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3 years ago

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Consider a car engine running at constantspeed. That is, the crankshaft of the en-gine rotates at constant angular velocity whil

eeach piston moves back-and-forth in its cylin-der according to the rules of simple harmonicmotion.2030 rpm6.95 cmSuppose the two extremal positionsxmaxandxminof a piston are 6.95 cm from eachother.When the crankshaft of the engine rotatesat 2030 rpm (revolutions per minute), what isthe maximal speed|v|maxof the piston?Answer in units of m/s.

1 answer:

Irina-Kira [14]3 years ago

7 0

Answer:

The value is $|v| = 7.39 \ m/s$

Explanation:

From the question we are told that

The angular speed is $w = 2030 \ rpm = \frac{2030 * 2 * \pi }{ 60} = 212.61 \ rad/s$

The distance between the minimum and maximum external position is $d = 6.95 \ cm = 0.0695 \ m$

Generally the amplitude of the crank shaft is mathematically represented as

$A = \frac{d}{2}$

=> $A = \frac{0.0695}{2}$

=> $A = 0.03475 \ m$

Generally the maximum speed of the piston is mathematically represented as

$|v| = A * w$

=> $|v| = 0.03475 * 212.61$

=> $|v| = 7.39 \ m/s$

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During a demonstration of the gravitational force on falling objects to her class, Sarah drops an 11 lb. bowling ball from the t

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1.A) 4.9 m

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The instant it was dropped, the ball had zero speed.

After falling for 1 second, its speed was 9.8 m/s straight down (gravity).

Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.

Falling for 1 second at an average speed of 4.9 m/s, is covered 4.9 meters.

ANYTHING you drop does that, if air resistance doesn't hold it back.

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Answer:

Answers in solutions.

Explanation:

<u>Question 6:</u>

The density of gold is 19.3 g/cm³

The density of silver is 10.5 g/cm³

The density of the substance in Crown A;

Density = mass ÷ volume = $\frac{1930}{100}$ = 19.3 g/cm³

Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.

The density of the substance in Crown B;

Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043 g/cm³ ≈ 10.5 g/cm³ (answer rounded up to one decimal place)

Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.

The density of substance in Crown C;

Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)

<h3><u>The density of the mixture:</u></h3><h3 />

For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g

∴ for 1 cm³ of the mixture, its mass equal to $\frac{29.8}{2}$ = 14.9 g

Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.

* Crown C is not made up of a mixture of gold and silver.

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An empty masuring cylinder has a mass of 500 g.

Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.

The total mass is now 850 g

The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g

Density of the liquid (water) poured into the container = mass ÷ volume = 350 g ÷ 100 cm³ = 3.5g/cm³

<u>Question 8:</u>

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A tank filled with water has a volume of 0.02 m³

(a) 1 liter = 0.001 m³

How many liters? = 0.02 m³ ?

Cross multiplying gives:

$\frac{0.02 * 1}{0.001}$ = 20 liters

(b) 1 m³ = 1,000,000 cm³

0.02 m³ = how many cm³ ?

Cross-multiplying gives;

$\frac{0.02 * 1,000,000}{1}$ = 20,000 cm³

(c) 1 cm³ = 1 ml

∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml

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Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

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A stone is gently and completely immersed in a liquid of density 1.0 g/cm³
in a displacement can
The mass of liquid which overflow is 20 g

The mass of the liquid which overflow = mass of the stone = 20 g

1 gram of the liquid occupies 1 cm³ of space.

20 g of the liquid will occupy; $\frac{20 * 1}{1}$ = 20 cm³

(a) Since the volume of the water displaced is equal to the volume of the stone.

∴ The volume of the stone = 20 cm³

(b) Mass = density × volume

Density of the stone = 5.0 g/cm³

Volume of the stone = 20 cm³

Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g

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A uniform electric field has a magnitude 1.80 kV/m and points in the +x direction. (a) What is the electric potential difference

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Answer:

a)ΔV = 6.48 KV

b)ΔU =18.79 mJ

Explanation:

Given that

E= 1.8 KV/m

a)

We know that

Electric potential difference ΔV given as

ΔV = E .d

Here

E= 1.8 KV/m

d= 3.6 m

ΔV = E .d

ΔV = 1.8 x 3.6 KV

ΔV = 6.48 KV

b)

Given that

q=+2.90 µC

Change in electric potential energy ΔU given as

ΔU = q .ΔV

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Which formula can be used to solve problems related to the first law of thermodynamics?

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Answer: The formula used to solve the problems related to first law of thermodynamics is $\Delta U=Q+W$

Explanation:

First law of thermodynamics states that the total energy of the system remains conserved. Energy can neither be destroyed, nor be created but it can only be transformed into one form to another.

Its implication is any change in the internal energy will be either due to heat energy or work energy.

Mathematically,

$\Delta U=Q+W$

where, Q = heat energy

W = work energy

$\Delta U$ = Change in internal energy

Sign convention for these energies:

For Q: Heat absorbed will be positive and heat released will be negative.

For W: Work done by the system is negative and work done on the system is positive.

For $\Delta U$ : When negative, internal energy is decreasing and when positive, internal energy is increasing.

Hence, the formula used to solve the problems related to first law of thermodynamics is $\Delta U=Q+W$

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Blababa [14]

Answer:

11.23%

Explanation:

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Speed of flow of water = v=0.52 m/s

When swims in downward direction then speed of man = u + v

When swims in upward direction then speed of man = u - v

Lets time taken by man when he swims in downward direction is $t_1$ and when he swims in downward direction is $t_2$

Lets distance is d and it will be remain constant in both the case

$d=(u+v)t_1$

$d=(u-v)t_2$

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$t_2=1.85t_1$

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2 d= t x 1.73

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$t_2=0.82d\ sec$

total time in current = 0.82 +0.44 d=1.26 d sec

So the percentage time

$percentage\ time =\dfrac{1.28-1.15}{1.15}$

Percentage time =11.32%

So it will take 11.32% more time as compare to still current.

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