A car initially at rest, accelerates at a constant rate of 4.0 m/s for 6s. How fast will the car be traveling at 6s

A liquid with a density of 900kg/m 3 is stored in a pressurized, closed storage tank. The tank is cylindrical with a 10m diamete

Orlov [11]

Answer:

18.62 m/s

Explanation:

Given that:

A liquid with a density of 900 kg/m 3 is stored in a pressurized, closed storage tank.

Diameter of the tank = 10 m

The absolute pressure in the tank above the liquid is 200 kPa = 200, 000 Pa

At pressure of 200 kPa ; the final velocity = 0

Atmospheric pressure at 5cm = 101325 Pa

We are to calculate the initial velocity of a fluid jet when a 5cm diameter orifice is opened at point A?

By using Bernoulli's theorem between the shaded portion in the diagram;

we have:

$Pa \ + \ \frac{1}{2} \ \delta \ v^2_1 \ + \ \delta gy_1 = P + \delta gy_2$

$\frac{1}{2} \ \delta \ v^2_1 \ = P + \delta gy_2 - \ \delta gy_1 - Pa$

$\frac{1}{2} \ \delta \ v^2_1 \ = \delta g(y_2 -y_1 )+ ( P - Pa )$

$v_1 \ = \sqrt{ \frac {2 \ \delta g(y_2 -y_1 )+ 2 ( P - Pa )}{\delta}}$

where;

Pa = atmospheric pressure = 101325 Pa

$\delta$ = density of liquid = 900 kg/m³

$v_1$ = initial velocity = ???

g = 9.8 m/s²

$y_1$ = height of the hole from the buttom

$y_2$ = height of the liquid surface from the button

$v_1 \ = \sqrt{ \frac {2*900*9.8(7 - 0.5 )+ 2 ( 200,000 - 101325 )}{900}}$

$v_1 = 18.62 \ m/s$

Thus, the initial velocity of the fluid jet = 18.62 m/s

3 0

3 years ago

1. A giri rides her bike at 15 m/s for 20 s. How far does she travel in that time?

Svetach [21]

Answer:

300m

Explanation:

distance=speed x time

=15 x 20

= 300m

6 0

3 years ago

Read 2 more answers

The tibia is a lower leg bone (shin bone) in a human. The maximum strain that the tibia can experience before fracturing corresp

IRISSAK [1]

Answer:

a

$k = 11600000 N/m$

b

$\Delta L = 3.2323 *10^{-5} \ m$

c

$F = 3750.28 \ N$

Explanation:

From the question we are told that

The Young modulus is $E = 1.4 *10^{10} \ N/m^2$

The length is $L = 0.35 \ m$

The area is $2.9 \ cm^2 = 2.9 *10^{-4} \ m ^2$

Generally the force acting on the tibia is mathematically represented as

$F = \frac{E * A * \Delta L }{L}$ derived from young modulus equation

Now this force can also be mathematically represented as

$F = k * \Delta L$

So

$k = \frac{E * A }{L}$

substituting values

$k = \frac{1.4 *10^{10} * 2.9 *10^{-4} }{ 0.35}$

$k = 11600000 N/m$

Since the tibia support half the weight then the force experienced by the tibia is

$F_k = \frac{750 }{2} = 375 \ N$

From the above equation the extension (compression) is mathematically represented as

$\Delta L = \frac{ F_k * L }{ A * E }$

substituting values

$\Delta L = \frac{ 375 * 0.35 }{ (2.9 *10^{-4}) * 1.4*10^{10} }$

$\Delta L = 3.2323 *10^{-5} \ m$

From the above equation the maximum force is

$F = \frac{1.4*10^{10} * (2.9*10^{-4}) * 3.233*10^{-5} }{ 0.35}$

$F = 3750.28 \ N$

4 0

3 years ago

(SCIENCE) 22 POINTS AND CROWN ASAP

gayaneshka [121]

Answer:

A or D

Explanation:

Net force includes addition or subtraction.

5 0

3 years ago

During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapu

ella [17]

Answer:

The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules

Explanation:

The given spring constant of the of the spring, k = 88.0 N/m

The length by which the hose is stretched, x = 4.20 m

For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose

The elastic potential energy, P.E., of a compressed spring is given as follows;

P.E. = 1/2·k·x²

∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²

1/2 × 88.0 N/m × (4.20 m)² = 776.16 J

The work done on the hose = The potential energy given to hose, P.E. = 776.16 J

5 0

3 years ago