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worty [1.4K]
2 years ago
14

"A hole is punched in a full milk carton, 10 cm below the top. What is the initial velocity of outflow?"

Physics
2 answers:
lesya692 [45]2 years ago
8 0
P 1 = 101,325 Pa (atmospheric pressure)
Milk has almost same density as water: (Rho)= 1,000 kg /m³ 
P 2 = 101,325 Pa + 1,000 kg/m³ · 9.81 m/s² · 0.1 m = 102,306 Pa
The hydrostatic equation:
P 1 + (Rho)v1² / 2 = P 2 + (Rho)·g·h2
101,325 + 1,000 v1²/2  = 102,306 + 1000 · 9.81 · 0.1
500 v 1² = 102,306 + 981 - 101,325
v 1² = 3.924
v 1 = √ 3.924
v 1 = 1.98 m/s
The initial velocity of outflow is 1.98 m/s.

Lelechka [254]2 years ago
8 0

Answer:

The initial velocity of outflow is 1.4 m/s.

Explanation:

As a hole is punched in a full milk carton, and we have to calculate the initial velocity of outflow 10 cm below the top. We use the concept of conservation of energy.  

Further Explanation:

Using the conservation of energy, below the top the potential energy converted in kinetic energy

P.E =K.E

mgh=\frac{1}{2} mv^{2}

As mass of milk does not change at the top and 10 cm below the top, therefore  

v=\sqrt{2gh}

Here, v is the initial velocity,  

g is acceleration due to gravity  

h is height.  

Given: h = 10 cm and g = 9.8 m/s2

Substituting the given values, we get

v=\sqrt{2×9.8×0.1m}  

v=\sqrt{1.96}=1.4 m/s

Lear more:

brainly.com/question/1581227

Key word:  

Conservation of energy, Kinetic energy, potential energy.

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An automobile approaches a barrier at a speed of 20 m/s along a level road. The driver locks the brakes at a distance of 50 m fr
AleksAgata [21]

Answer:

μ = 0.408

Explanation:

given,

speed of the automobile (u)= 20 m/s

distance = 50 m

final velocity  (v) = 0 m/s

kinetic friction = ?

we know that,

v² = u² + 2 a s

0 = 20² + 2 × a × 50

a = \dfrac{400}{2\times 50}

a = 4 m/s²

We know

F = ma = μN

ma = μ mg

a = μ g

\mu = \dfrac{a}{g}

\mu = \dfrac{4}{9.81}

μ = 0.408

hence, Kinetic friction require to stop the automobile before it hit barrier is 0.408

5 0
2 years ago
According to the nebular theory of solar system formation, what key difference in their early formation explains why the jovian
svp [43]

Answer:

The Jovian planets formed beyond the Frostline while the terrestrial planets formed in the Frostline in the solar nebular

Explanation:

The Jovian planets are the large planets namely Saturn, Jupiter, Uranus, and Neptune. The terrestrial planets include the Earth, Mercury, Mars, and Venus. According to the nebular theory of solar system formation, the terrestrial planets were formed from silicates and metals. They also had high boiling points which made it possible for them to be located very close to the sun.

The Jovian planets formed beyond the Frostline. This is an area that can support the planets that were made up of icy elements. The large size of the Jovian planets is as a result of the fact that the icy elements were more in number than the metal components of the terrestrial planets.

3 0
2 years ago
A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air
Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, v_{o} = 20.0 m/s

Angle made by the ramp, \theta = 22.0^{\circ}

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH

where

H = dsin22^{\circ} = 5sin22^{\circ}

g = 9.8 m/s^{2}

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}

v = \sqrt{400 - 19.6\times 5sin22^{\circ}} = 19.06 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m

(b) now, considering the coefficient of friction bhetween ramp and the ball, \mu = 0.150:

velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH - \mu gd

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5

v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5} = 18.7 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m

6 0
2 years ago
3. A 75kg man sits at one end of a uniform seesaw pivoted at its center, and his 24kg son sits at the
bulgar [2K]

Answer:

The wife have to sit at 0.46 L from the middle point of the seesaw.

Explanation:

We need to make a sketch of the seesaw and the loads acting over it.

And by the studying of the Newton's law we can find the equation useful to find the distance of the mother sitting on the seesaw with respect to the center ot the pivot point.

A logical intuition will give us the idea that the mother will be on the side of her son to make the balance.

The maximum momentum with respect to the pivot point (0) will be:

M=75 *\frac{L}{2}

Where L/2 is the half of the distance of the seesaw

Therefore the other loads ( mom + son) must be create a momentum equal to the maximum momentum.

7 0
2 years ago
1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r
DENIUS [597]

Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle \theta with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

Where y_o is the initial height above the ground level, v_{oy} is the vertical component of the initial velocity and t is the time

The y-component of the speed is

v_y=v_{oy}-gt

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of v_o

The object will reach the maximum height when v_y=0. It allows us to compute the time to reach that point

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum heigh is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Solving for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Replacing the known values

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

y=7.39\ m

5 0
2 years ago
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