Answer:
0.8 x 10^-9 kg
Explanation:
Given,
Distance ( R ) = 10 m
Force ( F ) = 3.2 x 10^-9 N
Mass ( m1 ) = 40 kg
To find : Mass ( m2 ) = ?
Formula : -
F = m1.m2 / R^2
m2 = FR^2 / m1
= 3.2 x 10^-9 x 10 / 40
= 3.2 x 10^-9 / 4
= ( 3.2 / 4 ) x 10^-9
m2 = 0.8 x 10^-9 kg
Answer:
(A) The speed just as it left the ground is 30.25 m/s
(B) The maximum height of the rock is 46.69 m
Explanation:
Given;
weight of rock, w = mg = 20 N
speed of the rock at 14.8 m, u = 25 m/s
(a) Apply work energy theorem to find its speed just as it left the ground
work = Δ kinetic energy
F x d = ¹/₂mv² - ¹/₂mu²
mg x d = ¹/₂m(v² - u²)
g x d = ¹/₂(v² - u²)
gd = ¹/₂(v² - u²)
2gd = v² - u²
v² = 2gd + u²
v² = 2(9.8)(14.8) + (25)²
v² = 915.05
v = √915.05
v = 30.25 m/s
B) Use the work-energy theorem to find its maximum height
the initial velocity of the rock = 30.25 m/s
at maximum height, the final velocity = 0
- mg x H = ¹/₂mv² - ¹/₂mu²
- mg x H = ¹/₂m(0) - ¹/₂mu²
- mg x H = - ¹/₂mu²
2g x H = u²
H = u² / 2g
H = (30.25)² / 2(9.8)
H = 46.69 m
Answer:
b
Explanation:
the ability for gases to compress is extremely helpful it allows tanks of oxygen to hold enough air for up to two hours and the strange thing about compression is that it allows some liquids to stay liquid at their boiling point allowing liquid nitrogen to stay liquid at room temperature
In an uniform circular motion, the direction of the net force on the object is radially inward, passing through the center of the circle.
Answer:
v₂ = 70 m / s
Explanation:
For this exercise let's use Bernoulli's equation
where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂
indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute
ρ g y₁ = ½ ρ v₂²
v₂ = 
let's calculate
v₂ = √( 2 9.8 250)
v₂ = 70 m / s
