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Gnesinka [82]
2 years ago
14

. What happens to the wasted energy in the kettle?

Physics
1 answer:
7nadin3 [17]2 years ago
7 0

Answer:

It evaporate

Explanation:

Pls i need brainliest answer

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g You shine orange laser light that has a wavelength of 600 nm through a narrow slit. The slit forms a diffraction pattern on a
zimovet [89]

Answer:

 λ = 3 10⁻⁷ m,   UV laser

Explanation:

The diffraction phenomenon is described by the expression

         a sin θ = m λ

let's use trigonometry

         tan θ = y / L

as in this phenomenon the angles are small

        tan θ = \frac{sin \ \theta}{cos \ \theta} = sin θ

        sin θ = y / L

we substitute

      a y / L = m  λ

let's apply this equation to the initial data

       a  0.04 / L = 1 600 10⁻⁹

       a / L = 1.5 10⁻⁵

now they tell us that we change the laser and we have y = 0.04 m for m = 2

      a 0.04 / L = 2  λ

       a / L = 50  λ

we solve the two expression is

         1.5 10⁻⁵ = 50  λ

          λ = 1.5 10⁻⁵ / 50

          λ = 3 10⁻⁷ m

    UV laser

3 0
2 years ago
What is the answer to this question
leva [86]
The answer is A 0 degrees is the freezing point of water



Plz mark brainliest
7 0
3 years ago
An object with a mass of 10 kg is rolled down a frictionless ramp from a height of 3 meters. If a factory worker at the bottom o
ruslelena [56]

Answer:

The amount of work the factory worker must to stop the rolling ramp is 294 joules

Explanation:

The object rolling down the frictionless ramp has the following parameters;

The mass of the object = 10 kg

The height from which the object is rolled = 3 meters

The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp

Where;

The potential energy P.E. = m × g × h

m = The mass of the ramp = 10 kg

g = The acceleration due to gravity = 9.8 m/s²

h = The height from which the object rolls down = 3 m

Therefore, we have;

P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules

The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules

8 0
2 years ago
Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of
max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

Q =\frac{KA(\delta T)}{L}

Q =\frac{25.87*1*20}{1}

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0
3 years ago
Read 2 more answers
We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

F_D = c_D A \frac{\rho V^2}{2}

Where,

F_D = Drag Force

c_D = Drag coefficient

A = Area

\rho= Density

V = Velocity

Our values are given by,

c_D = 0.5 (That is proper of a cone-shape)

A = 9m^2

\rho = 1.2Kg/m^3

V = 6.5m/s

Part A ) Replacing our values,

F_D = 0.5*9*\frac{1.2*6.5^2}{2}

F_D = 114.075N

Part B ) To find the torque we apply the equation as follow,

\tau = F*d

\tau = (114.075N)(7)

\tau = 798.525N.m

3 0
3 years ago
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