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Vladimir [108]
3 years ago
11

An atom emission of light with a specific amount of energy confirms that ?

Physics
1 answer:
marusya05 [52]3 years ago
8 0

The correct answer is B, electrons emit and absorb energy based on their position around the nucleus.

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What type of rock this is
seraphim [82]
Sedimentary was the right one
8 0
3 years ago
Calculate (a) the torque, (b) the energy, and (c) the average power required to accelerate Earth in 4.0 days from rest to its pr
natima [27]
<h2>Answer:</h2>

Torque = <em>2.05 x 10²⁸ Nm</em>

Energy = <em>3.54 x 10³³ J</em>

Average power = <em>1.02 x 10²⁸ W</em>

<h2>Explanation:</h2>

(a) Torque (τ) is the rotational effect of a given force.  

It is given by

τ = I x α          -------------(i)

Where;

I = rotational inertia of the object

α = angular acceleration of the object.

In this case, the object is the Earth. Therefore,

I = 9.71 x 10³⁷ kg m²

α = ω / t

Where;

ω = angular velocity of earth = 2π rad / day

<em>Since </em>

<em>1 day = 24 hours and 1 hour = 3600seconds</em>

<em>1 day = 24 x 3600 seconds = 86400seconds</em>

<em>=> ω = 2π rad / 86400seconds</em>

<em>=> ω = 7.29 × 10⁻⁵ rad/s</em>

<em />

t = 4 days = 4 x 24 x 3600 seconds = 345600 seconds

=> α = ω / t

=> α = 7.29 × 10⁻⁵ / 345600

=> α = (7.29 × 10⁻⁵) / (3.456 x 10⁵)

=> α = (7.29 × 10⁻⁵⁻⁵) / (3.456)

=> α = (7.29 × 10⁻¹⁰) / (3.456)

=> α = 2.11 × 10⁻¹⁰ rad/s²

Now substitute the values of I and α into equation (i)

τ = 9.71 x 10³⁷ x 2.11 × 10⁻¹⁰

τ = 9.71 x 10²⁷ x 2.11

τ = 20.5 x 10²⁷ Nm

τ = 2.05 x 10²⁸ Nm

(ii) The energy (rotational energy) E is given by;

E = \frac{1}{2} x I x ω

E = \frac{1}{2} x 9.71 x 10³⁷ x 7.29 × 10⁻⁵

E = 35.4 x 10³² J

E = 3.54 x 10³³ J

(iii) The average power P, is given by;

P = E / t

P = 3.54 x 10³³ / 345600

P = 1.02 x 10²⁸ W

5 0
3 years ago
A 2kg bowling ball is 2.5 meter off the ground on a post when it falls. Just before it reaches the ground, it is traveling 7m/s.
shutvik [7]
<span>The mechanical energy is conserved.

I hope this helps, good luck! :)</span>
7 0
3 years ago
The near point of an eye is 110 cm. A corrective lens is to be used to allow this eye to focus clearly on objects 26.0 cm in fro
irga5000 [103]

Answer:

The focal length of the lens is 34.047 cm

The power of the needed corrective lens is 2.937 diopter.

Explanation:

Distance of the object from the lens,u = 26 cm

Distance of the image from the lens ,v= -110 cm

(Image is forming on the other side of the lens)

Since ,lens of the human eye is converging lens,convex lens.

Using a lens formula:

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

\frac{1}{f}=\frac{1}{26 cm}+\frac{1}{-110 cm}

f = 34.047 cm = 0.3404 m

Power of the lens = P

P=\frac{1}{f}=\frac{1}{0.34047 m}=2.937 diopter

6 0
3 years ago
Is a spring stores 5 J of energy when its conpresses by 0.5 m what is the spring constant of the spring?
d1i1m1o1n [39]

Answer:

k = 40 N/m

Explanation:

A spring's energy is given:

U = 0.5kx^2

U is the energy in the spring, k is the spring constant and x is the spring displacement.

We are told that the spring stores 5J of energy, therefore, U = 5J. We are also told that the spring is compressed by 0.5m, so the spring x = 0.5m

5 J = 0.5k(0.5m)^2\\5J = 0.5k(0.25m^2)\\5J = k*(0.125)m^2\\\\k = 5J/(0.125)m^2\\k = 40 N/m

k = 40 N/m

Hope this helps!

3 0
3 years ago
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