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SCORPION-xisa [38]

4 years ago

11

A 2 kg block is pushed by an external force against a spring with spring constant 131 N/m until the spring is compressed by 2.1

m from its uncompressed length ( 0).
The block rests on a horizontal plane that has a coefficient of kinetic friction of 0.74 but is NOT attached to the sprinr 2 kg 2.1 m 2 kg 131 N/m μ 0.74

After all the external forces are removed (so the compressed spring releases the mass) how far D along the plane will the block move before coming to a stop?
The acceleration due to gravity is 9.8 m/s .
Answer in units of m.

1 answer:

Andrej [43]4 years ago

7 0

Answer:

d = 19.92 m

Explanation:

As in this exercise there is friction we must use the relationship between work and energy

W = ΔEm

Look for energy in two points

Initial. Fully compressed spring

Em₀ = $K_{e}$ = ½ k x²

Final. When the block stopped

$Em_{f}$ = 0

Let's look for the work of the rubbing force

W = fr d cos θ

Since rubbing is always contrary to movement, θ = 180

W = - fr d

Let's use Newton's second Law, to find the force of friction

Y Axis

N- w = 0

N = mg

The equation for the force of friction is

fr = μ N

fr = μ mg

We substitute in the work equation

W = - μ m g d

We write the relationship of work and energy

-μ m g d = 0 - ½ k x²

d = ½ k x² / μ m g

Let's calculate

d = ½ 131 2.1 2 / (0.74 2 9.8)

d = 19.92 m

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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o

matrenka [14]

Answers:

(a) $0.0073kg$

(b) Volume gold: $3.79(10)^{-7}m^{3}$ , Volume cupper: $7.6(10)^{-8}m^{3}$

(c) $17633.554kg/m^{3}$

Explanation:

<h2>(a) Mass of gold </h2><h2 />

We are told the total mass $M$ of the coin, which is an alloy of gold and copper is:

$M=m_{gold}+m_{copper}=7.988g=0.007988kg$ (1)

Where $m_{gold}$ is the mass of gold and $m_{copper}$ is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats $K$ and mass is:

$K=24\frac{m_{gold}}{M}$ (2)

Finding ${m_{gold}$ :

$m_{gold}=\frac{22}{24}M$ (3)

$m_{gold}=\frac{22}{24}(0.007988kg)$ (4)

$m_{gold}=0.0073kg$ (5) This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density $\rho$ of an object is given by:

$\rho=\frac{mass}{volume}$

If we want to find the volume, this expression changes to: $volume=\frac{mass}{\rho}$

For gold, its volume $V_{gold}$ will be a relation between its mass $m_{gold}$ (found in (5)) and its density $\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}$ :

$V_{gold}=\frac{m_{gold}}{\rho_{gold}}$ (6)

$V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}$ (7)

$V_{gold}=3.79(10)^{-7}m^{3}$ (8) Volume of gold in the coin

For copper, its volume $V_{copper}$ will be a relation between its mass $m_{copper}$ and its density $\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}$ :

$V_{copper}=\frac{m_{copper}}{\rho_{copper}}$ (9)

The mass of copper can be found by isolating $m_{copper}$ from (1):

$M=m_{gold}+m_{copper}$

$m_{copper}=M-m_{gold}$ (10)

Knowing the mass of gold found in (5):

$m_{copper}=0.007988kg-0.0073kg=0.000688kg$ (11)

Now we can find the volume of copper:

$V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}$ (12)

$V_{copper}=7.6(10)^{-8}m^{3}$ (13) Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density $\rho_{coin$ will be a relation between its total mass $M$ and its total volume $V$ :

$\rho_{coin}=\frac{M}{V}$ (14)

Knowing the total volume of the coin is:

$V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}$ (15)

$\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}$ (16)

Finally:

$\rho_{coin}=17633.554kg/m^{3}}$ (17) This is the total density of the British sovereign coin

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What are the three primary particles that make up an atom, and

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Answer:

Protons, Electrons, and Neutrons are the 3 primary particles in an atom.

Protons - (+1)

Electrons - (-1)

Neutrons - (0)

<h3>Your answer would be C</h3>

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3 years ago

A uniform 1.0-N meter stick is suspended horizontally by vertical strings attached at each end. A 2.0-N weight is suspended from

fgiga [73]

Answer:

3.5 N

Explanation:

Let the 0-cm end be the moment point. We know that for the system to be balanced, the total moment about this point must be 0. Let's calculate the moment at each point, in order from 0 to 100cm

- Tension of the string attached at the 0cm end is 0 as moment arm is 0

- 2 N weight suspended from the 10 cm position: 2*10 = 20 Ncm clockwise

- 2 N weight suspended from the 50 cm position: 2*50 = 100 Ncm clockwise

- 1 N stick weight at its center of mass, which is 50 cm position, since the stick is uniform: 1*50 = 50 Ncm clockwise

- 3 N weight suspended from the 60 cm position: 3*60 = 180 Ncm clockwise

- Tension T (N) of the string attached at the 100-cm end: T*100 = 100T Ncm counter-clockwise.

Total Clockwise moment = 20 + 100 + 50 + 180 = 350Ncm

Total counter-clockwise moment = 100T

For this to balance, 100 T = 350

so T = 350 / 100 = 3.5 N

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Air will break down (lose its insulating quality) and sparking will result if the field strength is increased to about 3 × 106 N

MrRissso [65]

Answer:

Acceleration, $a=5.27\times 10^{17}\ m/s^2$

Explanation:

Given that,

Electric field strength, $E=3\times 10^6\ N/C$

Mass of the electron, $m=9.109\times 10^{-31}\ kg$

Charge on electron, $q=1.602\times 10^{-19}\ C$

Let a is the acceleration experienced by an electron. It can be calculated as :

$ma=qE$

$a=\dfrac{qE}{m}$

$a=\dfrac{1.602\times 10^{-19}\times 3\times 10^6}{9.109\times 10^{-31}}$

$a=5.27\times 10^{17}\ m/s^2$

Hence, this is the required solution.

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3 years ago

A 2.1 kg steel ball strikes a massive wall at 13.2 m/s at an angle of 64.8 ◦ with the perpendicular to the plane of the wall. It

solong [7]

Answer:

112 N

Explanation:

going through the question you would notice that some detail is missing, using search engines u was able to find a similar question on "https://socratic.org/questions/a-2-3-kg-steel-ball-strikes-a-wall-with-a-speed-of-8-5-m-s-at-an-angle-of-64-wit"

and here is the question i found

"A 2.3 kg steel ball strikes a wall with a speed of 8.5 m/s at an angle of 64⁰ with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.448 s. What is the average force exerted by the ball?"

you would notice that there is a change in the values from the question posted, hence we would only take the following part to complete our question, " If the ball is in contact with the wall for 0.448 s what is the average force exerted by the ball?" while retaining all original detail.

solution

mass of ball (m) = 2.1 kg

speed of ball (v) = 13.2 m/s

angle of contact (p) = 64.8°

time of contact (t) = 0.448 s

What is the average force exerted by the ball?

The average force exerted by the ball = $\frac{change in momentum}{change in time}$

where

The momentum changes only the direction perpendicular to the wall, hence the component of momentum perpendicular to the wall = m x v x sin (p) = 2.1 x 13.2 x sin 64.8 = 25.1 kg.m/s

since the ball strikes the wall and bounces off it with the same speed

and at the same angle, the component of momentum acting

perpendicular to the wall remains the same while hitting and leaving

the wall but in opposite directions.

Hence the component of momentum acting perpendicular to the wall while hitting and leaving the wall will be 25.1 kg.m/s and -25.1 kg.m/s respectively.

change in momentum = 25.1 - (-25.1) = 25.1 + 25.1 = 50.2 kg.m/s

change in time = 0.448 s

now substituting the above into the equation we have

The average force exerted by the ball = $\frac{50.2}{0.448}$ = 112 N

6 0

4 years ago