To solve this problem we will apply the concepts of acceleration according to the laws of kinetics. Acceleration can be defined as the change in velocity per unit time, that is,
Here,
v = Final velocity
u = Initial velocity
t = Time
Rearranging to find the initial velocity,
Now the acceleration is equal to the gravity, then,
The velocity of the shell at 5.5s after the launch is,
Therefore the magnitude of the velocity of the shell at 5.5s is -19.26m/s
Answer:
I believe the answer is 5718.75. Respond if it is wrong please.
Explanation:
I used a calculator.
For this case we have that by definition, the momentum equation is given by:
Where:
m: It is the mass
v: It is the velocity
According to the data we have:
Substituting:
On the other hand, if we clear the variable "mass" we have:
According to the data we have:
Thus, the mass is 
Answer:
Answer:
834 milli-volts
Explanation:
Data provided in the question:
Current = 139 milliamps = 139 × 10⁻³ A
resistance = 6 ohms
Now,
The relation between the current , resistance and voltage is given as:
Voltage = Current × Resistance
on substituting the respective values, we get
Voltage = 139 × 10⁻³ × 6
or
Voltage = 834 milli-volts
"increments of 8" means the major divisions are 0,8,16,24 ?
<span>x axis, calculate the moment arms from 0 </span>
<span>3x4, 2x12, 1x20 </span>
<span>from an arbitrary C </span>
<span>3(c-4) + 2(c-12) + (c-20) = 0 </span>
<span>3c - 12 + 2c -24 + c - 20 = 0 </span>
<span>6c = 56 </span>
<span>c = 9.33 </span>
<span>y axis </span>
<span>3x3, 1x12, 2x20 </span>
<span>3(c-4) + 1(c-12) +2 (c-20) = 0 </span>
<span>3c - 12 + c - 12 + 2c - 40 = 0 </span>
<span>6c = 64 </span>
<span>c = 10.67 </span>
<span>so center is x = 9.33, y = 10.67 </span>
