Ask question

Ask question

All categories

3 years ago

11

A thermostat is used for which of the following? maintaining the temperature in a room turning a heating system on turning a hea

ting system off all of the above

2 answers:

Molodets [167]3 years ago

4 0

All of the above, it's a very useful device :P

Vlada [557]3 years ago

3 0

Thermostat can do all the three.

So answer is D.

You might be interested in

Suppose a car is traveling at +25.0 m/s, and the driver sees a traffic light turn red. After 0.340 s has elapsed (the reaction t

scoundrel [369]

First, we will get the distance traveled before the driver applied the brakes.
distance = velocity * time
distance = 25*0.34 = 8.5 m

Now, we will calculated the distance that the car traveled after the driver applied the brakes. To do this, we will use the equation of motion:
<span>vf^2 = vi^2 + 2*a*d where:
</span>vf = zero, vi = 25 m/s and a = -7 m/s^2
Note: The negative sign is only to show deceleration
d = <span> 1/2*(625) /(7) = 44.6428 m

The total stopping distance =</span> 8.5 + 44.6428 = 53.1428 m

3 0

3 years ago

A 16.0 kg child on roller skates, initially at rest, rolls 2.0 m down an incline at an angle of 20.0° with the horizontal. If th

enyata [817]

The kinetic energy of the child at the bottom of the incline is 106.62 J.

The given parameters:

<em>Mass of the child, m = 16 kg</em>
<em>Length of the incline, L = 2 m</em>
<em>Angle of inclination, θ = 20⁰</em>

The vertical height of fall of the child from the top of the incline is calculated as;

$sin(20) = \frac{h}{2} \\\\h = 2 \times sin(20)\\\\h = 0.68 \ m$

The gravitational potential energy of the child at the top of the incline is calculated as;

$P.E = mgh\\\\P.E = 16 \times 9.8 \times 0.68\\\\P.E = 106.62 \ J$

Thus, based on the principle of conservation of mechanical energy, the kinetic energy of the child at the bottom of the incline is 106.62 J since no energy is lost to friction.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

7 0

2 years ago

A 1.50-kg iron horseshoe initially at 550°C is dropped into a bucket containing 25.0 kg of water at 20.0°C. What is the final te

Ber [7]

Answer:

Te = 23.4 °C

Explanation:

Given:-

- The mass of iron horseshoe, m = 1.50 kg

- The initial temperature of horseshoe, Ti_h = 550°C

- The specific heat capacity of iron, ci = 448 J/kgC

- The mass of water, M = 25 kg

- The initial temperature of water, Ti_w = 20°C

- The specific heat capacity of water, cw = 4186 J/kgC

Find:-

What is the final temperature of the water–horseshoe system?

Solution:-

- The interaction of horseshoe and water at their respective initial temperatures will obey the Zeroth and First Law of thermodynamics. The horseshoe at higher temperature comes in thermal equilibrium with the water at lower temperature. We denote the equilibrium temperature as (Te) and apply the First Law of thermodynamics on the system:

m*ci*( Ti_h - Te) = M*cw*( Te - Ti_w )

- Solve for (Te):

m*ci*( Ti_h ) + M*cw*( Ti_w ) = Te* (m*ci + M*cw )

Te = [ m*ci*( Ti_h ) + M*cw*( Ti_w ) ] / [ m*ci + M*cw ]

- Plug in the values and evaluate (Te):

Te = [1.5*448*550 + 25*4186*20 ] / [ 1.5*448 + 25*4186 ]

Te = 2462600 / 105322

Te = 23.4 °C

7 0

3 years ago

Read 2 more answers

Which is the correct scientific notation of the number 0.000681?

Natasha2012 [34]

In scientific notation", that number would be written as

6.81 x 10⁻⁴ .

3 0

3 years ago

Read 2 more answers

The Hubble Space Telescope has a mass of 1.16*10^ 4 kg and orbits the Earth at an altitude of 5.68 * 10 ^ 5 above Earth's surfac

andrezito [222]

Answer:

$E=8.13\times 10^{12}\ J$

Explanation:

Given that,

The mass of a Hubble Space Telescope, $m_1=1.16\times 10^4\ kg$

It orbits the Earth at an altitude of $5.68\times 10^5\ m$

We need to find the potential energy the telescope at this location. The formula for potential energy is given by :

$E=\dfrac{Gm_1m_e}{r}$

Where

$m_e$ is the mass of Earth

Put all the values,

$E=\dfrac{6.67\times 10^{-11}\times 1.16\times 10^4\times 5.97\times 10^{24}}{5.68\times 10^5}\\\\E=8.13\times 10^{12}\ J$

So, the potential energy of the telescope is $8.13\times 10^{12}\ J$ .

5 0

2 years ago