When a photon hits an electron <span>the photon will be partially </span>absorbed<span>and the electron emits another photon with lower energy.</span>
Answer: hi your question is incomplete attached below is the complete question
answer :
magnitude of acceleration : | a | = g = 9.81 m/s^2
direction : a = - g j
Explanation:
Neglecting Air resistance
magnitude of acceleration :
| a | = g = 9.81 m/s^2
Direction of acceleration
a = - g j ( given that the direction of acceleration is against the acceleration due to gravity i.e. in the opposite direction )
Answer:
a:it speed up
b:it should be positive since final
velocity is larger than initial velocity
c:acceleration is approximately 4.5
m/s^2
Explanation:
initial velocity=u=4.47m/s
Final velocity=v=17.9m/s
Time=t=3 seconds
a:the car speed up since the velocity
increased
b:change in velocity is positive
because final velocity is larger than
initial velocity
17.9-4.47=13.43 m/s
c: acceleration=(v-u)/t
acceleration=(17.9-4.47)/3
acceleration=13.43/3
acceleration=4.5 m/s^2
Answer:
b
Explanation:
40+273
u have to choose option b
Answer:
Explanation:
We know that,
Neptune is 4.5×10^9 km from the sun
And given that,
Earth is 1.5×10^8km from sun
Then,
Let P be the orbital period and
Let a be the semi-major axis
Using Keplers third law
Then, the relation between the orbital period and the semi major axis is
P² ∝ a³
Then,
P² = ka³
P²/a³ = k
So,
P(earth)²/a(earth)³ = P(neptune)² / a(neptune)³
Period of earth P(earth) =1year
Semi major axis of earth is
a(earth) = 1.5×10^8km
The semi major axis of Neptune is
a (Neptune) = 4.5×10^9km
So,
P(E)²/a(E)³ = P(N)² / a(N)³
1² / (1.5×10^8)³ = P(N)² / (4.5×10^9)³
Cross multiply
P(N)² = (4.5×10^9)³ / (1.5×10^8)³
P(N)² = 27000
P(N) =√27000
P(N) = 164.32years
The period of Neptune is 164.32years