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2 years ago

Which of the following use the same units of measurement?

Physics

2 answers:

o-na [289]2 years ago

8 0

Meters, metric, body mass index.

Afina-wow [57]2 years ago

7 0

Metric
meters
body mass index

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The drawing shows three identical springs hanging from the ceiling. Nothing is attached to the first spring, whereas a 4.50-N bl

Pavlova-9 [17]

Answer:

a. 30 N / m

b. 9.0 N

Explanation:

Given that

Unstretched length of the spring, $L_o$ = 20.0cm = 0.2m

a) When the mass of 4.5N is hanging from the second spring, then extended length Is

$L_1$ = 35.0cm = 0.35m

So, the change in spring length when mass hangs is

$x = L_1 - L_o$

= (0.35 - 0.20) m

= 0.15m

As spring are identical

Let us assume that the spring constant be "k", so at equilibrium

Restoring Force on spring = Block weightage

kx = W = 4.50

$k= \frac{4.50}{x} = \frac{4.50}{0.15}$

= 30 N / m

b) Now for the third spring, stretched the length of spring is

$L_2$ = 50cm = 0.5m

So, the change in spring length is

$x'= L_2 - L_o$

= (0.5-0.20)m

= 0.30m

At equilibrium,

Restoring Force on spring = Block weightage

Now using all mentioned and computed values in above,

$W'= kx'$

= 30(0.3)

= 9.0 N

4 0

2 years ago

Which one of the following represents an acceptable set of quantum numbers for an electron in an atom? (arranged as n, l, m l ,

Vitek1552 [10]

Answer:

The correct option that represents an acceptable set of quantum numbers for an electron in an atom is;

(b) 4, 3, -3, 1/2.

Explanation:

To solve the question, we note that the available options where the set of quantum numbers for an electron in an atom are arranged as n, l, m l , and ms are;

4, 4, 4, 1/2

4, 3, -3, 1/2

4, 3, 0, 0

4, 5, 7, -1/2

4, 4, -5, 1/2

Let us label them as a to as follows

(a) 4, 4, 4, 1/2

(b) 4, 3, -3, 1/2

(d) 4, 5, 7, -1/2

(e) 4, 4, -5, 1/2

Next we note the rules for the assignment and arrangement of quantum numbers are as follows

Number Symbol Possible values

Principal Quantum Number .......n........................1, 2, 3, ......n

Angular momentum quantum

number...............................................l.........................0, 1, 2, .......(n - 1)

Magnetic Quantum Number........m₁......................-l, ..., -1, 0, 1,.....,l

Spin Quantum Number.................m $_s$ .....................+1/2, -1/2

We are meant to analyze each of the arrangement for acceptability.

Therefore for (a),

we note that the angular momentum quantum number, l =4 , is equal to the principal quantum number n =4 which violates the rule as the maximum value of the angular momentum quantum number is (n-1) where the maximum value of the principal quantum number is n.

Therefore (a) is not acceptable.

(b) Here we note that

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = -3 ∈ (-l, ..., -1, 0, 1,.....,l) → acceptable

The spin quantum number m $_s$ = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (b) 4, 3, -3, 1/2 represents an acceptable set of quantum numbers for an electron in an atom.

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = 0 ∈ (-l, ..., -1, 0, 1,.....,l) → acceptable

The spin quantum number m $_s$ = 0 ∉ (+1/2, -1/2) → not acceptable

Therefore (c) 4, 3, 0, 0 does not represents an acceptable set of quantum numbers for an electron in an atom.

(d) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 5 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = 7 ∉ (-l, ..., -1, 0, 1,.....,l) → acceptable

The spin quantum number m $_s$ = -1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (d) 4, 5, 7, -1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

(e) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 4 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = -5 ∉ (-l, ..., -1, 0, 1,.....,l) → acceptable

The spin quantum number m $_s$ = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (e) 4, 4, -5, 1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

3 0

2 years ago

A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the

mash [69]

Answer:

$\Delta x = v_0 t + \frac{1}{2}at^2$

Explanation:

To solve this problem, we can use the following suvat equation:

$\Delta x = v_0 t + \frac{1}{2}at^2$

where

$\Delta x$ is the vertical displacement of the frog

$v_0$ is the initial vertical velocity

t is the time

a is the acceleration

We have chosen this formula because apart from $v_0$ , all the other quantities are known. In fact:

$\Delta x =0.1 m$ is the vertical displacement

t = 2 s is the total time of flight

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

Therefore, solving for $v_0$ , we find the initial velocity of the frog:

$v_0 = \frac{\Delta x-\frac{1}{2}at^2}{t}=\frac{0.1-\frac{1}{2}(-9.8)(2)^2}{2}=9.85 m/s$

4 0

2 years ago

The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This sp

Delvig [45]

Answer:

Explanation:

Here is the full question and answer,

The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.

When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.

Part A: At what speed v should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60 degrees above the water surface.

Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60 degrees above the surface and with the same initial speed v as before. At what height h above the surface was the insect?

Answer

A.) The path of a projectile is horizontal and symmetrical ground. The time is taken to reach maximum height, the total time that the particle is in flight will be double that amount.

Calculate the speed of the archer fish.

The time of the flight of spitted water is,

$t = \frac{{2v\sin \theta }}{g}$

Substitute $9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for g and $60^\circ$ for $\theta$ in above equation.

$t = \frac{{2v\sin 60^\circ }}{{9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}}}\\\\ = \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\$

Spitted water will travel $0.80{\rm{ m}}$ horizontally.

Displacement of water in this time period is

$x = vt\cos \theta$

Substitute $\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}$ for $t\rm 60^\circ[tex] for [tex]\theta$ and $0.80{\rm{ m}}$ for x in above equation.

$\\0.80{\rm{ m}} = v\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\left( {\cos 60^\circ } \right)\\\\0.80{\rm{ m}} = {v^2}\left( {0.1767{\rm{ }}} \right)\frac{1}{2}{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\\\v = \sqrt {\frac{{2\left( {0.80{\rm{ m}}} \right)}}{{0.1767\;{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}}}} \\\\ = 3.01{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

B.) There are two component of velocity vertical and horizontal. Calculate vertical velocity and horizontal velocity when the angle is given than calculate the time of flight when the horizontal distance is given. Value of the horizontal distance, angle and velocity are given. Use the kinematic equation to solve the height of insect above the surface.

Calculate the height of insect above the surface.

Vertical component of the velocity is,

${v_v} = v\sin \theta$

Substitute $3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ for v and $60^\circ$ for $\theta$ in above equation.

$\\{v_v} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\sin 60^\circ \\\\ = 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

Horizontal component of the velocity is,

${v_h} = v\cos \theta$

Substitute $3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ for v and $60^\circ$ for $\theta$ in above equation.

$\\{v_h} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\cos 60^\circ \\\\ = 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

When horizontal $({0.60\;{\rm{m}}}$ distance away from the fish.

The time of flight for distance (d) is ,

$t = \frac{d}{{{v_h}}}$

Substitute $0.60\;{\rm{m}}$ for d and $1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$ for ${v_h}$ in equation $t = \frac{d}{{{v_h}}}$

$\\t = \frac{{0.60\;{\rm{m}}}}{{1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}}}\\\\ = 0.3987{\rm{ s}}\\$

Distance of the insect above the surface is,

$s = {v_v}t + \frac{1}{2}g{t^2}$

Substitute $2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$ for ${v_v}$ and $0.3987{\rm{ s}}$ for t and $- 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for g in above equation.

$\\s = \left( {2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {0.3987{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right){\left( {0.3987{\rm{ s}}} \right)^2}\\\\ = 0.260{\rm{ m}}\\$

7 0

2 years ago

The following represents a process used to assemble a chair with an upholstered seat. stationsa, b, and c make the seat; station

Mekhanik [1.2K]

There are three questions here:

A. The possible daily output of this process if there is 8 processing time each day?

the time it takes to assemble a chair in seconds = A + B + C + J + K + L + X + Y + Z

which is equal to = 38 + 34 + 35 + 32 + 30 + 34 + 22 + 18 + 20 =263 per chair

Processing hours = 8 hours x 60 minutes x 60 seconds

= 8 x 60 x 60

= 480 x 60 = 28,800 seconds is available in an 8 hr day.

28,800 / 263 =109.5057034220532

Therefore, It is possible to make 109 chairs in an 8 hour day.

B. Given your output rate in above, what is the efficiency of the process?

The time it takes to assemble a chair in seconds = A+ B + C + J + K + L + X + Y + Z.

= 34 + 34 + 34 + 30 + 30 + 30 + 22 + 18 + 20 = 252.6315789473684 per chair

A total of 252 per chair

Processing hours = 8 hours x 60 minutes x 60 seconds

= 8 x 60 x 60

= 480 x 60 = 28,800 seconds is available in an 8 hour day

= 28,000 / 252 = 114.2857142857143114 chairs

= 109/114 x 100 = 95.6140350877193

Therefore, the efficiency of making the chair is 95.61%

C. What is the flow time of the process?

Flow time is the period that it takes a completed chair to flow through the process from the beginning assemblage step to the last step. Take note that ABC and JKL are parallel legs in the process, and as a result both do not include to flow time to the procedure. In addition, Flow time comprises both pass time and run time at each position in the procedure.

7 0

2 years ago