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den301095 [7]
3 years ago
15

A triangular plate with base 4 m and height 5 m is submerged vertically in water so that the tip is even with the surface. Expre

ss the hydrostatic force against one side of the plate as an integral and evaluate it. (Round your answer to the nearest whole number. Use 9.8 m/s2 for the acceleration due to gravity. Recall that the weight density of water is 1000 kg/m3.)

Physics
2 answers:
8090 [49]3 years ago
6 0

Answer:

The hydro-static force is 326666.67 N.

Explanation:

Given that,

Base = 4 m

Height = 5 m

We need to calculate the area of strip

using similar triangle,

\dfrac{y}{x}=\dfrac{B}{H}\Rightarrow y=\dfrac{B}{H}x

Area of strip, dA =y\ dx

dA =\dfrac{b}{H}\times x\ dx

Where, b = base

dx = width of strip

H = height

Put the value of base into the formula

dA=\dfrac{4}{5}\times x dx

We need to calculate the force

Using formula of force

dF= P\times dA

Where, P = pressure

A = area

Put the value into the formula

dF=\rho g h\times dA

dF=\rho g x\times\dfrac{4}{5}xdx

On integration

\int_{0}^{F}dF=\int_{0}^{5}{\rho g x}\times\dfrac{4}{5}\times xdx

F = \rho g\int_{0}^{5}(\dfrac{4}{5}x^2)dx

F=\rho g\times\dfrac{4}{5}(\dfrac{x^3}{3})_{0}^{5}

F=1000\times9.8\times\dfrac{4}{5}\times\dfrac{5^3}{3}

F=326666.67\ N

Hence, The hydro-static force is 326666.67 N.

Artyom0805 [142]3 years ago
4 0

Answer:

hydrostatic force is 327000.00 N

Explanation:

given data

base =  4 m

height =  5 m

density of water = 1000 kg/m3

to find out

hydrostatic force

solution

we know this is a triangle so we consider here a small strip PQ whoes  area da in with length x and width dy

so area will be

da = base/H × height = base/H × y

so da = x dy = base/H × y dy

and we know pressure = ρ × g × h

here h = y

hydrostatic force = pressure × area

df = (ρ × g × h) × base/H × y dy

now integrate it from 0 to 5 height

f = ρ × g \int_{0}^{5} (h) (base/H)ydy

f = ρ × g \int_{0}^{5} (y) (4/5)ydy

f = ρ × g  ×  4/5  ×  (y^{3} /3)^{5} _0

now put value ρ = 1000 and g = 9.81

f = ρ × g  ×  4/5  ×  (y^{3} /3)^{5} _0

f = 1000 × 9.81  ×  4/5  × (5³/3)

force = 7848 × 41.666667

force = 327000.00 N

hydrostatic force is 327000.00 N

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wlad13 [49]

Answer:

Power output = 2\times 10^4\ W

Explanation:

Given:

Mass of the elevator is, m=1000\ kg

Height to which it is raised is, h=10\ m

Acceleration due to gravity is, g=10\ m/s^2(Approximately)

Time taken by the motor to raise the elevator is, t=5.0\ s

Now, work done on the elevator by the motor is equal to the increase in the gravitational potential energy of the elevator.

Increase in gravitational potential energy is given as:

\Delta U=mgh=(1000)(10)(10)=10\times 10^4\ J

Therefore, work done by motor is, W=10\times 10^4\ J

Now, we know that, power is work done in unit time. So, power output is given as:

Power=\frac{W}{t}\\\\Power=\frac{10\times 10^4\ J}{5.0\ s}\\\\Power=2\times 10^4\ J/s\\\\Power=2\times 10^4\ W..........[1 W = 1\ J/s]

Therefore, the power output of the first motor is 2\times 10^4\ W

7 0
3 years ago
Which of the following is the best definition of an isotope?
ad-work [718]

Answer:

A. elements of the same kind with different numbers of neutrons

Explanation:

As we know that an atom is represented by

_z^AX

here we know that

z = atomic number

A = atomic number + number of neutrons

now if the number of neutrons in an atom is different but having same number  atomic number then the combination of such group of atoms is known as isotopes.

So here we have

_z^{A_1}X, _z^{A_2}X

so above is the example of isotopes

8 0
2 years ago
Read 2 more answers
What’s the answer to this
ladessa [460]
Choices  1,  2,  and 4 . . . . . Yes

Choices  3  and 5 . . . . . No
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A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t
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Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

v_{avg} = \frac{6000}{20*60}

v_{avg} = 5m/s

now since effect of wind is in opposite direction then we can say

V_{net} = v_{bird} - v_{wind}

5 = 9 - v_{wind}

v_{wind}= 4 m/s

Part b)

now if bird travels in the same direction of wind then we will have

v_{net}= v_{bird} + v_{wind}

v_{net} = 9 + 4 = 13 m/s

now we can find the time to go back

time = \frac{distance}{speed}

time = \frac{6000}{13}

time = 7.7 minutes

Part c)

Total time of round trip when wind is present

T = t_1 + t_2

T = 20 + 7.7 = 27.7 min

now when there is no wind total time is given by

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4 0
3 years ago
A certain spring has a spring constant k1 = 660 N/m as the spring is stretched from x = 0 to x1 = 35 cm. The spring constant the
pantera1 [17]

(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².

(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.

(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.

<h3>Work done in the spring</h3>

The work done in stretching the spring is calculated as follows;

W = ¹/₂kx²

W(1 to 2) = ¹/₂K₂Δx²

W(1 to 2)  =  ¹/₂(250)(0.65 - 0.35)²

W(1 to 2)  = 11.25 J

W(0  to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃

W(0  to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)

W(0  to 3) = 64.28 J

Learn more about work done here: brainly.com/question/25573309

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