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Nutka1998 [239]
3 years ago
15

Find the wavelength of a photon which has 500.0 eV of energy.

Physics
1 answer:
jekas [21]3 years ago
5 0

Answer:

2.5 x 10⁻⁹ m

Explanation:

E = Energy of photon = 500 eV = 500 x 1.6 x 10⁻¹⁹ J

c = speed of photon = 3 x 10⁸ m/s

λ = wavelength of photon = ?

Energy of photon is given as

E = \frac{hc}{\lambda }

inserting the values

500\times 1.6\times 10^{-19} = \frac{(6.63\times 10^{-34})(3\times 10^{8})}{\lambda }

λ = 2.5 x 10⁻⁹ m

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A new restaurant is interested in determining the best time-temperature combination for roasting a five-pound cut of lamb. The t
Leto [7]

Answer:

C

Explanation:

(c) The two cuts that are being roasted for each time-temperature combination are an example of replication.

In the question it is given that  From 10 identical cuts of lamb, 2 are randomly selected to roast using each of the time-temperature combinations in the same oven. Here it is an act of copying the exact sahpe size of the lamb in all cuts, which is nothing but replication. Moreover, this replication can help in proper comparision.

7 0
3 years ago
What is the function of the wire in
Naily [24]

Answer:

I think the right answer is option B.

3 0
4 years ago
7. A car moving at 10m/s (about 22.4 mph) crashes into a barrier and stops in 0.25 m.
Galina-37 [17]

Answer:

a) 0.05s

b) 4000N

Explanation:

a)When car is stopped its final velocity become zero

U- 10 m/s

V- 0 m/s

S - 0.25 m

t -?

S = (v+u)*t/2

0.25 =(10+0)*t/2

t = 0.05s

b) If we happened to calculate the avarage force we have to consider about acceleration

V= 0

U = 10

t = 0.05 s

a =?

V = U + at

0 = 10 -a * 0.05

a = 200 m/s2

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= 20 * 200

= 4000N

6 0
3 years ago
Problem 1 Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed
ANTONII [103]

Answer:

markers are 29.76 m far apart in the laboratory

Explanation:

Given the data in the question;

speed of particle = 0.624c

lifetime = 159 ns = 1.59 × 10⁻⁷ s

we know that; c is speed of light which is equal to 3 × 10⁸ m/s

we know that

distance = vt

or s = ut

so we substitute

distance = 0.624c × 1.59 × 10⁻⁷ s

distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s

distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s

distance =  29.76 m

Therefore, markers are 29.76 m far apart in the laboratory

3 0
3 years ago
How deep can an object with 6360N hitting on a sponge get ???
borishaifa [10]

Answer:

The sponge must go \dfrac{636}{m}\ \text{meter} deep

Explanation:

If F = 6360 N, then it is required to find how deep can an object with this force hitting on a sponge get.

We know that, F = mgh

m is mass

g is acceleration due to gravity

h=\dfrac{F}{mg}\\\\h=\dfrac{6360}{10m}\\\\h=\dfrac{636}{m}\ \text{meter}

So, the sponge must go \dfrac{636}{m}\ \text{meter} deep.

5 0
4 years ago
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