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2 years ago

7

onsidering light at the two ends of the visible light spectrum, violet light has a _ wavelength and a _ photon energy th

an red light. Shorter, lower Longer, lower Longer, higher Shorter, higher

1 answer:

Alenkasestr [34]2 years ago

6 0

Answer:

Shorter, higher

Explanation:

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Help!! <br> A. 0.5 s <br> B. 1.0 s<br> C. 1.5 s<br> D. 2.0 s

algol13

Answer:

c. 1.5 s

Explanation:

i had this and i got it right

3 0

3 years ago

A person wishes to heat pot of fresh water from 20°C to 100°C in order to boil water for pasta. They calculate that their pot ho

nasty-shy [4]

Given:
Water, 2 kilograms
T1 = 20 degrees Celsius, T2 = 100 degrees Celsius.

Required:
Heat produced

Solution:
Q (heat) = nRT = nR(T2 = T1)
Q (heat) = 2 kilograms (4.184 kiloJoules per kilogram Celsius) (100 degrees Celsius – 20 degrees Celsius)
<u>Q (heat) = 669.42 Joules
</u>This is the amount of heat produced in boiling 2 kg of water.

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3 years ago

Forces and pres

Sliva [168]

Answer:

A. It is zero.

Explanation:

D Later in the day, more power is developed in lifting each box. 12 A manometer is used to indicate the pressure in a steel vessel, as shown in the diagram. What value does the liquid manometer give for the pressure in the vessel? It is zero

5 0

2 years ago

At what condition does a body becomes weightless at the equator?

ArbitrLikvidat [17]

Answer:

The decrease is due to the bulge at the equator (putting more distance between the rest of the planet and the surface

Explanation:

4 0

3 years ago

Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini

amid [387]

Answer:

Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

$E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}$

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

$E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}$

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

$V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r$

= $\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]$

∴NOTE: Graph is attached

8 0

3 years ago