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2 years ago

Velocity is an extensive property of a system. TRUE OR FALSE

Physics

2 answers:

Anna007 [38]2 years ago

7 0

Answer:

False

Explanation:

I hope this helped have a great day!!!

Snowcat [4.5K]2 years ago

5 0

$\: \: \:$

$\pink{false}$

because velocity is neither an intensive nor extensive property

<h2>hope it helps</h2>

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A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo

WITCHER [35]

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

$F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}$

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

$acceleration=\frac{30}{5}=6m/s^{2}$

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

$v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds$

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

5 0

3 years ago

A person throws a pumpkin at a horizontal speed of 4.0 — off a cliff. The pumpkin travels 9.5 m horizontally

emmainna [20.7K]

Complete Question

A person throws a pumpkin at a horizontal speed of 4.0 m/s off a cliff. The pumpkin travels 9.5m horizontally before it hits the ground. We can ignore air resistance.What is the pumpkin's vertical displacement during the throw? What is the pumpkin's vertical velocity when it hits the ground?

Answer:

The pumpkin's vertical displacement is $H = 27 .67 \ m$

The pumpkin's vertical velocity when it hits the ground is $v_v__{f}} = 23.298 \ m/s$

Explanation:

From the question we are told that

The horizontal speed is $v_h = 4 m/s$

The horizontal distance traveled is $d = 9.5 \ m$

The horizontal distance traveled is mathematically represented as

$S = v_h * t$

Where t is the time taken

substituting values

$9.5 = 4 * t$

=> $t = \frac{9.5}{4}$

$t = 2.38 \ sec$

Now the vertical displacement is mathematically represented as

$H = v_v t + \frac{1}{2} a_v t^2$

now the vertical velocity before the throw is zero

$H = 0 + \frac{1}{2} (9.8) * (2.375)^2$

$H = 27 .67 \ m$

Now the final vertical velocity is mathematically represented as

$v_v__{f}} = v_v + at$

substituting values

$v_v__{f}} = 0 + (9.8)* (2.375)$

$v_v__{f}} = 23.298 \ m/s$

7 0

2 years ago

What charge accumulates on the plates of a 2.0-μF air-filled capacitor when it is charged until the potential difference across

enot [183]

Answer:

0.0002 C.

Explanation:

Charge: This can be defined as the ratio of current to time flowing in a circuit. The S.I unit of charge is Coulombs (C)

Mathematically, charge can be expressed as

Q = CV ................................. Equation 1

Where Q = amount of charge, C = capacitance of the capacitor, V = potential difference across the plates.

Given: C = 2.0-μF = 2×10⁻⁶ F, V = 100 V.

Substitute into equation 1

Q = 2×10⁻⁶× 100

Q = 2×10⁻⁴ C

Q = 0.0002 C.

The amount of charge accumulated = 0.0002 C

7 0

2 years ago

Read 2 more answers

You observe two cars traveling in the same direction on a long, straight section of Highway 5. The red car is moving at a consta

Arte-miy333 [17]

Answer:

1) 3.66 s

2) 124.44 m

3) 3.12 s

Explanation:

Let's start by first listing down the information in the question.

Red Car : 34 m/s

Blue Car: 28 m/s

Distance between them : 22 m

The difference in speed between the cars is: 34 - 28 = 6 m/s

This means that the red car is catching up to the blue car at a speed of 6 m/s.

1) We can solve this by just dividing the distance by the difference in speed. This becomes: $\frac{Distance}{Speed}= \frac{22}{6} = 3.66$

Thus it takes 3.66 seconds for the red car to catch up to the blue car.

2) We know from (1) that it took 3.66 seconds for the red car to catch up. Since the speed it was travelling at is constant, we only need to multiply it by the time from (1) to get the distance.

This becomes: $Speed * Time = 34 * 3.66 = 124.44$

Thus the red car travels 124.44 m before catching up to the blue car.

3) If the red car starts to accelerate the moment we see it, the time taken to get to the blue car will be less than before. We can find this in a simple way.

We can use the motion equation : $s = u*t + \frac{1}{2}(a * t^2)$