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2 years ago

13

It takes a photon 8 minutes and 25 seconds ro reach earth. What is the distance (labeled 1 au) in meter between the sun and eart

h? The speed od light un a vacuum is 3e8 m/s

1 answer:

tangare [24]2 years ago

3 0

Answer:

$x=0.017 AU$

Explanation:

We can use the equation of speed in terms of distance. We know that the speed of light is constant value so we will have:

$v=\frac{x}{t}$

$x=v*t$

$x=3*10^{8}*8.417$

$x=2.53*10^{9}m$

Now we know that $1 AU = 1.496*10^{11}m$

$x=0.017 AU$

I hope it helps you!

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A sprinter must average 24.0 mi/h to win a 100-m dash in 9.30 s. What is his wavelength at this speed if his mass is 84.5 kg?

crimeas [40]

Answer:

Wavelength λ = 7.31 × 10^-37 m

Explanation:

From De Broglie's equation;

λ = h/mv

Where;

λ = wavelength in meters

h = plank's constant = 6.626×10^-34 m^2 kg/s

m = mass in kg

v = velocity in m/s

Given;

v = 24 mi/h

Converting to m/s

v = 24mi/h × 0.447 m/s ×1/(mi/h)

v = 10.73m/s

m = 84.5kg

Substituting the values into the equation;

λ = (6.626×10^-34 m^2 kg/s)/(84.5kg × 10.73m/s)

λ = 7.31 × 10^-37 m

7 0

3 years ago

Read 2 more answers

Consider a loop of wire with a resistor. In the same plane, (with switch S closed) a long wire has a current I flowing from left

Ulleksa [173]

Solution :

The given figure is a loop of a wire with a resistor.

When the switch S is closed for long time and is suddenly opened, the direction of the induced current can be find out by using the rule of right hand screw. According to the right hand screw rule, the direction of the magnetic field at the loop is in the direction that points outwards. The strength of the current rapidly decreases as it is switch off and the magnetic flux that is linked with the loop wire will also decrease.

According to the Lenz's law, the direction of the induced current must be such $\text{that it opposes}$ the decrease in the magnetic flux. It means the direction of the magnetic field must be outwards and also normal to the plane of the screen. The direction of the induced anti clockwise or from right to left in the resistance.

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2 years ago

Three energy changes that result in movements accuring in your body as sit down reading book?

Rom4ik [11]

Gravitational to Kinetic to Chemical

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2 years ago

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

2 years ago

A food department is kept at â12Â°c by a refrigerator in an environment at 30Â°c. the total heat gain to the food department is

slava [35]

As per energy conservation in the reversible engine we can say

$Q_2 + W = Q_1$

here we know that

$Q_2 = 3300 kJ/h$

$Q_1 = 4800 kJ/h$

now from above equation

$3300 + W = 4800$

$W = 1500 kJ/h$

now we can convert it into kW

$W = 1500\times \frac{kJ}{3600s}$

$W = 0.42 kW$

so above is the power input to the refrigerator

now to find COP we know that

$COP = \frac{Q_2}{W}$

$COP = \frac{3300}{1500} = 2.2$

so COP of refrigerator is 2.2

3 0

2 years ago