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.
Kinetic energy is the answer to your question.
Answer:
<h3><em>
28.01m/s.</em></h3>
Explanation:
Given maximum height reached by the ball as H = 40 metres
Since the ball rises straight up when hit by a ball, then the angle of launch will be perpendicular to the ground and that is 90°.
To determine the upward speed of the ball in meters per second after it got struck by the bat, we will use the formula for calculating the maximum height according to projectile motion;
Maximum Height H =
where;
u is the speed of the ball
is the angle of launch
g is the acceleration due to gravity = 9.81m/s²
Substituting the given parameters into the formula;

<em>Hence the upward speed of the ball in meters per second after it got struck by the bat is 28.01m/s.</em>
Answer:
The value is
Explanation:
From the question we are told that
The mass is
The needed oscillation time is
Generally the spring constant is mathematically represented as

=> 
=>
Answer:
56°
Explanation:
Brewsters angle can be simply derived from
n1sin theta1= n2sintheta2= n2costheta1
because the reflected light will be 100% polarized if it is reflected at an angle 90o to the refracted light. Hence, Brewsters angle is
Tan theta= n2/n1
1.66/1.11= 1.495
Theta = 56°
Explanation: