The heater in the thermostat control circuit in the cooling mode is active when the ____.

1 answer:

7 0

Here is the correct answer that would best complete the given statement above. The heater in the thermostat control circuit in the cooling mode is active when the heat anticipator in the thermostat control circuit is in the heating mode. Hope this is the answer that you are looking for.

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If a builder of mass 75kg climbs a vertical ladder of 25m how much energy has she gained ?

erica [24]

So,

GPE (graviational potential energy) = mass x g x height

GPE is depends on where zero height is defined. In this situation, we define h = 0 as the initial height.

$GPE = 75 \ kg*9.8 \ \frac{m}{s^2}*25 \ m$

$GPE = 18,375 \frac{kg*m^2}{s^2}$

$GPE = 18,375 \ joules(J) \ or \ 18.375 \ kilojoules(kJ)$

The builder has gained 18.375 kJ of PE.

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2 years ago

____ 1. A mineral is inorganic, which means that it contains

Dimas [21]

Answer:

no material that were once part of living things

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3 years ago

Topic Gravitational force amd firld strength.. help me please

I am Lyosha [343]

The gravitational force between m₁ and m₂ has magnitude

$F_{1,2} = \dfrac{Gm_1m_2}{x^2}$

while the gravitational force between m₁ and m₃ has magnitude

$F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}$

where x is measured in m.

The mass m₁ is attracted to m₂ in one direction, and attracted to m₃ in the opposite direction such that m₁ in equilibrium. So by Newton's second law, we have

$F_{1,2} - F_{1,3} = 0$

Solve for x :

$\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}$

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

$x \approx 6.74\,\mathrm m$

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2 years ago

Calculate the force experienced by a positive test charge of 8.1 × 10-9 coulombs if the electric field strength is 9.4 × 107 new

Arte-miy333 [17]

The electric force exerted by an electric field of intensity E on a charge q is equal to the product between E and q, so:
$F=qE=(8.1 \cdot 10^{-9}C)(9.4 \cdot 10^7 N/C)=0.76 N$

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3 years ago

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6. Aaron will run from home at y mph and walk back at x mph. how much distance does he need to travel to spend a total of t hour

Kisachek [45]

Answer:

Total distance, $d=\dfrac{xyt}{(x+y)}$