Answer:
Explanation:
<h2><u>Given</u> :-</h2>
<h2><u>To Find</u> :-</h2>
<h2><u>Formula to be used</u> :-</h2>
Where,
- K.E. = Kinetic energy possessed by the body
- M = Mass of the body
- V = Velocity of the body
<h2><u>Solution</u> :-</h2>








- Velocity of the vehicle at the instant is

It must gain an electron because if the proton number was to change it would no longer be the same element.
The acceleration of the car is 0.8049
.It takes 13.802s to travel the 230 m.
<h3>
What is acceleration?</h3>
In mechanics, acceleration refers to the rate at which an object's velocity with respect to time varies. Acceleration is a vector quantity (in that they have magnitude and direction). The direction of an object's acceleration is determined by the direction of the net force acting on it. Newton's Second Law states that the combined effect of two factors determines how much an item accelerates:
(i) It follows that the magnitude of the net balance of all external forces acting on the object is directly proportional to the magnitude of this net resulting force, and
(ii) the mass of the thing, depending on the materials out of which it is constructed, is inversely proportional to the mass of the thing.
Calculations:
40 km/hr ----- 11.11m/s
80 km/hr ----- 22.22m/s

Time taken
v-u=at
22.22-11.11= 0.8049 x t
t=13.802s
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The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as
- t=0.476v
- t=1.967v
- V2=4.323v
<h3>What is the potential across the capacitor?</h3>
Question Parameters:
A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
at
- t = 1.0 seconds
- 5.0 seconds
- 20.0 seconds.
Generally, the equation for the Voltage is mathematically given as
v(t)=Vmax=(i-e^{-t/t})
Therefore
For t=1
V=5(i-e^{-1/10})
t=0.476v
For t=5s
V2=5(i-e^{-5/10})
t=1.967
For t=20s
V2=5(i-e^{-20/10})
V2=4.323v
Therefore, the values of voltages at the various times are
- t=0.476v
- t=1.967v
- V2=4.323v
Read more about Voltage
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Complete Question
A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.
Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.
La longitud <em>final</em> del puente de acero es 100.018 metros.
Asumamos que la dilatación <em>térmica</em> experimentada por el puente de acero es <em>pequeña</em>, de modo que podemos emplear la siguiente aproximación <em>lineal</em> para determinar la longitud <em>final</em> del puente de acero (
), en metros:
(1)
Donde:
- Longitud inicial del puente, en metros.
- Coeficiente de dilatación, sin unidad.
- Temperatura inicial, en grados Celsius.
- Temperatura final, en grados Celsius.
Si tenemos que
,
,
y
, entonces la longitud final del puente de acero es:
![L = (100\,m)\cdot [1+(11.5\times 10^{-6})\cdot (24\,^{\circ}C - 8\,^{\circ}C)]](https://tex.z-dn.net/?f=L%20%3D%20%28100%5C%2Cm%29%5Ccdot%20%5B1%2B%2811.5%5Ctimes%2010%5E%7B-6%7D%29%5Ccdot%20%2824%5C%2C%5E%7B%5Ccirc%7DC%20-%208%5C%2C%5E%7B%5Ccirc%7DC%29%5D)

La longitud <em>final</em> del puente de acero es 100.018 metros.
Para aprender más sobre dilatación térmica, invitamos cordialmente a ver esta pregunta verificada: brainly.com/question/24953416