Answer:
H = V t - 1/2 g t^2 V and g are in different directions
-8.2 = V * 6.2 - 1/2 * 10 * 6.2*2 = 6.2 V - 192.2
V = 184 / 6.2 = 29.7 m/s
Check: find time for ball to return to zero height
0 = 29.7 T - 5 T^2
T = 29.7 / 5 = 5.94 sec
The ball must have fallen 8.2 m in (6.2 - 5.94) sec = .26 sec
S = 29.7 * .26 + 5 * .26^2 = 8.1
Answer:
Explanation:
Argon to potassium ratio after 1 half life = 1:1
After 2 half lives = 75/25= 3:1
After 3 half lives = 87.5/12.5= 7:1
After 4 half lives = 93.75/6.25 = 15:1
After 5 half lives = 96.875/3.125 = 31/1
Complete Question
The speed of a transverse wave on a string of length L and mass m under T is given by the formula
If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string
Answer:
Explanation:
From the question we are told that
Speed of a transverse wave given by
Maximum Tension is 
Generally making 
subject from the equation mathematically we have
Therefore the Linear mass in terms of Velocity is given by
Answer:
303 Ω
Explanation:
Given
Represent the resistors with R1, R2 and RT
R1 = 633
RT = 205
Required
Determine R2
Since it's a parallel connection, it can be solved using.
1/Rt = 1/R1 + 1/R2
Substitute values for R1 and RT
1/205 = 1/633 + 1/R2
Collect Like Terms
1/R2 = 1/205 - 1/633
Take LCM
1/R2 = (633 - 205)/(205 * 633)
1/R2 = 428/129765
Take reciprocal of both sides
R2 = 129765/428
R2 = 303 --- approximated
