Answer:
"1155 N" is the appropriate solution.
Explanation:
Given:
Acceleration,

Forces resisting motion,

Mass,

By using Newton's second law, we get
⇒ 
Or,
⇒ 
By putting the values, we get
⇒ 
⇒ 
⇒ 
Answer:
Explanation:
distance of shuttle from centre of the earth = radius of the orbit
= 6300 + 300 = 6600 km
= 6600 x 10³
Formula of time period of the satellite
T = 2π R /v₀ , v₀ is orbital velocity
v₀ = √gR , ( if height is small with respect to radius )
T = 2π R /√gR
= 2π√ R /√g
= 2 x 3.14 x √ 6600 x 10³ / √9.8
= 2 x 3.14 x 256.9 x 10 / 3.13
= 5154.41 s
= 5154.41 / 60 minutes
= 85.91 m
85.9 minutes.
2 ) No of sunrise per day = no of rotation per day
= 24 x 60 / 85.9
= 16.76
or 17 sunrises.
Answer:
No
Explanation:
Let the reference origin be location of ship B in the beginning. We can then create the equation of motion for ship A and ship B in term of time t (hour):
A = 12 - 12t
B = 9t
Since the 2 ship motions are perpendicular with each other, we can calculate the distance between 2 ships in term of t

For the ships to sight each other, distance must be 5 or smaller







Since
then

So our equation has no solution, the answer is no, the 2 ships never sight each other.
The energy of photon in kJ/mol is 329kJ/mol.
Wavelength of radiation is 370nm. The frequency of given wavelength is
ν = c / λ
ν = 3×10^8 / 370×10^-9
ν = 8.11 × 10^14 s^-1
Now the energy of photon is:
E = hν
E = 6.63×10^-34 J.s/photon × 8.11×10^14s^-1
E = 5.41× 10^-19 J/photon
To find in mole
E = 5.41× 10^-19 × 6.022×10^23
E = 3.29 ×10^ 5 J/mol
So, the energy of mole of photon is equal to 329 kJ/mol.
Learn more about radiation here:
brainly.com/question/18650102
#SPJ4
<span>U could compare them using the intensity
technique when bending waves are negligible in comparison with
quasi-longitudinal waves.</span>