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qwelly [4]
3 years ago
8

What is the energy of a photon with a frequency of 3. 6 × 1015 hz? planck’s constant is 6. 63 × 10–34 j•s.

Physics
1 answer:
GrogVix [38]3 years ago
3 0

Answer:

2.029×10^-18 J

Explanation:

E=hv

so

E=(3.06×10^15)*(6.63×10^-34)

E=2.029×10^-18 J

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Which sentence describes an example of sublimation?
k0ka [10]

answer C is the correct one

7 0
3 years ago
Two wires each carry 10.0 A of current (in opposite directions) and are 2.50 mm apart. What is the magnetic field 37.0 cm away a
lyudmila [28]

Answer:

see answer below

Explanation:

Before we do any kind of calculation, we need to convert the proper units of the exercise. All the units of distance must be in meters, so, let's change distance of the wire, and the magnetic field to meters:

Separation between the wires are 2.5 mm:

2.5 mm * (1 m / 1000 mm) = 0.0025 m

The distance of P from the bottom of the wires is 37 cm:

37 cm * (1 m/100 cm) = 0.37 m

The distance of P from the top of the wires is just the sum of the two distances:

R = 0.37 + 0.0025 = 0.3725 m

Now that we have the distance, we can determine the magnetic field, using the following expression:

B = B(bottom) - B(top)   or just B₂ - B₁

And B = μ₀ I / 2πR

Replacing in the above expression we have:

B = μ₀ I / 2π ( 1/R₂ - 1/R₁)

Now we can determine the magnetic field:

B = (4πx10⁻⁷ * 10 / 2π) (1/0.37 - 1/0.3725)

<h2>B = 3.63x10⁻⁸ T</h2><h2></h2>

Which means that the magnetic field is out of the page.

Hope this helps

4 0
3 years ago
How many ohms of resistance must be present in a circuit that has 240 volts and a current of 15 amps?
PIT_PIT [208]

Answer:

16 ohms

Explanation:

V= I ⋅ R

where,  V  is the net potential difference in the circuit,  I  is the current in the circuit and  R  is the net resistance of the circuit.  

In this case,  V = 240  volts,  I = 15  amperes.

240 = 15 ⋅ R  

⇒ R = 240/ 15 = 16  ohms

5 0
3 years ago
Read 2 more answers
Which of the following would decrease the resistance in a wire?
padilas [110]

<u>Increase the thickness of the wire</u> would decrease the resistance in a wire

Explanation:

Thicker wires have a larger cross-section that increases the surface area with which electrons can flow unimpeded.  The thicker the wire, therefore, the lower the resistance.

Thin wires have very high resistance the reason the thin tungsten in a bulb glows because it is heated from the high resistance of many electrons trying to pass through a very small cross-section.

6 0
3 years ago
A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest.
Airida [17]

Answer:

a) v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} ), b) \Delta K = 9.559\times 10^{-5}\,J

Explanation:

a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

(4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v

The final velocity is:

v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )

b) The change in the kinetic energy of the 13.5 g coin is:

\Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]

\Delta K = 9.559\times 10^{-5}\,J

4 0
3 years ago
Read 2 more answers
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