3 years ago

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