_c g V_n = \rho_w gV_f$

$\rho_c A h = \rho A h_f$

So $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

=> $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as

$W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

$= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

$= \frac{g A \rho}{2} [h^2 - h_f^2]$

$= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

$W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

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3 years ago

Audible beats are formed bu the interference of two waves

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<span>Audible beats are formed by the interferences of two waves in which are from the slightly different frequencies which is letter a. It is because having two waves that consist of slightly different frequency causes or produce the audible beats to be formed in which they are beats that causes sound to be either loud or soft.</span>

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3 years ago

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Vikentia [17]

Answer:

The speed of light in glycerol is 2.04×108 m/s .

Explanation:

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3 years ago