First, we write the SI prefixed. The SI unit for distance is meters.
Kilo = 10³
Mega = 10⁶
Giga = 10⁹
Terra = 10¹²
Because our value has ten to the power of 11, we will use the closest and lowest power prefix, which is giga.
1.5 x 10¹¹ / 10⁹
= 1.5 x 10² Gm or 150 Gm
Writing in kilometers, we simply repeat the procedure except we divide by 10³ this time.
1.5 x 10¹¹ / 10³
= 1.5 x 10⁸ km
Answer:
2046.37 kPa
Explanation:
Given:
Number of moles, n = 125
Temperature, T = 20° C = 20 + 273 = 293 K
Radius of the cylinder, r = 17 cm = 0.17 m
Height of the cylinder, h = 1.64 m
thus,
volume of the cylinder, V = πr²h
= π × 0.17² × 1.64
= 0.148 m³
Now,
From the ideal gas law
we have
PV = nRT
here,
P is the pressure
R is the ideal gas constant = 8.314 J / mol. K
thus,
P × 0.148 = 125 × 8.314 × 293
or
P × 0.148 = 304500.25
or
P = 2046372.64 Pa = 2046.37 kPa
Answer:
* The value of the magnetic field changes either in time or space
* The waxed area changes, the bow is fitting in size
* The angle between the field and the area changes
Explanation:
Magnetic flux is the scalar product of the magnetic field over the area
Ф = ∫ B. dA
where B is the magnetic field and A is the area
Let's look at stationary, for which factors affect flow
* The value of the magnetic field changes either in time or space
* The waxed area changes, the bow is fitting in size
* The angle between the field and the area changes
Answer:
The <em>correct</em> statements are:
- <em>A. The electric field is nonuniform.</em>
- <em>D. Charge Q is positive.</em>
- <em>E. If charge A moves toward charge Q, it must be a negative charge</em>
Explanation:
The answer choices are:
- A. The electric field is nonuniform.
- B. The electric field is uniform.
- E. If charge A moves toward charge Q, it must be a negative charge.
- F. If charge A moves toward charge Q, it must be a positive charge.
<h2>Solution</h2>
The <em>electric field</em> is the electrostatic force per unit of charge,

around around a charge, where another charge would experience the electrostatic force.
The electric field lines are shown in a diagram with arrows ditributed radially away from a positive charge and radially toward a negative charge.
Since the arrows are away from Q, Q is a positive charge: <em>statement D.</em>
Since the size of the arrows decreases as you move away from Q the stregth of the field is not uniform: <em>statement A.</em>
Since the charge Q is positive, a negative charge would be attracted toward it: <em>statement E.</em>
T=m1a+m2a
a=v^2/r v=r*w w=2πF--->a=r(2f)^2 <r1 for m1 and r2 for m2>
T=(m1r1+m2r2)(2πf)^2 <i factored out (2πf)^2