Answer:
9.0 m
Explanation:
Applying
V = kq/r......................... Equation 1
Where V = Electric potential, q = charge, r = distance, k = Coulomb's constant.
make r the subject of the equation
r = kq/V................... Equation 2
From the question,
Given: q = 1.00×10⁻⁵ C, V = 10000 V
Constant: k = 9.0×10⁹ N⋅m²/C²
Substitute these values into equation 2
r = (9.0×10⁹×1.00×10⁻⁵)/10000
r = 9.0 m
FIrst, just write down what we know :)
1. 1 mile = 1.6klm
2. There are 3600 seconds in an hr (<em>60s x 60mins = 3600</em>).
Now to get to <em>meters per second</em> (m/s) we need to first work out <em>meters per </em><em>hr</em> (simply convert miles to klm and then multiply by 1000)<em>.</em>
60 x 1.6 = 96 klm/hr x 1000 = 96,000 meters/hr
Now we know this, we can simply divide the meters per hr by <em>how many seconds there are in 1hr</em> to get meters per second:
96000 / 3600 = 26.67m/s
Answer:
Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, \sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}∑F=
r
1
mv
2
=
r
2
mu
2
................ii
form i and ii we can write
v^2= \frac{1}{2} u^2v
2
=
2
1
u
2
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v
Answer:
less than one light year. We would all be dead if it was more than a light year.
Answer:
Fm is the net force down on the metal in air
Fm / 2 is the net force down on the metal in liquid
Fl is the buoyant force on the metal due to liquid
Fm - Fl = Fm / 2 equating upward and downward forces
Fm / 2 = Fl
The specific gravity of the metal is twice that of the liquid
Note: F = M g = ρ V g since ρ = M / V
