<span>The surface charge density = q/A
So q = surface charge density x Area
The surface area of a sphere of radius R is 4*Pi*R^2. R = d/2 where d is diameter. This leaves us with 1.3/2 = 0.65. Area = 4 * pie * (0.65)^2 = 5.30998.
So the net charge q = 8.1 * 10^(-6) * 5.30998 = 42.47998 * 10^(-6)
The Total electric flux = Q/e_0 where , 8.854 Ă— 10â’12, e_0 is permitivity of free space.
So Flux = 42.47998 * 10^(-6) / 8.854 * 10(â’12) = 4.833 * 10^(-6 - (-12)) = 4.833 * 10^(6)</span>
Answer:
1) v = 7.70 10³ m/s
, 2) F = 115 N and 3) (F/W)% = 90.2%
Explanation:
1) To solve the problem let's use Newton's second law where force is gravitational force and acceleration is centripetal
F = ma.
F = G m M / r²
a = v² / r
G m M / r² = m v² / r
G M / r = v²
Let's look for the distance is the distance from the surface of the has to the station 345 103 m plus the radius of the Earth
r = Re + 345 103
r = 6.37 10⁶ + 3.45 10⁵
r = 6.715 10⁶ m
Let's calculate the speed
v = √ (6.67 10⁻¹¹ 5.98 10²⁴ / 6,715 10⁶) = √ (59,399 10⁶)
v = 7.70 10³ m/s
The speed module is constant, so we can use the uniform motion relationships
v = d / t
The distance is the length of the circle
d = 2π r
d = 2π 6.715 106
d = 42.2 10⁶ m
Let's calculate the time
t = d / v
t = 42.2 10⁶ / 7.70 10³
t = 5.48 10³ s
2) Let's use the universal gravitation equation
F = G m M / r²
F = 6.67 10⁻¹¹ 13.0 5.98 10²⁴ /(6.715 10⁶)²
F = 11.5 10¹ N
F = 115 N
3) in this for we are asked the relationship is out with the weight of the body on earth
F / W = F / mg
F / W = 115 / (13.0 9.8)
F / W = 0.902
F / W% = 90.2%
Answer:
Explanation:
Let the length of inclined plane be L .
work done by gravity on the block
= force x length of path
= mg sinθ x L , m is mass of the block , θ is inclination of path
This in converted into potential energy of compressed spring
1/2 k x² = mgL sin31 , k is force constant . x is compression
.5 x 3400 x .37² = 33 x9.8 x sin31 L
L = 1.4
Length of incline = 1.4 m .
Answer:
are you asking how the planets are arranged
Answer:
Explanation:
Distance between ship and enemy ship
= 500 + 610
= 3110 m
Range of projectile
R = u² sin2θ / g
= (250x 250 sin 150) / 9.8
= 3188m
The projectile falls within a distance of 3188 - 3110 = 78 m from enemy ship
Height of mountain = 1800 m
We shall find the height of projectile when its horizontal displacement is 2500m
x = 2500 , y = ?
u = 2500 ,
y = x / cos θ - .5 g x² /u²cos² θ
9660 - 7315 m
= 2345 m
It is within 545 m from mountain peak .
