The acceleration of the body is 2 m/s^2 while the deceleration is - 1.2 m/s^2.
<h3>
What is the acceleration?</h3>
Let us recall that the acceleration is the change in the speed of a body with time. We have been told that the body accelerates for 3s and then decelerates to 2s. This implies that the total time that the object spent in motion is 5 s.
Thus;
v = u + at
v = final velocity
u = initial velocity
a = acceleration
t = time taken
v - u/t = a
a = 6 - 0/3
= 2 m/s^2
Again;
v - u/t = a
a = 0 - 6/5
a = - 1.2m/s^2
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<h3 />
The wavelength of a photon of electromagnetic radiation with a frequency of 151.7 Mhz is 1.978 meter
According to the question
The frequency of a photon of electromagnetic radiation = 151.7 Mhz
By using the Formula of wavelength and frequency
λ = C/f
Where,
λ (Lambda) = Wavelength in meters
C = Speed of Light (c = 3x10⁸ m/s)
f = Frequency
Now, the wavelength of a photon of electromagnetic radiation with a frequency of 151.7 Mhz is
1 Mhz = 1000000 Hz
151.7 Mhz = 151.7 * 10⁶ Hz
λ = C/f
Substituting the value of C and f
λ =
λ = 1.978 meter
Hence, the wavelength of a photon of electromagnetic radiation with a frequency of 151.7 Mhz is 1.978 meter
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Answer:
Lower energy shell which will be nearer to the nucleus.
Explanation:
When electron move from one energy level to another, an electron must gain or lose just the right amount of energy.
When atoms releases energy, electrons move into lower energy levels. The electrons in the shells aways from the nucleus have more energy as compared to the electrons in the nearer shells.
Electrons with the lowest energy are found closest to the nucleus, where the attractive force of the positively charged nucleus is the greatest. Electrons that have higher energy are found further away
Friction I believe is the answer
hope this helps
Answer:
a) T = 2.26 N, b) v = 1.68 m / s
Explanation:
We use Newton's second law
Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string
sin 30 =
cos 30 =
Tₓ = T sin 30
T_y = T cos 30
Y axis
T_y -W = 0
T cos 30 = mg (1)
X axis
Tₓ = m a
they relate it is centripetal
a = v² / r
we substitute
T sin 30 = m
(2)
a) we substitute in 1
T =
T =
T = 2.26 N
b) from equation 2
v² =
If we know the length of the string
sin 30 = r / L
r = L sin 30
we substitute
v² =
v² =
For the problem let us take L = 1 m
let's calculate
v =
v = 1.68 m / s