Answer:
<em>The work done by the car is 363 kJ</em>
Explanation:
Work : Work is said to be done when a Force moves an object through a certain distance. Work and Energy are interchangeable because they have the same unit. The unit of work is Joules (J).
Mathematically work done can be expressed as,
E = W = 1/2mv²
W = 1/2mv²................................ Equation 1
Where E = Energy, W = work done, m = mass of the car, v = velocity of the car
<em>Given: m=1500 kg, v=22 m/s</em>
<em>Substituting these values into equation 1</em>
<em>W = 1/2(1500)(22)²</em>
<em>W = 750 × 484</em>
<em>W = 363000 J</em>
<em>W = 363 kJ</em>
<em>Thus the work done by the car is 363 kJ</em>
Answer:
9.8 m/s/s
Explanation:
The numerical value, in meters per second squared, of the acceleration of an object experiencing true free fall is 9.8 m/s/s. This is called the acceleration due to gravity.
It is known as silicon dioxide or silica!
Hope this helps!
Answer:
Hi Carter,
The complete answer along with the explanation is shown below.
I hope it will clear your query
Pls rate me brainliest bro
Explanation:
The magnitude of the magnetic field on the axis of a circular loop, a distance z from the loop center, is given by Eq.:
B
= NμοiR² / 2(R²+Z²)³÷²
where
R is the radius of the loop
N is the number of turns
i is the current.
Both of the loops in the problem have the same radius, the same number of turns, and carry the same current. The currents are in the same sense, and the fields they produce are in the same direction in the region between them. We place the origin at the center of the left-hand loop and let x be the coordinate of a point on the axis between the loops. To calculate the field of the left-hand loop, we set z = x in the equation above. The chosen point on the axis is a distance s – x from the center of the right-hand loop. To calculate the field it produces, we put z = s – x in the equation above. The total field at the point is therefore
B
= NμοiR²/2 [1/ 2(R²+x²)³÷² + 1/ 2(R²+x²-2sx+s²)³÷²]
Its derivative with respect to x is
dB
/dx= - NμοiR²/2 [3x/ (R²+x²)⁵÷² + 3(x-s)/(R²+x²-2sx+s²)⁵÷²
]
When this is evaluated for x = s/2 (the midpoint between the loops) the result is
dB
/dx= - NμοiR²/2 [3(s/2)/ (R²+s²/4)⁵÷² - 3(s/2)/(R²+s²/4)⁵÷²
] =0
independent of the value of s.
The Earth rotates due to the manner in which it was formed. Earth began its rotation from the spinning movement of the accretion disk. In short, the earth rotates because of angular momentum caused by the accretion disk.
