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Nutka1998 [239]
3 years ago
13

In doing a load of clothes, a clothes dryer uses 16 A of current at 240 V for 45 min. A personal computer, in contrast, uses 2.7

A of current at 120 V. With the energy used by the clothes dryer, how long (in hours) could you use this computer to "surf" the internet?
Physics
1 answer:
jasenka [17]3 years ago
6 0

Explanation:

Given that,

Current of clothes dryer, I = 16 A

Voltage, V = 240 V

Time, t = 45 min = 2700 s

Current of personal computer, I' = 2.7 A

Voltage, V' = 120 C

Energy used by clothes dryer is given by :

E=I^2Rt\\\\E=IVt\\\\E=16\times 240\times 2700\\E=1.03\times 10^7\ J

Let t' is the time for this computer to "surf" the internet. Again using formula of energy used as :

E=I'V't'\\\\t'=\dfrac{E}{I'V'}\\\\t'=\dfrac{1.03\times 10^7}{2.7\times 120}\\\\t'=31790.12\ s\\\\t'=8.83\ h

So, for 8.83 hours you could use this computer to "surf" the internet.

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I know 1, that is in the case of a burning of a candle.

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A beam of monochromatic light with a wavelength of 400 nm in air travels into water. what is the wavelength of the light in wate
slava [35]
The refractive index of water is n=1.33. This means that the speed of the light in the water is:
v= \frac{c}{n}= \frac{3 \cdot 10^8 m/s}{1.33 }=2.26 \cdot 10^8 m/s

The relationship between frequency f and wavelength \lambda of a wave is given by:
\lambda= \frac{v}{f}
where v is the speed of the wave in the medium. The frequency of the light does not change when it moves from one medium to the other one, so we can compute the ratio between the wavelength of the light in water \lambda_w to that in air \lambda as
\frac{\lambda_w}{\lambda}= \frac{ \frac{v}{f} }{ \frac{c}{f} } = \frac{v}{c}
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5 0
3 years ago
4. A 1,000 kg truck moving at 10 m/s runs into a concrete wall. It takes 0.5 seconds for the truck to conipietery
Maru [420]

Answer:

\large \boxed{\text{h. 20 000 N}}  

Explanation:

Force is the change in momentum over time

F = Δp/Δt

1. Calculate the change in momentum

p₁ = mv₁ = 1000 kg × 10 m/s = 10 000 kg·m·s⁻¹

p₂ = 0

Δp = p₂ - p₁= (0 - 10 000) kg·m·s⁻¹ = -10 000 kg·m·s⁻¹

2. Calculate the force

\begin{array}{rcl}F & = & \dfrac{\Delta p}{\Delta t}\\\\& = & \dfrac{-10 000 \text{ kg$\cdot$m$\cdot$ s}^{-1}}{\text{ 0.5 s}}\\\\& = & \textbf{-20 000 N}\\\end{array}\\\text{The negative sign shows that the force is exerted opposite to the direction of motion.}\\\text{The magnitude of the force is $\large \boxed{\textbf{20 000 N}}$}

3 0
3 years ago
Read 2 more answers
Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance
NemiM [27]

Answer:

Explanation:

Force between two charges of q₁ and q₂ at distance d is given by the expression

F = k q₁ q₂ / d₂

Here force between charge q₁ = - 15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = (1.66 - 1.24 ) = .42 mm

k = 1/ 4π x 8.85 x 10⁻¹²

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 35969.4 x 10⁻³ N .

force between charge q₂ =  34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = ( 1.24 - 0 ) = 1.24 mm .

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x  34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 82729.6  x 10⁻³ N

Both these forces will act in the same direction towards the left (away from the origin towards - ve x axis)

Total force = 118699 x 10⁻³

= 118.7 N.

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