Thank you for posting your question your question here. Below is the solution:
time = 3.00 / 1.75
<span>time = 1.714 </span>
<span>now </span>
<span>distance = (initial velocity + final velocity) / 2 * time/ 1 </span>
<span>Hence</span>
<span>3 = ( 1.75 + final velocity) / 2 * 1.714 </span>
<span>3.5 - 1.75 = final velocity </span>
<span>Therefore, the final velocity is 1.75 m /s</span>
C. Acceleration is the rate of change of velocity. So at the top of the path, while the velocity is zero, the CONSTANT GRAVITATIONAL ACCELERATION is about 10 m/s^2 (9.8)
Answer:
NO
Explanation:
The angular position of a diffraction null is given by
Asinθ=m λ=m c/f
Given that A is the slit width,
and m is an integer.
The wavelength of the second null satisfies
λ= A sinθ₁/2
of which if there were distinct third null it will satisfy
sinθ₂=3λ/A= 3* sin 45/2
λ=(3*0.7071)/2
λ=2.121/2
λ=1.060
Of which 1.060 is greater than 1
Answer:
The time required for sucrose transportation through the tube is 8.4319 sec.
Explanation:
Given:
L = 0.025 m
A = 6.5×10^-4 m^2
D = 5×10^-10 m^2/s
ΔC = 5.2 x 10^-3 kg/m^3
m = 5.7×10^-13 kg
Solution:
t = m×L / D×A×ΔC
t = (5.7×10^-13) × (0.025) / (5×10^-10)×(6.5×10^-4)×(5.2 x 10^-3)
t = 8.4319 sec.
Answer:
36.3 kgm2
Explanation:
According to the conservation of angular momentum, the angular momentum is preserved before and after the child grab. Since the moments of inertia increase, the angular velocity decreases.
Let I be the moment of inertia of the merry-go-round and treat the child as a point particle, his moment of inertia would be
Hence the moment inertia of the child-merry-go-round system is:
I + 55.5
From here we can apply the conservation theory