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3 years ago

Compare a change in temperature of 1°C to a change of 1°F.

Physics

2 answers:

NikAS [45]3 years ago

5 0

Answer:

A change of one degree Celsius = a change of one Kelvin, but a Celsius temperature is never equal to a Kelvin temperature. A change of 1 degree Fahrenheit equals a change of 5/9 = 0.56 degrees Celsius. To convert a Fahrenheit temperature to Celsius, subtract 32 and multiply by 5/9.

Explanation:

Aleks04 [339]3 years ago

5 0

Answer/Explanation:

1 degree Fahrenheit is equal to -16.77777... Celsius

C = F - 32 ÷ 1.8

For every increase in F, C will be 16.77777... degrees lower.

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Answer:

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2 years ago

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The heating element of a water heater in an apartment building has a maximum power output of 37 kW. Four residents of the buildi

kkurt [141]

Answer:

maximum possible temperature is 34.088°C

Explanation:

Given data

power output Q = 37 kW

volume flow rate = 11 x 10^-5 m3/s

temperature t = 14°C

to find out

maximum possible temperature T

solution

we can say

total water volume = 4 residents × volume flow

total water volume = 4 × 11 x 10^-5 = 44 × 10^-5 m³/s

so we say total water mass = 1000 × volume

total water mass = 1000 kg/m³ × 44 × 10^-5

total water mass = 0.44 kg

we know

dQ/dt = (dm/dt)× (S)× ( T - t)

so here we know specific heat of water S = 4.186 joule/gram °C

37 = 0.44 × 4.186 × (T-14)

T = 34.088°C

maximum possible temperature is 34.088°C

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2 years ago

10 Select the correct answer. What property of a wave remains unchanged when a wave enters a different medium? O A amplitude OB.

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2 years ago

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Two electrons are at rest and separated by a distance of 4.32 × 10-10 m. When they are released they accelerate away from each o

sasho [114]

Answer:

Speed of electron when their separation increased by a factor of 4.10 is 9.41 x 10⁵ m/s .

Explanation:

The electric potential energy is given by the relation :

$U = \frac{kq_{1}q_{2} }{r}$

Here q₁ and q₂ are the two charge particles and r is the distance between them and k is electric constant.

In this case, there are two electrons which are separated by the distance 4.32 x 10⁻¹⁰ m.

Let e be the electron charge and r₁ be the distance between them. Then, the initial electric potential energy is :

$U_{1} = \frac{ke^{2} }{r_{1} }$

Now, the distance between the electrons increases by the factor of 4.10. Let r₂ be the new distance between them i.e. r₂ = 4.10 r₁.

Thus, the new electric potential energy is :

$U_{2} = \frac{ke^{2} }{r_{2} }=\frac{ke^{2} }{4.10r_{1} }$

Applying law of conservation of energy :

ΔU = ΔK

Here ΔU is change in electric potential energy and ΔK is change in kinetic energy.

( U₁ - U₂ ) = ( K₂ - K₁ )

Here K₂ and K₁ are initial and final kinetic energy of electron.

Since, the electron initially is at rest, so its initial kinetic energy is zero. Thus, the above equation becomes:

K₂ = U₁ - U₂

$\frac{1}{2}mv^{2}=\frac{ke^{2} }{r_{1} }- \frac{ke^{2} }{4.10r_{1} }$

Here m and v are the mass and final speed of electron respectively.

$v^{2}=\frac{2}{m} \frac{ke^{2} }{r_{1} }(1- \frac{1 }{4.10 })$

Substitute 9.1 x 10⁻³¹ kg for m, 9 x 10⁹ N m² C⁻² for k, 1.6 x 10⁻¹⁹ C for e and 4.32 x 10⁻¹⁰ m for r₁ in the above equation.

$v^{2}=\frac{2}{9.1\times10^{-31} } \frac{9\times10^{9}\times(1.6\times10^{-19})^{2} }{4.32\times10^{-10} }(1- \frac{1 }{4.10 })$

$v^{2}=8.86\times10^{11}$

v = 9.41 x 10⁵ m/s

5 0

2 years ago

Circuit A in a house has a voltage of 208 V and is limited by a 40.0-A circuit breaker. Circuit B is at 120.0 V and has a 20.0-A

Setler79 [48]

Given Information:

Voltage of circuit A = Va = 208 Volts

Current of circuit A = Ia = 40 Amps

Voltage of circuit B = Vb = 120 Volts

Current of circuit B = Ib = 20 Amps

Required Information:

Ratio of power = Pa/Pb = ?

Answer:

Ratio of power = Pa/Pb = 52/15

Explanation:

Power can be calculated using Ohm's law

P = VI

Where V is the voltage and I is the current flowing in the circuit.

The power delivered by circuit A is

Pa = Va*Ia

Pa = 208*40

Pa = 8320 Watts

The power delivered by circuit B is

Pb = Vb*Ib

Pb = 120*20

Pb = 2400 Watts

Therefore, the ratio of the maximum power delivered by circuit A to that delivered by circuit B is

Pa/Pb = 8320/2400

Pa/Pb = 52/15

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3 years ago