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olchik [2.2K]
3 years ago
14

Lightning can sometimes occur on hot and humid summer evenings when there are no thunderstorms.

Physics
2 answers:
N76 [4]3 years ago
8 0
B False because facts
mart [117]3 years ago
4 0

Answer:

(B) False

Explanation:

No, it is not possible to have thunder without lightning. Thunder is a direct result of lightning.

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Calculate the wavelength of the electromagnetic radiation required to excite an electron from the ground state to the level with
anastassius [24]

Answer:

2.11 m

Explanation:

<u>Given data</u>

h=6.626×10^{-34\\}(plank constant)

L=45.7 pm

n=n2-n1

n=5-1

n=4

<em>where,</em>

<em>n2=5</em>

<em>n1=1</em>

<u>To find:</u>

E=?

λ=? (Wavelength)

<u>solution</u>

The energy stored in an electron at a specified level is given by;

E=h^{2}×n^{2}÷8ml^{2}..........(1)

<em>m=mass of electron(9.1×</em>10^{-31})

<em>l=length of box</em>

<u>To find E</u>

putting the value of given data in eq(1)

E=9.41×10^{-16}

<u>To find λ</u>

λ=hc/E............................(2)

c=3×10^{8}(speed of light)

putting the value in eq 2 to find wavelength

λ=2.11 m

<u></u>

<u>Note:</u>

There is a chance in calculation error. but the method is correct to solve the problem.

7 0
3 years ago
Read 2 more answers
The blades in a blender rotate at a rate of 6800 rpm . When the motor is turned off during operation, the blades slow to rest in
tangare [24]
The angular acceleration of the blade when it's switched off is (-6800 rev/min) divided by (2.8 sec) = -2,428.6 rev/(min-sec) = -40.5 rev/sec^2 .
5 0
3 years ago
What is the magnitude of the electric force of attraction between two point charges + 2.8 mc and -1.2 mc if the distance
Anit [1.1K]

Answer:

a

Explanation:

it just a

8 0
3 years ago
Hello help me pls! i need serious help
9966 [12]

Answer:c

Explanation:

7 0
2 years ago
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Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of
vredina [299]

Answer:

The angular speed after 6s  is \omega = 1466.67s^{-1}.

Explanation:

The equation

I\alpha  = Fd

relates the moment of inertia I of a rigid body, and its angular acceleration \alpha, with the force applied F at a distance d from the axis of rotation.

In our case, the force applied is F = 22N, at a distance d = 6cm =0.06m, to a ring with the moment of inertia of I =mr^2; therefore, the angular acceleration is

$\alpha =\frac{Fd}{I} $

$\alpha =\frac{22N*0.06m}{(1.5kg)*(0.06)^2} $

\alpha  = 244.44\: s^{-2}

Therefore, the angular speed \omega which is

\omega  = \alpha t

after 6 seconds is

\omega = 244.44$\: s^{-2}* 6s

\boxed{\omega = 1466.67s^{-1}}

7 0
3 years ago
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