Ask question

Ask question

All categories

3 years ago

10

1. A 2m-long and 3m-wide horizontal rectangular plate is submerged in water. The distance of the top surface from the free surfa

ce is 5m. The atmospheric pressure is 95kPa. Calculate the hydrostatic force acting on the top surface of this plate

1 answer:

ExtremeBDS [4]3 years ago

8 0

Answer:

864 KN

Explanation:

(Absolute pressure) = (Atmospheric pressure) + (Gauge Pressure)

Atmospheric pressure = 95 KPa = 95000 Pa

Gauge Pressure = ρgh

ρ = density of the fluid = 1000 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = depth below the fluid level that the object is at = 5 m

Gauge Pressure = 1000 × 9.8 × 5 = 49000 Pa

Absolute pressure = 95000 + 49000 = 144000 Pa.

Pressure = (Hydrostatic force)/(Area perpendicular to the force)

Hydrostatic force = (Pressure) × (Area perpendicular to the force)

Area perpendicular to the force = 2 × 3 = 6 m²

Hydrostatic force on the top of the plate = 144000 × 6 = 864000 N = 864 KN

Hope this Helps!!!

You might be interested in

A white pool ball of mass 1.0 kg moving at 10 m/s collides with

solniwko [45]

The velocity of the red ball after the collision is 5.8 m/s

Explanation:

In absence of external forces on the system, we can apply the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision, so we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 1.0 kg$ is the mass of the pool ball

$u_1 = 10 m/s$ is the initial velocity of the pool ball

$v_1 = 3.0 m/s$ is the final velocity of the pool ball

$m_2 = 1.2 kg$ is the mass of the red ball

$u_2 = 0$ is the initial velocity of the red ball

$v_2$ is the final velocity of the red ball

Solving the equation for v2, we find the final velocity of the red ball after the collision:

$v_2 = \frac{m_1 u_1-m_1v_1}{m_2}=\frac{(1.0)(10)-(1.0)(3.0)}{1.2}=5.8 m/s$

Learn more about collisions:

brainly.com/question/13966693#

brainly.com/question/6439920

#LearnwithBrainly

7 0

3 years ago

If the mass of the earth and all objects on it were suddenly doubled, then the ________

krok68 [10]

Answer:

The force of universal gravitation between earth and all objects in it will be quadrupled

Explanation:

Newtons law of universal gravitation tell us the force of attraction between two bodies on the earth surface . This force is directly proportional to the product of the masses of the bodies and inversely proportional to their distance apart.

f is directly proportional to m1*m2

doubling these masses will be 2m1*2m2 = 4m1m2

if mas of earth is Me and mass of all objects is Mo

the f is proportional to Me*Mo

doubling the masses of the earth and all objects we have, 2Me*2Mo= 4MeMo

This means that doubling the masses of the earth and all objects on it will cause the force of gravitational attraction to be quadrupled.

3 0

3 years ago

A hiker walks due east for a distance of 25.5 km from her base camp. On the second day, she walks 41.0 km northwest till she dis

GarryVolchara [31]

Resultant displacement is 29.2 km at $83.1^{\circ}$ north of west

Explanation:

To solve the problem, we have to use the rules of vector addition, resolving first each vector along the x- and y- direction.

Taking east as positive x direction and north as positive y- direction, we have:

- First displacement is 25.5 km east, therefore its components are

$A_x = 25.5 km\\A_y = 0 km$

- Second displacement is 41.0 km northwest, so its components are

$B_x = (41.0)cos(135^{\circ})=-29.0 km\\B_y =(41.0)sin(135^{\circ})=29.0 km$

So, the components of the resultant displacement are

$R_x=A_x+B_x=25.5+(-29.0)=-3.5 km\\R_y=A_y+B_y=0+29.0=29.0 km$

And so, the magnitude is calculated using Pythagorean's theorem:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(-3.5)^2+(29.0)^2}=29.2 km$

And the direction is given by

$\theta=tan^{-1}(\frac{R_y}{|R_x|})=tan^{-1}(\frac{29.0}{3.5})=83.1^{\circ}$

Where the angle is measured from the west direction, since Rx is negative.

Learn more about displacement:

brainly.com/question/3969582

#LearnwithBrainly

3 0

3 years ago

Acceleration due to gravity on the moon is 1.6 m/s^2 or about 16% of the value of g on Earth. If an astronaut on the moon threw

Digiron [165]

Answer:

easy the ansewr is 3

Explanation:

because i calculated

7 0

3 years ago

Energy of a SpacecraftVery far from earth(at R=\infty), a spacecraft has run out of fuel and its kineticenergy is zero. If only

Firdavs [7]

Answer:

$s_e=\sqrt{\frac{2GM_e}{R_e^2}}$

Explanation:

In this case mechanical energy is conserved, which means that the sum of the initial kinetic energy and initial potential gravitational energy will be equal to the sum of the final kinetic energy and final potential gravitational energy:

$K_i+U_i=K_f+U_f$

Which in our case will be:

$\frac{mv_i^2}{2}+\frac{-GM_em}{r_i^2}=\frac{mv_f^2}{2}+\frac{-GM_em}{r_f^2}$

Which, since $v_i=0m/s$ , $r_i=infinity$ , $r_f=R_e$ , $v_f=s_e$ and canceling <em>m</em> means that:

$\frac{s_f^2}{2}=\frac{GM_e}{R_e^2}$

Solving for the final velocity we get:

$s_e=\sqrt{\frac{2GM_e}{R_e^2}}$

6 0

3 years ago