Ask question

Ask question

All categories

4 years ago

7

A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac

e of the water. Calculate the fluid force F(in Newtons) on one side of the plate (assume the water weight density isW=9800 N/m3).

1 answer:

Tpy6a [65]4 years ago

7 0

Answer:

The force on one side of the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

$y=\dfrac{\Delta y}{\sin60}$

$y=\dfrac{2\Delta y}{\sqrt{3}}$

The area of strip is

$dA=\dfrac{9\times2\Delta y}{\sqrt{3}}$

We need to calculate the force on one side of the plate

Using formula of pressure

$P=\dfrac{dF}{dA}$

$dF=P\times dA$

On integrating

$\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}$

$F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}$

$F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})$

$F=3093529.3\ N$

Hence, The force on one side of the plate is 3093529.3 N.

You might be interested in

Two 125 kg bumper cars are moving toward each other in opposite directions. Car X is moving at 10 m/s and Car Z at −12 m/s when

nignag [31]

<h2>Given that,</h2>

Mass of two bumper cars, m₁ = m₂ = 125 kg

Initial speed of car X is, u₁ = 10 m/s

Initial speed of car Z is, u₂ = -12 m/s

Final speed of car Z, v₂ = 10 m/s

We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

v₁ is the final speed of car X.

$m_1u_1+m_2u_2-m_1v_1=m_2v_2\\\\m_2v_2=m_1u_1+m_2u_2-m_2v_2\\\\m_1v_1=125\times 10+125\times (-12)-125\times 10\\\\v_1=\dfrac{-1500}{125}\\\\v_1=-12\ m/s$

So, car X will move with a velocity of -12 m/s.

3 0

3 years ago

12) A neutral atom has two, eight, eight electrons in its first, second, and third energy levels.This information ________.

pentagon [3]

Answer:

C.

Explanation:

Option C

The information given in the question tells us about the number of electrons in an atom and also the number of shells in the atom. So, we will come to know about the atomic number, size and chemical properties of the atom. But we cannot determine atomic mass. Atomic mass is a function of number of neutrons and protons.

3 0

3 years ago

A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$

The friction forces are computed by

$\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g$

$\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g$

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

$\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g$

Simplifying

$\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)$

Plugging in the values

$\displaystyle F_{a}=0.25(9.8)[400+2(100)]$

$\boxed{F_a=1470\ N}$

8 0

3 years ago

A 100 kg box is suspended from two ropes. The "left rope makes an angle of 20" degrees with the vertical, and the right rope mak

GarryVolchara [31]

Explanation:

It is given that,

Mass of the box, m = 100 kg

Left rope makes an angle of 20 degrees with the vertical, and the right rope makes an angle of 40 degrees.

From the attached figure, the x and y component of forces is given by :

$T_{1x} =-T_1 cos (20)$

$T_{2x} = T_2 cos (40)$

$mg_x = 0$

$T_{1y} = T_1 sin (20)$

$T_{2y} = T_2 sin (40)$

$mg_y= -mg$

Let $R_x$ and $R_y$ is the resultant in x and y direction.

$R_x=-T_1 cos (20)+T_2 cos (40)+0$

$R_y=T_1 sin(20)+T_2 sin(40)-mg$

As the system is balanced the net force acting on it is 0. So,

$-T_1 cos (20)+T_2 cos (40)+0=0$ .............(1)

$T_1 sin(20)+T_2 sin(40)-100\times 9.8=0$ ..................(2)

On solving equation (1) and (2) we get:

$T_1=866.86\ N$ (tension on the left rope)

$T_2=1063.36\ N$ (tension on the right rope)

So, the tension on the right rope is 1063.36 N. Hence, this is the required solution.

7 0

3 years ago

A speeding car collides with an unlucky bug flying across the road

alex41 [277]

<span>The car has more momentum. (A)*
Also, the car is made out of more resilient material than the bug is.

* Unless the bug is much bigger than the car.</span>

7 0

3 years ago

Read 2 more answers