Ball A has a mass of 50 kg. Ball B has a mass of 75 kg. Both move at the same velocity. Which has a greater momentum?

Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diamete

Sonbull [250]

Answer:

work done in pumping the entire fuel is 466587 J

Explanation:

weight of the gasoline per volume = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 2 m

height of the tractor tank above the top of the tank = 5 m

work done in pumping fuel to this height = ?

First, we find the volume of the fuel

since the tank is cylindrical, we assume that the fuel within also takes the cylindrical shape.

Also, we assume that the fuel completely fills the tank.

volume of a cylinder = $\pi r^{2}l$

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x $1.5^{2}$ x 2 = 14.139 m^3

we then find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 14.139 = 93317.4 N

work done = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 93317.4 x 5 = 466587 J

8 0

3 years ago

Which energy transformation occurs in an

Mnenie [13.5K]

A motor takes in electricity from a cell or power source and then uses it to move components.

Therefore the answer is (1) Electrical to Mechanical

5 0

3 years ago

Read 2 more answers

A parallel plate capacitor is connected to a DC battery supplying a constant DC voltage V0= 600V via a resistor R=1845MΩ. The ba

tensa zangetsu [6.8K]

Answer:

See explanation

Explanation:

Given:-

- The DC power supply, Vo = 600 V

- The resistor, R = 1845 MΩ

- The plate area, A = 58.3 cm^2

- Left plate , ground, V = 0

- The right plate, positive potential.

- The distance between the two plates, D = 0.3 m

- The mass of the charge, m = 0.4 g

- The charge, q = 3*10^-5 C

- The point C = ( 0.25 , 12 )

- The point A = ( 0.05 , 12 )

Find:-

What is the speed, v, of that charge when it reaches point A(0.05,12)?

How long would it take the charge to reach point A?

Solution:-

- The Electric field strength ( E ) between the capacitor plates, can be evaluated by the potential difference ( Vo ) of the Dc power supply.

E = Vo / D

E = 600 / 0.3

E = 2,000 V / m

- The electrostatic force (Fe) experienced by the charge placed at point C, can be evaluated:

Fe = E*q

Fe = (2,000 V / m) * ( 3*10^-5 C)

Fe = 0.06 N

- Assuming the gravitational forces ( Weight of the particle ) to be insignificant. The motion of the particle is only in "x" direction under the influence of Electric force (Fe). Apply Newton's equation of motion:

Fnet = m*a

Where, a : The acceleration of the object/particle.

- The only unbalanced force acting on the particle is (Fe):

Fe = m*a

a = Fe / m

a = 0.06 / 0.0004

a = 150 m/s^2

- The particle has a constant acceleration ( a = 150 m/s^2 ). Now the distance between (s) between two points is:

s = C - A

s = ( 0.25 , 12 ) - ( 0.05 , 12 )

s = 0.2 m

- The particle was placed at point C; hence, velocity vi = 0 m/s. Then the velocity at point A would be vf. The particle accelerates under the influence of electric field. Using third equation of motion, evaluate (vf) at point A:

vf^2 = vi^2 + 2*a*s

vf^2 = 0 + 2*0.2*150

vf = √60

vf = 7.746 m/s

- Now, use the first equation of motion to determine the time taken (t) by particle to reach point A:

vf - vi = a*t

t = ( 7.746 - 0 ) / 150

t = 0.0516 s

- The charge placed at point C, the Dc power supply is connected across the capacitor plates. The capacitor starts to charge at a certain rate with respect to time (t). The charge (Q) at time t is given by:

$Q = c*Vo*[ 1 - e^(^-^t^/^R^C^)]$

- Where, The constant c : The capacitance of the capacitor.

- The Electric field strength (E) across the plates; hence, the electrostatic force ( Fe ) is also a function of time:

$E = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D} \\\\Fe = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D}*q\\\\$

- Again, apply the Newton's second law of motion and determine the acceleration (a):

Fe = m*a

a = Fe / m

$a = \frac{Vo*q*[ 1 - e^(^-^t^/^R^C^)]}{m*D}$

- Where the acceleration is rate of change of velocity "dv/dt":

$\frac{dv}{dt} = \frac{Vo*q}{m*D} - \frac{Vo*q*[ e^(^-^t^/^R^C^)]}{m*D}\\\\B = \frac{600*3*10^-^5}{0.0004*0.3} = 150, \\\\\frac{dv}{dt} = 150*( 1 - [ e^(^-^t^/^R^C^)])\\\\$

- Where the capacitance (c) for a parallel plate capacitor can be determined from the following equation:

$c = \frac{A*eo}{d}$

Where, eo = 8.854 * 10^-12 .... permittivity of free space.

$K = \frac{1}{RC} = \frac{D}{R*A*eo} = \frac{0.3}{1845*58.3*8.854*10^-^1^2*1000} = 315\\\\$

- The differential equation turns out ot be:

$\frac{dv}{dt} = 150*( 1 - [ e^(^-^K^t^)]) = 150*( 1 - [ e^(^-^3^1^5^t^)]) \\\\$

- Separate the variables the integrate over the interval :

t : ( 0 , t )

v : ( 0 , vf )

Therefore,

$\int\limits^v_0 {dv} \, = \int\limits^t_0 {150*( 1 - [ e^(^-^3^1^5^t^)])} .dt \\\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^)}{315} )^t_0\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^) - 1}{315} )$