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cupoosta [38]
3 years ago
10

a milk carton filled with water is hanging from a string. what will happen if you punch two holes in opposite sides of the carto

n at the opposite corners?
Physics
1 answer:
marin [14]3 years ago
3 0
When you punch holes in opposite corners of a milk carton filled with water which is hanging from a string, water will start to flow outside with a certain force. There will be a reaction equal and opposite to this force that will make the milk carton start to spin.

This is a typical example of water turbine that is used to demonstrate the third Newton's law.
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If you were standing at the center of curvature in front of a concave mirror, what image would be projected?
matrenka [14]
"<span>The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are" is the type of image among the choices given in the question that would be projected. The correct option among all the options that are given in the question is the first option. I hope it helps you.</span>
6 0
2 years ago
Read 2 more answers
A Scooter travelling at 10m/s speed up to 20m/s in 4 sec.find the acceleration of scooter​
stira [4]

Answer:

2.5 m/s²

Explanation:

Given,

Initial speed ( u ) = 10 m/s

Final speed ( v ) = 20 m/s

Time ( t ) = 4 seconds

To find : Acceleration ( a ) = ?

Formula : -

a = ( v - u ) / t

a = ( 20 - 10 ) / 4

= 10 / 4

= 5 / 2

a = 2.5 m/s²

Therefore,

The acceleration of the scooter is 2.5 m/s²

7 0
2 years ago
A certain ideal gas has molar heat capacity at constant volume CV. A sample of this gas initially occupies a volume V0 at pressu
ANTONII [103]

Answer:

Explanation:

The processes are described on the image attached below. The isobaric process consists of an horizontal line, the adiabatic expansion is described by a polytropic curve:

P_{2} \cdot V_{2}^{\gamma} = P_{3} \cdot V_{3}^{\gamma}

Where:

\gamma = \frac{c_{p}}{c_{v}}

\gamma = 1 + \frac{R}{c_{v}}

Final pressure is:

P_{3} = P_{2}\cdot \left(\frac{V_{2}}{V_{3}}  \right)^{\gamma}

P_{3} = P_{o}\cdot \left(\frac{1}{2}\right)^{\gamma}

P_{3} = \frac{P_{o}}{2^{\gamma}}

8 0
2 years ago
Explain what major mistake producers made when editing the battle scenes for the movie Star Wars
Artyom0805 [142]

Answer: Air

Explanation:

Roffns dn b bdd

4 0
2 years ago
(b) The distance of mass from mass A if there is no gravitational force acted on C
shepuryov [24]

Answer:

(a) The force, acting on object 'C' is approximately 2.66972 × 10⁻¹⁰ Newtons

(b) The distance of 'C' from 'A', in the direction particle 'B' if there is no  meters gravitational force acting on 'C' is appromimately 0.829 meters or 1.877 meters

Explanation:

The given parameters are;

The mass of particle, A, m₁ = 2 kg

The mass of particle, B, m₂ = 0.3 kg

The mass of particle, C, m₃ = 0.05 kg

The distance between particle 'A' and particle 'B', r₁ = 0.15 m

The distance between particle 'B' and particle 'C', r₂ = 0.05 m

(a) The gravitational force, 'F', is given as follows;

F =G \times \dfrac{m_{1} \times m_{2}}{r^{2}}

Where;

F = The force between the two masses

G = The gravitation constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m₁ = The mass of object 1

m₂ = The mass of object 2

If 'C' is placed at 0.05 m from 'B', we have;

F₂₃ =  6.67430 × 10⁻¹¹ × 0.05 × 0.3/(0.05²) ≈ 4.00458 × 10⁻¹⁰

The gravitational force between force between particle 'B' and particle 'C', F₂₃ = 4.00458 × 10⁻¹⁰ N (towards the right)

F₁₃ =  6.67430 × 10⁻¹¹ × 0.05 × 2/(0.1²) ≈ × 10⁻¹⁰

The gravitational force between force between particle 'A' and particle 'B', F₁₃ = 6.6743 × 10⁻¹⁰ N (towards the left)

The force, 'F', acting on object 'C' = F₁₃ - F₂₃

F = (6.6743 - 4.00458) × 10⁻¹⁰ = 2.66972 × 10⁻¹⁰ N

The force, acting on object 'C' ≈ 2.66972 × 10⁻¹⁰ N

(b), When there is no gravitational force acting on 'C', let the distance of 'C' from 'A' = x

We have;

F₂₃ = F₁₂

F_{23} =G \times \dfrac{m_{1} \times m_{2}}{r_1^{2}} = F_{13} =G \times \dfrac{m_{1} \times m_{3}}{r_2^{2}}

By plugging in the values and removing like terms, we get;

\dfrac{0.3 \times 0.05}{(1.15 - x)^{2}}  = \dfrac{2 \times 0.05}{x^2}

(1.15 - x)² × 2 × 0.05 = 0.3 × 0.05 × x²

0.1·x² - 0.23·x + 1.3225 = 0.015·x²

0.1·x² - 0.23·x + 1.3225 - 0.015·x² = 0

0.085·x² - 0.23·x + 0.13225= 0

x = (0.23± √((-0.23)² - 4 × 0.085 × ( 0.13225)))/(2 × 0.085))

x ≈ 0.829, or x ≈ 1.877

Therefore, the distance of 'C' from 'A', if there is no gravitational force acting on 'C', x ≈ 0.829 m, or x = 1.877 m, in the direction of 'B'

7 0
2 years ago
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