Answer:
Explanation:
The cannonball goes a horizontal distance of 275 m . It travels a vertical distance of 100 m
Time taken to cover vertical distance = t ,
Initial velocity u = 0
distance s = 100 m
acceleration a = 9.8 m /s²
s = ut + 1/2 g t²
100 = .5 x 9.8 x t²
t = 4.51 s
During this time it travels horizontally also uniformly so
horizontal velocity Vx = horizontal displacement / time
= 275 / 4.51 = 60.97 m /s
Vertical velocity Vy
Vy = u + gt
= 0 + 9.8 x 4.51
= 44.2 m /s
Resultant velocity
V = √ ( 44.2² + 60.97² )
= √ ( 1953.64 + 3717.34 )
= 75.3 m /s
Angle with horizontal Ф
TanФ = Vy / Vx
= 44.2 / 60.97
= .725
Ф = 36⁰ .
Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m
Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²
Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s
The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s
Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.
Answer: The speed will be 30 m/s .
Explanation:
Given: Initial velocity of the car: u = 0 m/s
Constant Acceleration: a = 5 m/s²
Time: t= 6 seconds
To find: Final velocity(v)
Formula: v = u+at
Substitute values in the formula, we get
v= 0+(5)(6) m/s
⇒ v= 30 m/s
i.e. Final velocity = 30 m/s
Hence, the speed will be 30 m/s .
Answer:
The pressure will be transmitted equally to all other parts of the confined fluid causing a general increase in pressure throughout the container.
Explanation:
This is in line with pascal's law of pressure which states that the pressure exerted on a given mass of fluid is transmitted undiminished to other parts of the fluid.
Answer:
ФE = 9.403W
Explanation:
In order to calculate the magnitude of the electric flux trough the sheet, you use the following formula:
(1)
A: area of the rectangular sheet = (0.400m)(0.600m) = 0.24m^2
E: magnitude of the electric field = 95.0N/C
θ: angle between the direction of the electric field and the normal to the surface of the sheet
You replace the values of the parameters in the equation (1):
The magnitude of the electric flux is trough the sheet is 9.403W
