Answer:
1.53seconds
Explanation:
Using first equation of motion :
V=U + at
Where final velocity (V) =+8.3m/s
Initial velocity (U) =+4.4m/s
Acceleration (a) = 0.65m/s^2
time(s)=?
V=U + at
+8.3^2 = +4.4 + 0.65 * t
Making t the subject of the formula :
Therefore, t= ( +8.3 - 4.4)/0.65 = 1.53seconds
<span>1.
</span><span>Efficiency is the
measure of how efficient a process is. It is used to assess the ability of a
process in avoiding waste energy, materials, money and time in doing a
desirable output. It is calculated as;
Efficiency = useful energy ouput / total energy input</span>
<span>.40 = useful work / 200</span>
<span>useful work = 80 joules</span>
Answer:
40m/s
Explanation:
The horizontal component of velocity remains constant because there are no external forces in that direction
By applying motion equations, V= U+ at
where ,
- v - final velocity
- u - initial velocity
- a-acceleration
- t - time
v = u +at
As no force act on the ball ( we neglect air resistance here) no acceleration is seen,
So v = u = 40m/s
Answer:
The shortest braking distance is 35.8 m
Explanation:
To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down
On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis
Y axis
N- W = 0
N = W = mg
X axis
-Fr = m a
-μ N = m a
-μ mg = ma
a = μ g
a = - 0.32 9.8
a = - 3.14 m/s²
We calculate the distance using the kinematics equations
Vf² = Vo² + 2 a x
x = (Vf² - Vo²) / 2 a
When the train stops the speed is zero (Vf = 0)
Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s
x = ( 0 - 15²) / 2 (-3.14)
x= 35.8 m
The shortest braking distance is 35.8 m