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Elena-2011 [213]
3 years ago
11

Why hydrogen atom does not emit X-rays??

Physics
1 answer:
goblinko [34]3 years ago
8 0
b) it has narrow energy levels  is your answer.
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An object is 39 cm away from a concave mirrors surface along the principles axis. If the mirrors focal length is 9.50 cm, how fa
Tatiana [17]

Answer:

12.6 cm

Explanation:

We can use the mirror equation to find the distance of the image from the mirror:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where here we have

f = 9.50 cm is the focal length

p = 39 cm is the distance of the object from the mirror

Solving the equation for q, we find:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{9.50 cm}-\frac{1}{39 cm}=0.080 cm^{-1}\\q = \frac{1}{0.080 cm}=12.6 cm

5 0
3 years ago
Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work? Move the machine to a
Lady bird [3.3K]
Reduce the friction. Since the total energy is conserved, the only way to improve its work capacity is by reducing energy that doesnt go into work.
7 0
3 years ago
Read 2 more answers
Can someone explain which of Newton’s Law is demonstrated in part 1 and which is demonstrated in part 2? (Picture)
rewona [7]

Answer:

Every action has an equal and opposite reaction. If the student doesn't push, nothing moves, is one student pushes, both move which is an example of newtons third law.

Explanation:

3 0
3 years ago
How much time will it take for a coyote to travel 48 meters across the field to get to the unsuspecting rabbit eating grass in t
kkurt [141]
Formula for time
t=d/s
so…
t= 48m/4m/s
the two ms cancel each other out and ur left with s

t=12s
4 0
3 years ago
A van has a weight of 4000 lb and center of gravity at Gv. It carries a fixed 900 lb load which has a center of gravity at Gl. I
natulia [17]

Answer:

 x = 25 / μ     [ ft]

Explanation:

To solve this exercise we can use Newton's second law.

Let's set a reference system where the x axis is parallel to the road

Y axis  

       N_B + N_A - W_van - W_load = 0

       N_B + N_A = W_van + W_load

X axis

     fr = ma

     a = fr / m

the total mass is

        m = (W_van + W_load) / g

the friction force has the expression

      fr = μ N_{total}

      fr = μy (W_van + W_load)

we substitute

      a = μ (W_van + W_load)    \frac{g}{W_van + W_load}

      a = μ g

taking the acceleration let's use the kinematic relations where the final velocity is zero

       v² = v₀² - 2 a x

       0 = v₀² -2a x

        x = \frac{v_o^2}{2a}

        x = \frac{v_o^2}{2 \mu g}

        x = \frac{40^2}{2 \ 32 \  \mu}

        x = 25 / μ     [ ft]

5 0
3 years ago
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