The mass of the block is 920 kg.
The angle of inclination of the slope with respect to the plane is 30 degree.
The initial speed of the block [u] =50 cm/s
=50×0.01 m/s
=0.5 m/s
The block is once pushed and it moves downward on the slope. The distance of the block from the bottom is 2.3 m .
Hence the distance travelled by the block [s]=2.3 m
We are asked to calculate the final velocity of the block when it touches the ground.
First we have to calculate the acceleration of the block.
Resolving g into horizontal and vertical component we get,
![[1] gsin\theta\ is\ along\ the\ slope\ and\ downward](https://tex.z-dn.net/?f=%5B1%5D%20gsin%5Ctheta%5C%20is%5C%20along%5C%20the%5C%20slope%5C%20and%5C%20downward)
![[2] gcos\theta\ is\ along\ the\ normal\ drawn\ to\ the\ block](https://tex.z-dn.net/?f=%5B2%5D%20gcos%5Ctheta%5C%20is%5C%20along%5C%20the%5C%20normal%5C%20drawn%5C%20to%5C%20the%5C%20block)
Now putting the equation of kinematics we get-
Here v is called final velocity and a is the acceleration.
![=[0.5]^2+2*gsin\theta *2.3](https://tex.z-dn.net/?f=%3D%5B0.5%5D%5E2%2B2%2Agsin%5Ctheta%20%2A2.3)
![=0.25+[2*9.8*0.5*2.3]](https://tex.z-dn.net/?f=%3D0.25%2B%5B2%2A9.8%2A0.5%2A2.3%5D)
[sin30=0.5]
[ans]
Answer:
Explanation:
Conceptual analysis
We apply the kinematic formula for an object that moves vertically upwards:
Where:
: final speed in ft/s
: initial speed in ft/s
g: acceleration due to gravity in ft/s²
y: vertical position at any time in ft
Known data
For 
,
; where h is the maximum height
for y=h, 
Problem development
We replace , 
in the formula (1),
[
Equation (1)
in maximum height(h): , Then we replace in formula (1):
Equation(2)
We replace (h) of Equation(2) in the Equation (1) :
Explanation:
time period= 1/frequency.
= 1/50 = 0.02 second.
hope this helps you.
The amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.
<h3>
Total capacitance of the circuit</h3>
The total capacitance of the circuit is calculated as follows;
Capacitors in series;
1/Ct = 1/8 + 1/7.5
1/Ct = 0.25833
Ct = 3.87 mF
Capacitors is parallel;
Ct = 3.87 mF + 12 mF + 15 mF
Ct = 30.87 mF
Ct = 0.03087 F
<h3>Charge in each capacitor</h3>
Q = CV
Q = 0.03087 x 12
Q = 0.37 C
Thus, the amount of electric charge that resides on each capacitor once it is fully charged is 0.37 C.
Learn more about capacitors here: brainly.com/question/13578522
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