Complete question:
A train has an initial velocity of 44m/s and an acceleration of -4m/s². calculate its velocity after 10s ?
Answer:
the final velocity of the train is 4 m/s.
Explanation:
Given;
initial velocity of the train, u = 44 m/s
acceleration of the train, a = -4m/s² (the negative sign shows that the train is decelerating)
time of motion, t = 10 s
let the final velocity of the train = v
The final velocity of the train is calculated using the following kinematic equation;
v = u + at
v = 44 + (-4 x 10)
v = 44 - 40
v = 4 m/s
Therefore, the final velocity of the train is 4 m/s.
For E = 200 gpa and i = 65. 0(106) mm4, the slope of end a of the cantilevered beam is mathematically given as
A=0.0048rads
<h3>What is the slope of end a of the cantilevered beam?</h3>
Generally, the equation for the is mathematically given as
Therefore
A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}
A=0.00288+0.00192=0.0048rads
A=0.0048rads
In conclusion, the slope is
A=0.0048rads
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Since my givens are x = .550m [Vsub0] = unknown
[Asubx] = =9.80
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0])
Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in
0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)
0 = [Vsub0x]^2 -10.78
10.78 = [Vsub0x]^2
Sqrt(10.78) = 3.28 m/s
Answer:
<u>1.8kJ</u>
Explanation:
Formula :
<u>Energy used = Power x time</u>
<u />
===============================================================
Given :
⇒ Power = 30 W
⇒ Time = 1 minute = 60 seconds
=============================================================
Solving :
⇒ Energy used = 30 W × 60 s
⇒ Energy used = 1,800 J
⇒ Energy used = <u>1.8kJ</u>
