Answer:
F = -2527.35 N
Explanation:
Given,
mass per unit time = 58.1 Kg/s
Speed of the water = 43.5 m/s
Final momentum = 0 kg/s
Force = Change in momentum
F = Final momentum - initial momentum
F = 0 - initial momentum
F = -2527.35 N
Hence, the Force exerted is equal to F = -2527.35 N
Answer:
Explanation:
Using the formula to calculate the maximum height
H = u²/2g
u is the initial velocity = 18m/s
g is the acceleration due to gravity = 9.81
H = 18²/2(9.81)
H = 324/19.62
H = 16.51m
The maximum height in which the ball reached from the ground = 85m + 16.51m = 101.5m
b) Time of fight = 2u/g
T = 2(18)/g
T = 36/9.81
T = 3.67s
It took the ball 3.67s later to reach the ground
3) To get the final velocity of the ball as it hits the ground, we need to calculate the horizontal component of the velocity,
Ux = Ucosθ
ux = 18cos 90 (angle of launch is 90 since the ball is thrown vertically upwards)
Ux = 18(0)
Ux = 0m/s
Hence the final velocity of the ball as it hits the ground is 0m/s
Answer:
a) = 72.75 kg
, b) V = 0.07275 m³
, c) ρ₂ = 1030.9 km / m³
, d) ρ₂= 996 kg / m³ float
Explanation:
This is an exercise that we must solve using the Archimedes principle, where the thrust is
B = ρ g V
a) Let's use Newton's second law
F = B - W
F = ρ g V - mg
The force is the apparent weight (m₂ = 2.25 kg) is directed downwards so it is negative
F = m₂ g
-m₂ g = ρ g V - m g
-m₂ g = g - mg
= -m₂ + m
= -2.25 + 75.0
= 72.75 kg
b) let's use the definition of density
ρ = m / V
V = m /ρ
V = 72.75 / 1000
V = 0.07275 m³
c) the density of man is
ρ₂ = m / V
ρ₂ = 75.0 / 0.07275
ρ₂ = 1030.9 km / m³
d) the volume of man increases because the lungs are full of air, as they are half full, with a capacity of 2.5 liters and two lungs the volume is
V’= ½ 2.5 2
V’= 2.5 liters
V’= 2.5 10⁻³ m³
The total volume of man is
Vt = 0.07275 + 0.0025
Vt = 0.07525
Let's calculate the density
ρ₂ = 75.0 / 0.07525
ρ₂= 996 kg / m³
As this is less than the density of water (1000 kg / m3) man must float
Answer:
Please find the answer in the explanation.
Explanation:
Mention three disadvantages of friction between the parts of a machine
Three disadvantages of friction between the part of a machine are;
1. Friction causes wear and tear of machine parts.
2. Friction generates heat between the part of a machine.
3. Friction reduces the efficiency of a machine.
How does.
1 . oiling
2 using ball bearing
To reduce friction, you need to apply a lubricant. Oil and grease are examples of a lubricant. A lubricant helps to reduce friction by making the surface of the machine parts slippery thereby reducing tear and wear.
Also, friction can be reduced by using ball bearing which will allow the rolling of machine parts.