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I am Lyosha [343]
3 years ago
12

A dairy farmer notices that a circular water trough near the barn has become rusty and now has a hole near the base. The hole is

0.19 m below the level of the water that is in the tank. If the top of the trough is open to the atmosphere, what is the speed of the water as it leaves the hole? Assume that the trough is large enough that the velocity of the water at the top is zero. The acceleration of gravity is 9.81 m/s 2 . Answer in units of m/s.
Physics
1 answer:
Crazy boy [7]3 years ago
5 0

Answer:

v=1.93m/s

Explanation:

From the concept of fluids mechanics we know that if a tank has a hole at the bottom, the equation that we need to use is:

v=\sqrt{2gh}

Since we know gravity and its hight

v=\sqrt{2*(9.81m/s^{2})(0.19m) }=1.93m/s

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An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 ​7 ​​ m/s over
stich3 [128]

Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :  

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron  

• a is the acceleration of the electron  

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.  

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m  

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

          a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

  = 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m  

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .  

xf =xi+vi*t+1/2*a*t

xf =xi+vi*t+1/2*a*t

  t=√2(xf-xi)/a

 t=3.765×10^-10 s

now we can use the power P Eq.  

 P=W/Δt => ΔK/Δt  

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

6 0
3 years ago
If the moon rises around 3<br> a.m., its phase must be
Luden [163]
New moon or cresant moon i believe
4 0
3 years ago
Which of the following ranching practices contributes to soil erosion?
Natasha2012 [34]

A. overgrazing of livestock

Explanation:

Overgrazing of livestock is the ranch practice that contributes to soil erosion.

Soil erosion is the washing away of the top layer of the earth on which plant grows.

  • Plants are covers of the earth and they prevent agents of denudation from actively washing away the soil.
  • When livestock are introduced into a place, continuous grazing leaves the surface bare.
  • The plants becomes depleted and the surface is exposed.
  • This fast tracks the process of overgrazing.

learn more:

Soil erosion brainly.com/question/2473244

#learnwithBrainly

6 0
3 years ago
Two cars are moving in the same direction at the same speed of (72.0 kmh). What is
Licemer1 [7]

Answer:

0 km/h

Explanation:

Relative speed is the speed of a moving body with respect to another.

When two bodies move in the same direction then the relative speed is calculated as difference of their speeds.

In this case;

The two cars have the same speed. The relative speed will be;

72 km/h -72 km/h = 0 km/h

6 0
2 years ago
A proton and an electron are fixed in space with a separation of 859 nm. Calculate the electric potential at the midpoint betwee
makkiz [27]

Answer:

The electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

Explanation:

Electric potential is given as;

V = E*r

where;

E is the electric field strength, = kq/r²

V = ( kq/r²)*r

V = kq/r

k is coulomb's constant = 8.99 X 10⁹ Nm²/C²

q is the charge of the particles = 1.6 X 10⁻¹⁹ C

r is the distance between the particles = 859 nm

At midpoint, the distance = r/2 = 859nm/2 = 429.5 nm

V = (8.99 X 10⁹  * 1.6 X 10⁻¹⁹)/ (429.5 X 10⁻⁹)

V = 3.349 X 10⁻³ Volts

Therefore, the electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

3 0
3 years ago
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