A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then the speed of efflux of non viscous water through the opening will be

Answers

Answer 1

Answer:

Answer:

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Explanation:

Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ( $P_(atm)$ ), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

$P_(1) + \rho\cdot (v_(1)^(2))/(2) + \rho\cdot g \cdot z_(1) = P_(2) + \rho\cdot (v_(2)^(2))/(2) + \rho\cdot g \cdot z_(2)$

Where:

$P_(1)$ , $P_(2)$ - Water total pressures inside the tank and at ground level, measured in pascals.

$\rho$ - Water density, measured in kilograms per cubic meter.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_(1)$ , $v_(2)$ - Water speeds inside the tank and at the ground level, measured in meters per second.

$z_(1)$ , $z_(2)$ - Heights of the tank and ground level, measured in meters.

Given that $P_(1) = P_(2) = P_(atm)$ , $\rho = 1000\,(kg)/(m^(3))$ , $g = 9.807\,(m)/(s^(2))$ , $v_(1) = 0\,(m)/(s)$ , $z_(1) = 6.9\,m$ and $z_(2) = 4.9\,m$ , the expression is reduced to this:

$\left(9.807\,(m)/(s^(2)) \right)\cdot (6.9\,m) = (v_(2)^(2))/(2) + \left(9.807\,(m)/(s^(2)) \right)\cdot (4.9\,m)$

And final speed is now calculated after clearing it:

$v_(2) = \sqrt{2\cdot \left(9.807\,(m)/(s^(2)) \right)\cdot (6.9\,m-4.9\,m)}$

$v_(2) \approx 6.263\,(m)/(s)$

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

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It's nighttime, and you've dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 0.90m above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.2m from the edge.-How far are the goggles from the edge of the pool?

Answers

Answer:

the googles are 5.3 m from the edge

Explanation:

Given that

depth of pool , d = 3.2 m

Now, let i be the angle of incidence

a laser pointer 0.90 m above the edge of the pool and laser beam enters the water 2.2 m from the edge

⇒tan i = 2.2/0.9

$i=arctan(2.2/.90)$

solving we get

i = 67.8°

Using snell's law ,

n1 ×sin(i) = n2 ×sin(r)

n1= refractive index of 1st medium= 1

n2= refractive index of 2nd medium = 1.33

r= angle of reflection

therefore,

$1* sin(67.8) = 1.33* sin(r)$

r = 44.1°

Now,

distance of googles = 2.2 + d×tan(r)

distance of googles = 2.2 + 3.2×tan(44.1)

distance of googles = 5.3 m

the googles are 5.3 m from the edge

Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k. Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward. Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use g for the magnitude of the acceleration due to gravity.

Answers

Answer:

Point motion will eventually stops due to action of g exactly perpendicular...

Explanation:

If ignoring the air resistance, the magnitude of gravitational acceleration is already strong enough to stops the acceleration. As we know that, the spring constant of a bungee spring cord will be F = -k/x, where x is the stretched length and k is the spring constant of bungee cord. If F = ma = w = mg, the g = -m k/x. Now we can clearly see that the value of g remains constant due to the fluctuating length of the cord as the motion progresses back and forth in SHM say from x1 to x2 and x2 to x1.

Why are certain things obligations of citizenship instead of responsibilities? atleast 5 sentences please

Answers

Answer:

Please find the answer in the explanation

Explanation:

Responsibilities of citizens are those things citizens are to take care of.

While obligations are those things that are compulsory for the citizens to observe and adhere to.

Why are certain things obligations of citizenship instead of responsibilities?

1.) Because of law and order of the community. It is mandatory for all citizens to obey the law of the land.

2.) Because of the progress and peaceful coexistence of the citizens in the community.

3.) Because of the protection of constitution of the land

4.) To support and defend the constitution

5.) To maintain orderliness and eschew violence.

Talia is on a road trip with some friends. In the first 2 hours, they travel 100 miles. Then they hit traffic and go only 30 miles in the next hour. The last hour of their trip, they drive 75 miles.Calculate the average speed of Talia’s car during the trip. Give your answer to the nearest whole number.

Answers

Answer:

51 mph

Explanation:

Since Speed, V = Distance/Time
Average speed = Total Distance/Total Time

From the given data, Total Distance = 100 + 30 + 75 miles
and Total Time = 2 + 1 + 1 hours

Average Speed = 205/4
Average Speed = 51.25 mph ( or 51mph to the nearest whole number)

A manufacturer claims that a carpet will not generate more than 5.8 kV of static electricity What magnitude of charge would have to be transferred between a carpet and a shoe for there to be a 5.8 kV potential distance d = 2.8 mm ? Approximate the area of a shoe as 30 cm x 8 cm. Express your answer using two significant figures.

Answers

Answer:

4.4×10⁻⁷ Coulomb

Explanation:

V = Voltage = 5.8 kV

d = Potential distance = 2.8 mm = 0.0028 m

A = Area = 0.3×0.08 = 0.024 m²

ε₀ = permittivity constant in a Vacuum= 8.85×10⁻¹² F/m

$(Q)/(V)=(A\epsilon_0)/(d)\n\Rightarrow \Q=V(A\epsilon_0)/(d)\n\Rightarrow Q=5.8* 10^3(0.024* 8.85* 10^(-12))/(0.0028)\n\Rightarrow Q=4.4* 10^(-7)\ C$

Magnitude of charge transferred between a carpet and a shoe is 4.4×10⁻⁷ Coulomb.

A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out. As the man reaches the horizontal section, he grabs a 20.0-kg backpack and skis off a 2.00-m-high ledge. At what horizontal distance from the edge of the ledge does the man land (the man starts at rest)?

Answers

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

$KE_(initial) + PE_(initial) = KE_(final) + PE_(final)$