determine the minimum gauge pressure needed in the water pipe leading into a building if water is to come out of a faucet on the
1 answer:
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Answer: = 16.49N
Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,
= horizontal component of the weight = mgsinФ
= vertical component of weight = mgcosФ
Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.
by using newton's law of motion, we have that
mgsinФ - = ma
where m = mass of object=5kg
a = acceleration= unknown
Ф = angle of inclination = 37°
g = acceleration due to gravity = 9.8
= frictional force = unknown
we need to first get the acceleration before the frictional force which is gotten by using the equation below
where v = final velocity = 2m/s
u = initial velocity = 0m/s (because the object started from rest)
a= unknown
S= distance covered = length of plane = 5m
we slot in a into the equation below to get frictional force
mgsinФ - = ma
3 * 9.8 * sin 37 - = 3* 0.4
17.9633 - = 1.2
= 17.9633 - 1.2
= 16.49N
Answer:
a) r₁ = 3.836 10⁷ m, b) F = - 3,390 10⁸ N , c) R = 120.3 m
Explanation:
a) This is a problem of equilibrium where the force is gravitational, we call F1 the force of the Moon and F2 the force from the earth.
F₁ -F₂ = 0
F₁ = F₂
G m / r₁² = G m
/ r₂²
Let's look for the measured distance from a Coordinate System located on the Moon,
r₂ = D - r₁
Where D is the average distance from Terra to the Moon 3.84 10⁸ m
/ r₁² =
/ (D - r₁)²
(D² - 2 D r₁ + r₁²) = /
r₁²
(1 - /
)r₁² - 2D r₁ + D² = 0
Let's replace and solve the second degree equation
(1 - 5.98 10²⁴ / 7.36 10²²) r₁² - 2 3.84 10⁸ r₁ + (3.84 10⁸)² = 0
-80.25 r₁² - 7.68 10⁸ r₁ + 14.75 10¹⁶ = 0
r₁² + 9.57 10⁶ r₁ - 1.838 10¹⁵ = 0
r₁ = [-9.57 10⁶ ±√ (91.58 10¹² + 7.352 10¹⁵)] / 2
r₁ = [-9.57 10⁶ + - 86.28 10⁶] / 2
The results are:
r₁’= 38.355 10 6 m
r₁ ’’ = -47.915 10 6 m
We take the positive result that a distance between the moon and the Earth, the equalization point is r₁ = 3.836 10⁷ m
b) The force at point R = 2 r₁
R = 2 3.8355 10⁷ = 7.671 10⁷ m
F = F₁ - F₂
F = G m / R² - G m
/ (D- R)²
F = G m ( / R² -
/ (D-R)²)
F = m 6.67 10⁻¹¹ (7.36 10²² / (7.671 10⁷)² - 5.98 10²⁴ / (3.84 10⁸ - 0.7671 10⁸)²
F = m 6.67 10⁻¹¹ (0.125076 10⁸ - 0.63329 10⁸)
F = m (-3.3897 10⁸) N
The mass m of the rocket must be known, suppose it is worth 1 kg (m = 1 kg)
F = - 3,390 10⁸ N
c) let's use gravitational force from the moon
F = G m / R²
R =√ G m / F
R = √ (6.67 10⁻¹¹ 1 7.36 10²² / 3.3897 10⁸)
R = √ (1.4482 10⁴)
R = 1.2034 10² m
R = 120.3 m