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zhenek [66]
3 years ago
7

Can a flashlight cause skin cancer

Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
4 0

Answer:

No

Explanation:

It doesn't use the same UV rays as the sun, but it can still damage other cells.

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What does special relativity reveal about the speed of light relative to its source?
Vladimir [108]

Regardless of the source's mobility, light travels at the same speed.

<h3>What makes special relativity so crucial?</h3>

In the calculating and interpretation of high-velocity phenomena, as well as on our methods of thinking, Einstein's special relativity has had a significant influence on the area of physics. Today, we have a considerably better knowledge of space and time than we did at the start of the century.

<h3>Why is special relativity thus named?</h3>

Because it exclusively uses inertial frames to apply the concept of relativity, the theory is known as "special". General relativity, which Einstein created, applies the principle broadly, that is, to any frame, and this theory takes the gravitational forces into account.

learn more about relativity here

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5 0
2 years ago
The following items are connected in a circuit. in which one would you find a voltage gain?
timurjin [86]

Answer:

A

Explanation:

7 0
3 years ago
A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3
Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,  

\text { Electric power }(p)=\frac{V^{2}}{R}

\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}

\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}

For calculating number of turns

\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}

Given that,

80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }

\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}

\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .

We need to find the number of turns in the secondary winding \left(N_{S}\right) to run the bulb at 120W \left(P_{2}\right)

Firstly find the secondary voltage in the transformer use, \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}

\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}

V_{2}^{2}=\frac{120^{2} \times 120}{80}

V_{2}^{2}=\frac{1728000}{80}

V_{2}^{2}=21600

V_{2}=\sqrt{21600}

V_{2}=146.9 \mathrm{V}=V_{S}

Now, finding the number of turns in secondary coil. Use, \frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}

\frac{30}{N_{S}}=\frac{65}{146.9}

N_{S}=\frac{30 \times 146.9}{65}

N_{S}=\frac{4407}{65}N_{S}=67.8

The number of turns in the secondary winding are 67.8 turns.

6 0
3 years ago
A comparison of a machine’s work output with its work input is called __??__.
Aloiza [94]
Ohhhhh its called a input machine
7 0
3 years ago
The magnetic field perpendicular to a single 16.7-cm-diameter circular loop of copper wire decreases uniformly from 0.750 T to z
sammy [17]

Answer:

1.24 C

Explanation:

We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.

The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m

So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.

Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.

So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²

So,  4iρD/d² = 0.750πD²/4Δt.

iΔt = 0.750πD²/4 ÷ 4iρD/d²

iΔt = 0.750πD²d²/16ρ.

So the charge Q = iΔt

= 0.750π(Dd)²/16ρ

= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)

= 123.76 × 10⁻² C

= 1.2376 C

≅ 1.24 C

4 0
4 years ago
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