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

Answers

Answer 1

Answer:

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

$mgh=(1)/(2) mv^(2)$

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

$v=√(2gh)$

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

$v=√(2×9.8×0.1m)$

$v=√(1.96)=1.4 m/s$

Lear more:

brainly.com/question/1581227

Key word:

Conservation of energy, Kinetic energy, potential energy.

Answer 2

Answer: 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.

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The momentum of a bald eagle in flight is calculated to be 345. The mass of the eagle is 5.0 kg. What is the magnitude of the velocity of the eagle?

Answers

Answer : v = 69 m/s.

Explanation :

It is given that,

The momentum of a bald eagle in flight, $p = 345\ Kgm/s$

Mass of the eagle, m = 5 kg

We know that the momentum of any object is defined as the product of mass and the velocity with which it is moving.

$p=mv$

$v=(p)/(m)$

$v=(345\ kgm/s)/(5\ kg)$

$v=69\ m/s$

The velocity of the eagle is 69 m/s.

Hence, this is the required solution.

The formula to find the magnitude of velocity is:

▲v= ▲M/m

▲v=Velocity Change
▲M=Momentum Change
m=Mass

Plug in the information;

▲v=345/5

The answer is 69

To test the quality of a tennis ball, you drop it onto a floor from a height of 4.00m . It rebounds to a height of 2.00 m. If the ball is in contact with the floor for 12.0ms,(a) What is the magnitude of its average acceleration during the contact?
(b) is the acceleration up or down?

Answers

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A 2.00 × 103 kg car rounds a circular turn of radius 20.0 m. If the road is flat and the coefficient of static friction between the tires and the road is 0.70, how fast can the car go without skidding?

Answers

Answer:

11.7 m/s

Explanation:

In order for the car to remain in circular motion along the road and not to skid, the frictional force between the tires and the road must be equal to the centripetal force.

Therefore, we can write:

$\mu mg = m(v^2)/(r)$

where:

m = 2000 kg is the mass of the car

$\mu=0.70$ is the coefficient of friction

$g=9.8 m/s^2$ is the acceleration due to gravity

v is the speed of the car

r = 20.0 m is the radius of the turn

Substituting and solving for v, we find the speed of the car:

$v=√(\mu gr)=√((0.70)(9.8)(20.0))=11.7 m/s$

ΡΟ. . When preparing for any of the golf variation games (golf foot golf, disc golf), a person should consider the following before they start. O Distance
O Terrain
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Answers

Answer:

All of the above

Explanation:

To ensure the ball go to where we expected it go

Where does the material for new mountain ranges come from

Answers

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Answers

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