V = u + at
0 = u -9.8 x 5.5
u = 9.8 x 5.5 = 53.9 m/s
Answer:
C = 4,174 10³ V / m^{3/4}
, E = 7.19 10² / ∛x, E = 1.5 10³ N/C
Explanation:
For this exercise we can calculate the value of the constant and the electric field produced,
Let's start by calculating the value of the constant C
V = C
C = V / x^{4/3}
C = 220 / (11 10⁻²)^{4/3}
C = 4,174 10³ V / m^{3/4}
To calculate the electric field we use the expression
V = E dx
E = dx / V
E = ∫ dx / C x^{4/3}
E = 1 / C x^{-1/3} / (- 1/3)
E = 1 / C (-3 / x^{1/3})
We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E
E = 3 / C (0- (-1 / x^{1/3}))
E = 3 / 4,174 10³ (1 / x^{1/3})
E = 7.19 10² / ∛x
for x = 0.110 cm
E = 7.19 10² /∛0.11
E = 1.5 10³ N/C
Energy to lift something =
(mass of the object) x (gravity) x (height of the lift).
BUT ...
This simple formula only works if you use the right units.
Mass . . . kilograms
Gravity . . . meters/second²
Height . . . meters
For this question . . .
Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms
Gravity (on Earth) = 9.8 m/second²
Height = 500 cm = 5.0 meters
So we have ...
Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)
= 2,696,925 joules .
That's quite a large amount of energy ... equivalent to
straining at the rate of 1 horsepower for almost exactly an
hour, or burning a 100 watt light bulb for about 7-1/2 hours.
The reason is the large mass that's being lifted.
On Earth, that much mass weighs about 61 tons.
The car’s velocity at the end of this distance is <em>18.17 m/s.</em>
Given the following data:
- Initial velocity, U = 22 m/s
- Deceleration, d = 1.4

To find the car’s velocity at the end of this distance, we would use the third equation of motion;
Mathematically, the third equation of motion is calculated by using the formula;

Substituting the values into the formula, we have;

<em>Final velocity, V = 18.17 m/s</em>
Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>
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Read more: brainly.com/question/8898885
Data Analysis and Conclusion