Answer:
Explanation:
Let m be mass of each sphere and θ be angle, string makes with vertex in equilibrium.
Let T be tension in the hanging string
T cosθ = mg ( for balancing in vertical direction )
for balancing in horizontal direction
Tsinθ = F ( F is force of repulsion between two charges sphere)
Dividing the two equations
Tanθ = F / mg
tan17 = F / (7.1 x 10⁻³ x 9.8)
F = 21.27 x 10⁻³ N
if q be charge on each sphere , force of repulsion between the two
F = k q x q / r² ( r is distance between two sphere , r = 2 x .7 x sin17 = .41 m )
21.27 x 10⁻³ = (9 X 10⁹ x q²) / .41²
q² = .3973 x 10⁻¹²
q = .63 x 10⁻⁶ C
no of electrons required = q / charge on a single electron
= .63 x 10⁻⁶ / 1.6 x 10⁻¹⁹
= .39375 x 10¹³
3.9375 x 10¹² .
Area of a circle is
A= pi r^2
so
500m^2 = 3.14 r2
500/pi = r ^2
152.1549...=r^2
square root both sides
r=12.61566...
d=2r
d=25.2
to 3 sig fig
Answer:
400 trips
Explanation:
Mechanical energy needed to climb 14 m by a man of 68 kg
= mgh
= 68 x 9.8 x 14
= 9330 J
1 Kg of fat releases 3.77 x 10⁷ J of energy
.45 kg of fat releases 1.6965 x 10⁷ J of energy
22% is converted into mechanical energy
so 22% of 1.6965 x 10⁷ J
= 3732.3 x 10³ J of mechanical energy will be available for mechanical work.
one trip of climbing of 14 m requires 9330 J of mechanical energy
no of such trip possible with given mechanical energy
= 3732.3 x 10³ / 9330
= 400 trips
Answer:to past school duhh
Explanation:
