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Amiraneli [1.4K]
3 years ago
8

_____________ and ____________ can fly in the stratosphere. Fill in blanks

Physics
2 answers:
soldi70 [24.7K]3 years ago
7 0
Jets and planes can fly in the stratophere


Kay [80]3 years ago
6 0
Some commercial aircraft jet and parachutes flies in the stratosphere. Systematically, time also flies.
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The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el
LUCKY_DIMON [66]

Answer:

e.)At twice the distance, the strength of the field is E/4.

Explanation:

The strength of the electric field at a certain distance from a point charge is given by:

E=k\frac{Q}{r^2}

where

k is the Coulomb's constant

Q is the charge

r is the distance from the point charge

In this problem, the distance from the point charge is doubled:

r' = 2r

So the new electric field strength is

E'=k\frac{Q}{(2r)^2}=k \frac{Q}{4 r^2}=\frac{1}{4} (k\frac{Q}{r^2})=\frac{E}{4}

so, at twice the distance the strength of the field is E/4.

4 0
3 years ago
PHYSICAL SCIENCE HELP PLEASE !!!!! Picture below is for question 4
Eva8 [605]

Question 1) excited and moving around a lot.

Question 2) heat

Question 3) specific heat

Question 4) As temperature increases, energy transferred increases.

Hope these answers help!

3 0
4 years ago
Two protons are 1 × 10−10 m apart (about one atomic radius). Which interaction between two protons is stronger, the gravitationa
anastassius [24]

Answer: The electric repulsion between the two protons is stronger than the gravitational attraction.

Explanation: Please see the attachments below

4 0
4 years ago
What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?
adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 10973731.6 m⁻¹

n_f = Higher energy level = 3

n_i= Lower energy level = 2

Putting the values, in above equation, we get:

\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )

\lambda=6.56\times 10^{-7}m

Now we have to calculate the energy.

E=\frac{hc}{\lambda}

where,

h = Planck's constant = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 6.56\times 10^{-7}m

Putting the values, in this formula, we get:

E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}

E=3.03\times 10^{-19}J

Therefore, the energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

3 0
3 years ago
The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz. What is the period of the crystal's motion?
Elan Coil [88]

Answer:

Time period, T=3.05\times 10^{-5}\ s

Explanation:

Given that,

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz

We need to find the period of the crystal's motion. The relationship between the frequency and the time period is given by :

T=\dfrac{1}{f}

T is the time period of the crystal's motion.

Time period is given by :

T=\dfrac{1}{32768}

T=3.05\times 10^{-5}\ s

So, the time period of the crystal's motion is 3.05\times 10^{-5}\ s. Hence, this is the required solution.

8 0
3 years ago
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