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3 years ago

A hole of 1 mm² is punched in a full 1-liter milk carton, 5 cm from the bottom. What is the initial speed of the outflow?

1 answer:

6 0

Use Bernoulli equation with the following assumptions.

1) velocity of the milk on the top is near 0, because the hole is so small that the level of milk will descend slowly, compared with the velocity of the outflow.

2) Pressure on the top = pressure on the bottom. This means that we consider that the top is open to the atmosphere.

Then, the terms of pressure cancel, and Bernoulli's equation leads to:

Z1 + p1/(d*g) + (V1)^2/(2g) = Z2 + p2 /(d*g) + (V2)^2 /(2g)

=> (V2)^2 = 2g (Z1 - Z2)

g = 9.8 m/s^2

Z1 - Z2 = 5cm

(V2)^2 = 2*(9.8 m/s^2)*(0.05m) = 0.98 m^2/s^2

V2 = 0.99 m/s

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Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in an elastic, glancing collision. The yello

klemol [59]

Answer:

3.62m/s and 2.83m/s

Explanation:

Apply conservation of momentum

For vertical component,

Pfy = Piy

m* Vof (sin38) - m*Vgf (sin52) = 0

Divide through by m

Vof(sin38) - Vgf(sin52) = 0

Vof(sin38) = Vgf(sin52)

Vof (sin38/sin52) = Vgf

0.7813Vof = Vgf

For horizontal component

Pxf= Pxi

m* Vof (cos38) - m*Vgf (cos52) = m*4.6

Divide through by m

Vof(cos38) + Vgf(cos52) = 4.6

Recall that

0.7813Vof = Vgf

Vof(cos38) + 0.7813 Vof(cos52) = 4.6

0.7880Vof + 0.4810Vof = 4.

1.269Vof = 4.6

Vof = 4.6/1.269

Vof = 3.62m/s

Recall that

0.7813Vof = Vgf

Vgf = 0.7813 * 3.62

Vgf = 2.83m/s

3 0

3 years ago

Help plzzzzzzzzzzzzzzzz!

aksik [14]

concave <span>ray diagrams were constructed in order to determine the general location, size, orientation, and type of image formed by concave mirrors. Perhaps you noticed that there is a definite relationship between the image characteristics and the location where an object placed in front of a concave mirror. but, convex</span><span>ray diagrams were constructed in order to determine the location, size, orientation, and type of image formed by concave mirrors. The ray diagram constructed earlier for a convex mirror revealed that the image of the object was virtual, upright, reduced in size and located behind the mirror. </span>

3 0

3 years ago

If you are asked to modify the robots or drones that are currently used , what kind of modifications you would do and suggest an

Eddi Din [679]

Explanation:

In recent times of pandemic, robots can be use as replacement of labor in the industries. Mundane tasks can be programmed in their system so that they can used readily.

Drones can used delivery for essential goods and services, so that human interference can be least and the spread of virus can be curbed.

In a recent example, Argentina where aerial data has reportedly been used to accelerate the construction of emergency hospitals.

4 0

4 years ago

Una partícula se mueve en el plano XY efectúa un desplazamiento mientras actúa sobre ella una fuerza constante. X= (4i + 3j) m,

dsp73

Answer:

a) La magnitud del desplazamiento es de 5 m

La magnitud de la fuerza es 20 N

b) El trabajo realizado por la fuerza es de 100 J

c) El ángulo entre la fuerza y el plano es 0 °

Explanation:

a) La magnitud del desplazamiento se encuentra por la relación;

$\left | X \right | = \sqrt{X_{x}^{2}+X_{y}^{2}}$

Lo que da;

$\left | X \right | = \sqrt{4^{2}+3^{2}} = 5 \ m$

De manera similar, la magnitud de la fuerza, F, se encuentra como sigue;

$\left | F \right | = \sqrt{F_{x}^{2}+F_{y}^{2}}$

Lo que da;

$\left | F \right | = \sqrt{16^{2}+12^{2}} = 20 \ N$

b) El trabajo, W, realizado por la fuerza = Fuerza, F × Distancia, X

∴ Ancho = 20 N × 5 m = 100 N · m = 100 J

c) La dirección de la fuerza viene dada por la siguiente fórmula;

$tan^{-1} \left (\dfrac{F_y}{F_x} \right ) = tan^{-1} \left (\dfrac{12}{16} \right ) = 38.9^{\circ}$

La dirección del plano viene dada por la siguiente fórmula;

$tan^{-1} \left (\dfrac{X_y}{X_x} \right ) = tan^{-1} \left (\dfrac{3}{4} \right ) = 38.9^{\circ}$

Por tanto, el ángulo entre la fuerza y el plano = 0 °

La fuerza actúa a lo largo del plano.

6 0

3 years ago

A 0.5 kg rock is dropped from a height of 1.0 m above the ground. Approximately how much kinetic energy will be stored in the ro

irina1246 [14]

Answer:

2.45 J

Explanation:

The following data were obtained from the question:

Mass (m) = 0.5 kg

Height (h) = 1 m

Kinetic energy (KE) =?

Next, we shall determine the velocity of the rock after it has fallen half way. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 1/2 = 0.5 m

Final velocity (v) =?

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 0.5)

v² = 9.8

Take the square root of both side

v = √9.8

v = 3.13 m/s

Finally, we shall determine the kinetic energy of the rock after it has fallen half way. This can be obtained as follow:

Mass (m) = 0.5 kg

Velocity (v) = 3.13 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.5 × 3.13²

KE = 0.25 × 9.8

KE = 2.45 J

Therefore, the kinetic energy of the rock after it has fallen half way is 2.45 J

8 0

3 years ago