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2 years ago

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Calculate the applied force of the washers on the car. First, convert the mass you recorded for one, two, three, and four washer

s from grams to kilograms. Record your calculations, to four decimal places, in Table D. The mass of one washer is kg. The mass of two washers is kg. The mass of three washers is kg. The mass of four washers is kg.

2 answers:

irga5000 [103]2 years ago

4 0

Answer:

mass of one washer is 0.0049 kg.

mass of two washers is 0.0098 kg.

mass of three washers is 0.0147 kg.

mass of four washers is 0.0196 kg.

Explanation:

hope it helped :)

Andrej [43]2 years ago

3 0

The mass of one washer is 0.0049 kg.

The mass of two washers is 0.0098 kg.

The mass of three washers is 0.0147 kg.

The mass of four washers is 0.0196 kg.

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Answer:

option (b)

Explanation:

According to the Pascal's law

F / A = f / a

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Diameter of ram, D = 20 cm, R = 20 / 2 = 10 cm

A = π R^2 = π x 100 cm^2

F = 3 tons = 3000 kgf

diameter of plunger, d = 3 cm, r = 1.5 cm

a = π x 2.25 cm^2

Use Pascal's law

3000 / π x 100 = f / π x 2.25

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2 years ago

You use a slingshot to launch a potato horizontally from the edge of a cliff with speed v0. The acceleration due to gravity is g

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Answer:

$\displaystyle t=\frac{2v_o}{g}$

Explanation:

<u>Horizontal Launch</u>

When an object is launched horizontally at a speed vo, it describes a curved called parabola as the speed in the x-direction does not change and the speed in the y-direction increases with time because the gravity makes it return to the ground.

The vertical distance the object (potato) travels downwards is:

$\displaystyle y=\frac{gt^2}{2}$

The horizontal distance is

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We need to find the time when both distances are equal, thus

$\displaystyle \frac{gt^2}{2}=v_ot$

Simplifying by t

$\displaystyle \frac{gt}{2}=v_o$

Solving for t

$\displaystyle \boxed{t=\frac{2v_o}{g}}$

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2 years ago

Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m

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Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

$m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s$

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

$V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

Substituting,

$V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}$

$V_{cm} = 1.034m/s$

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where here $v_f$ is the velocity after the collision.

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2 years ago

A boy uses a slingshot to launch a pebble straight up into the air. The pebble reaches a height of 37.0 m above the launch point

denis-greek [22]

Answer:

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$v_0=\frac{y_f+0.5gt^2}{t}=\frac{37+4.9*2.3^2}{2.3}=27.4m/s$

Now, using the same expression we estimated time to first reach 18.5 m :

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2 years ago

If the newton is the product of kilograms and meters/second2 what units comprise the pound?

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Answer:

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Therefore,

1 pound= $1 slug\times fts^{-2}$

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2 years ago