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sergij07 [2.7K]
3 years ago
13

Which statement best describes the difference between acceleration and deceleration?

Physics
1 answer:
natima [27]3 years ago
7 0
Acceleration is the rate at which an object picks up speed. deceleration is the rate at which an object loses speed.
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According to Newton's 2nd law of motion, Jim can throw - with equal amounts of force applied - a _____ farther than a _____.
kodGreya [7K]

The answer is All of the above

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9. What is the equivalent resistance of the circuit shown in the figure below?
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2 years ago
The current movement of air or water is a result of what action
Lostsunrise [7]

Answer:

Convection currents are the result of differential heating. Lighter (less dense), warm material rises while heavier (more dense) cool material sinks. It is this movement that creates circulation patterns known as convection currents in the atmosphere, in water, and in the mantle of Earth.

Explanation:

Hope this helps

7 0
3 years ago
Applying Kirchhoff’s Junction Rule, what happens to the power source and current when the parallel circuit has two branches, eac
slamgirl [31]

The power source voltage remains the same in a  parallel circuit,

And we'll have equal current in both lines

<h3>Kirchhoff's junction rule</h3>

Generally, Kirchhoff's junction rule states that when there is current flow at any junction of a circuit, the total sum of this current rushing into the junction amount to the same amount of current out of the Node.

Therefore, when the parallel circuit has two branches

i=i1+12

Since we have an equal resistor therefore we'll have equal current in both lines i.e i1=i2

And Voltage remains the same in a  parallel circuit

More on Voltage

brainly.com/question/14883923

7 0
2 years ago
A 4.00-kg block hangs by a light string that passes over a massless, frictionless pulley and is connected to a 6.00-kg block tha
Anna007 [38]

Answer:

v = 2.82 m/s

Explanation:

For this exercise we can use the conservation of energy relations.

We place our reference system at the point where block 1 of m₁ = 4 kg

starting point. With the spring compressed

        Em₀ = K_e + U₂ = ½ k x² + m₂ g y₂

final point. When block 1 has descended y = - 0.400 m

        Em_f = K₂ + U₂ + U₁ = ½ m₂ v² + m₂ g y₂ + m₁ g y

as there is no friction, the energy is conserved

       Em₀ = Em_f

       ½ k x² + m₂ g y₂ = ½ m₂ v² + m₂ g y₂ + m₁ g y

       ½ k x² - m₁ g y = ½ m₂ v²

 

       v² = \frac{k}{m_2} x^2 - 2 \frac{m_1}{m_2} \ g y

let's calculate

        v² = \frac{180}{6.00} \ 0.300^2 - 2 \ \frac{4.00}{6.00} \ 9.8 \ (- 0.400)

        v² = 2.7 + 5.23

        v = √7.927

        v = 2,815 m / s

using of significant figures

        v = 2.82 m/s

5 0
3 years ago
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