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Scorpion4ik
3 months ago
13

The wavelength of light is 5000 angstrom. Express it in nm and m.

Physics
1 answer:
serg [3.5K]3 months ago
7 0

Answer:

1 angstrom equals 0.1 nm.

To convert 5000 angstroms: 5000 angstrom = 5000 / 1 × 0.1 nm.

= 500 nm

1 \:  angstrom = 1 \times  {10}^{ - 10} m

To express 5000 angstroms in meters: 5000 angstrom = 5000 × 1 × 10^-10.

= 5 × 10^-7 m

Hope this explanation is useful for you.

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A crate on a motorized cart starts from rest and moves with a constant eastward acceleration ofa= 2.60 m/s^2. A worker assists t
Keith_Richards [3271]

Response:

To find power, we must first determine the work done by the force.

1) We will use the following equation to calculate work:

\int\limits {F} \, dx

The force is provided by the problem; our goal is to express 'dx' in terms of 't'

2) It's known that:

\frac{dV}{dt} = a = 2.6

Thus, we have:

v = 2.6t

Then:

\frac{dx}{dt} = V = 2.6t\\ \\dx = 2.6t*dt

3) Finally, substituting all known values gives us:

\int\limits^{4.7}_{0} {5.4t*2.6t} \, dt

After some calculations, the resulting work is:

161.9638 J.

4) To find power, we will use the following equation:

P = \frac{W}{t}

Thus

P = 161.9638/4.7 = 34.46 W

8 0
2 months ago
A person who weighs 625 N is riding a 98-N mountain bike. Suppose that the entire weight of the rider and bike is supported equa
Keith_Richards [3271]

Answer:

A = 4.76 x 10⁻⁴ m²

Explanation:

Given data:

Person's weight = 625 N

Bike's weight = 98 N

Pressure per tire = 7.60 x 10⁵ Pa

Find: Contact area per tire

Total system weight = 625 + 98 = 723 N

Let F represent the force supported by each tire

2F = 723 N

Therefore, F = 361.5 N

Using the formula F = P × A

A = \dfrac{F}{P}

A = \dfrac{361.5}{7.60 \times 10^5}

Contact area, A = 4.76 x 10⁻⁴ m²

7 0
3 months ago
In general, how do you find the average velocity of any object falling in a vacuum?
Softa [3030]
To determine the average velocity of an object descending in a vacuum, assuming you know its final speed, multiply the final result by the overall time. 3. The equation d = v • t expresses how distance, average velocity, and time relate to each other.
5 0
3 months ago
A car accelerates from rest to a velocity of 5 meters/second in 4 seconds. What is its average acceleration over this period of
Ostrovityanka [3204]

The average acceleration is

\bar a=\dfrac{5\,\frac{\mathrm m}{\mathrm s}-0\,\frac{\mathrm m}{\mathrm s}}{4\,\mathrm s}=1.25\,\dfrac{\mathrm m}{\mathrm s^2}

5 0
2 months ago
Read 2 more answers
Water is boiled at 300 kPa pressure in a pressure cooker. The cooker initially contains 3 kg of water. Once boiling started, it
inna [3103]

Answer:

The cooker receives an average energy transfer rate of 1.80 kW.

Explanation:

Given that,

Pressure of the boiling water = 300 kPa

Mass of water = 3 kg

Duration = 30 min

Dryness fraction of the water = 0.5

What is the average energy transfer rate to the cooker?

We know that,

The specific enthalpy of vapor at 300 kPa is

h_{f}=561.47\ kJ/kg

h_{fg}=2163.8\ kJ/kg

We need to determine the initial state enthalpy of water

h_{1}=h_{f}

h_{1}=561.47\ kJ/kg

We must find the enthalpy of water in the final state

Using the enthalpy formula

h_{2}=h_{f}+xh_{fg}

Insert the value into the formula

h_{2}=561.47+0.5\times2163.8

h_{2}=1643.37\ kJ/kg

Next, we calculate the energy transfer rate to the cooker

Using the energy rate formula

Q=\dfrac{m(h_{2}-h_{1})}{t}

Substituting the value into the formula

Q=\dfrac{3\times(1643.37-561.47)}{30\times60}

Q=1.80\ kW

Therefore, the average energy transfer rate to the cooker stands at 1.80 kW.

3 0
2 months ago
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