Given:
u = 10⁵ m/s, the entrance velocity
v = 2.5 x 10⁶ m/s, the exit velocity
s = 1.6 cm = 0.016 m, distance traveled
Let a = the acceleration.
Then
u² + 2as = v²
(10⁵ m/s)² + 2*(a m/s²)*(0.016 m) = (2.5 x 10⁶ m/s)²
0.032a = 6.25 x 10¹² - 10¹⁰ = 6.24 x 10¹²
a = 1.95 x 10¹⁴ m/s²
Answer: 1.95 x 10¹⁴ m/s²
K= (1/2) m v^2
so 0.5*0.145*<span>1281.64=92.92J of KE</span>
Answer:
angular range is ( 0.681 rad , 0.35 rad )
Explanation:
given data
wavelength λ = 380 nm = 380 × 
m
wavelength λ = 700 nm = 700 × m
to find out
angular range of the first-order
solution
we will apply here slit experiment equation that is
d sinθ = m λ ...........1
here m is 1 for single slit and d is = 
so put here value in equation 1 for 380 nm
we get
d sinθ = m λ
sinθ = 1 × 380 ×
θ = 0.35 rad
and for 700 nm
we get
d sinθ = m λ
sinθ = 1 × 700 ×
θ = 0.681 rad
so angular range is ( 0.681 rad , 0.35 rad )
Answer:
1.76m/s²
Explanation:
Acceleration is the time rate of change in velocity of a body. It is a vector quantity that is it has both magnitude and direction
From newton's second law of motion which states that the rate of change of momentum of a body is proportional to the applied force which takes place in the direction of force applied.
This law gives a formula which relate force, mass and acceleration.
Force = mass x acceleration
Given that force = 9.7N , mass = 5.5kg
Since force(F)= mass(m) x acceleration(a)
Therefore F = ma
Divide both sides by m
F/m = ma/m
Therefore,
Acceleration (a) = F/m
Acceleration = 9.7N ➗ 5.5kg
Acceleration = 1.76m/s²
The S. I unit of acceleration is m/s²
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