in which states of matter will a substance have a fixed volume

Answer:

Explanation:

Given that:

Focal length for the objective lens = 20 mm, 4 mm, 1.4 mm

For objective lens of focal length f₁ = 20 mm

s₁' = 120 mm + 20 mm = 140 mm

∴

Magnification [tex]m_1 = \dfrac{s'_1}{f_1}[/tex]

[tex]m_1 = \dfrac{140}{20}[/tex]

[tex]m_1 = 7 \ m[/tex]

For objective lens of focal length f₁ = 4 mm

s₁' = 120 mm + 4 mm = 124 mm

[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]

[tex]m_1 = \dfrac{124}{4}[/tex]

[tex]m_1 = 31 \ m[/tex]

For objective lens of focal length f₁ = 1.4 mm

s₁' = 120 mm + 1.4 mm = 121.4 mm

[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]

[tex]m_1 = \dfrac{121.4}{1.4}[/tex]

[tex]m_1 = 86.71 \ m[/tex]

The magnification of the eyepiece is given as:

[tex]m_e = 5X \ and \ m_e = 15X[/tex]

Thus, the largest angular magnification when  [tex]m_1 \ and \ m_e \ are \ large \ is:[/tex]

[tex]M_{large}= (m_1)_{large} \times (m_e)_{large}[/tex]

= 86.71 × 15

= 1300.65

The smallest angular magnification derived when [tex]m_1 \ and \ m_e \ are \ small \ is:[/tex]

[tex]M_{small}= (m_1)_{small} \times (m_e)_{small}[/tex]

= 7 × 5

= 35

The largest magnification will be 1300.65 and the smallest magnification will be 35.

Magnification is defined as the ratio of the size of the image of an object to the actual size of the object.

Now for objective lens and eyepieces, it is defined as the ratio of the focal length of the objective lens to the focal length of the eyepiece.

It is given in the question:

Focal lengths for the objective lens is = 20 mm, 4 mm, 1.4 mm

now we will calculate the magnification for all three focal lengths of the objective lens.

Also, each objective forms an image 120 mm beyond its second focal point.

(1) For an objective lens of focal length   [tex]f_1=20 \ mm[/tex]

[tex]s_1'=120\ mm +20 \ mm =140\ mm[/tex]

Magnification will be calculated as

[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{140}{20} =7[/tex]

(2) For an objective lens of focal length [tex]f_1= \ 4 \ mm[/tex]

s₁' = 120 mm + 4 mm = 124 mm

[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{124}{4} =31[/tex]

(3) For an objective lens of focal length [tex]f_1=1.4\ mm[/tex]

s₁' = 120 mm + 1.4 mm = 121.4 mm

[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{121.4}{1.4} =86.71[/tex]

Now the magnification of the eyepiece is given as:

[tex]m_e=5x\ \ \ & \ \ m_e=15x[/tex]

Thus, the largest angular magnification when  

[tex]m_1 = 86.17\ \ \ \ m_e=15x[/tex]

[tex]m_{large}= (m_1)_{large}\times (m_e)_{large}[/tex]

[tex]m_{large}=86.71\times 15=1300.65[/tex]

The smallest angular magnification derived when

[tex]m_1=7\ \ \ \ m_e=5[/tex]

[tex]m_{small}=(m_1)_{small}\times (m_e)_{small}[/tex]

[tex]m_{small}=7\times 5=35[/tex]

Thus the largest magnification will be 1300.65 and the smallest magnification will be 35.

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Answer:

0.32 m.

Explanation:

To solve this problem, we must recognise that:

1. At the maximum height, the velocity of the ball is zero.

2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.

With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:

Mass (m) = constant

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) = 2.5 m/s

Height (h) =?

PE = KE

Recall:

PE = mgh

KE = ½mv²

Thus,

PE = KE

mgh = ½mv²

Cancel m from both side

gh = ½v²

9.8 × h = ½ × 2.5²

9.8 × h = ½ × 6.25

9.8 × h = 3.125

Divide both side by 9.8

h = 3.125 / 9.8

h = 0.32 m

Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.

Answer:

Maria Corazon Sumulong Cojuangco Aquino, popularly known as Cory Aquino, was a Filipino politician who served as the 11th President of the Philippines, the first woman to hold that office.

Answer:

Former President of the Philippines

Explanation:

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = [tex]e^{kt}[/tex]

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = [tex]e^{-0.01t}[/tex]

V = V₀[tex]e^{-0.01t}[/tex]

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀[tex]e^{-0.01t}[/tex]

0.1V₀ = V₀[tex]e^{-0.01t}[/tex]

0.1V₀/V₀ = [tex]e^{-0.01t}[/tex]

0.1 = [tex]e^{-0.01t}[/tex]

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

Answer:

Newton's law of cooling says that the temperature of a body changes at a rate proportional to the difference between its temperature and that of the surrounding medium (the ambient temperature); dT/dt = -K(T - Tₐ) where T = the temperature of the body (°C), t = time (min), k = the proportionality constant (per minute),

Explanation:

Answer:

The tub turns 37.520 revolutions during the 25-second interval.

Explanation:

The total number of revolutions done by the tub of the washer ([tex]\Delta n[/tex]), in revolutions, is the sum of the number of revolutions done in the acceleration ([tex]\Delta n_{1}[/tex]), in revolutions, and deceleration stages ([tex]\Delta n_{2}[/tex]), in revolutions:

[tex]\Delta n = \Delta n_{1} + \Delta n_{2}[/tex] (1)

Then, we expand the previous expression by kinematic equations for uniform accelerated motion:

[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} - \ddot n_{2} \cdot t_{2}^{2})[/tex] (1b)

Where:

[tex]\ddot n_{1}, \ddot n_{2}[/tex] - Angular accelerations for acceleration and deceleration stages, in revolutions per square second.

[tex]t_{1}, t_{2}[/tex] - Acceleration and deceleration times, in seconds.

And each acceleration is determined by the following formulas:

Acceleration

[tex]\ddot n_{1} = \frac{\dot n}{t_{1}}[/tex] (2)

Deceleration

[tex]\ddot n_{2} = -\frac{\dot n}{t_{2} }[/tex] (3)

Where [tex]\dot n[/tex] is the maximum angular velocity of the tub of the washer, in revolutions per second.

If we know that [tex]\dot n = 3\,\frac{rev}{s}[/tex], [tex]t_{1} = 13\,s[/tex] and [tex]t_{2} = 12\,s[/tex], then the quantity of revolutions done by the tub is:

[tex]\ddot n_{1} = \frac{3\,\frac{rev}{s} }{13\,s}[/tex]

[tex]\ddot n_{1} = 0.231\,\frac{rev}{s^{2}}[/tex]

[tex]\ddot n_{2} = -\frac{3\,\frac{rev}{s} }{12\,s}[/tex]

[tex]\ddot n_{2} = -0.25\,\frac{rev}{s^{2}}[/tex]

[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} + \ddot n_{2} \cdot t_{2}^{2})[/tex]

[tex]\Delta n = \frac{1}{2}\cdot \left[\left(0.231\,\frac{rev}{s^{2}} \right)\cdot (13\,s)^{2}-\left(-0.25\,\frac{rev}{s^{2}} \right)\cdot (12\,s)^{2}\right][/tex]

[tex]\Delta n = 37.520\,rev[/tex]

The tub turns 37.520 revolutions during the 25-second interval.

I don't know

because I don't know

Answer:

Δd = 7.22 10⁻² m

Explanation:

For this exercise we must use the dispersion relationship of a diffraction grating

           d sin θ = m λ

let's use trigonometry

           tan θ = y / L

how the angles are small

           tant θ = sinθ  /cos θ = sin θ

we substitute  

           sin θ = y / L

          d y / L = m λ

          y = m λ L / d

let's use direct ruler rule to find the distance between two slits

If there are 500 lines in 1 me, what distance is there between two lines

         d = 2/500

        d = 0.004 me = 4 10⁻⁶ m

diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1

let's calculate for each wavelength

λ = 656 nm = 656 10⁻⁹ m

         d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶

         d₁ = 2.788 10⁻¹ m

λ = 486 nm = 486 10⁻⁹ m

         d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶

         d₂ = 2.066 10⁻¹ m

the distance between the two lines is

         Δd = d1 -d2

         Δd = (2,788 - 2,066) 10⁻¹

         Δd = 7.22 10⁻² m

Answer:

12

Explanation:

I'm a beginner so am not sureeeeee

Explanation:

The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum

Answer:

Answer:

a) the most probable speed of the molecules is 409.2 m/s

b) required temperature of xenon is 1322 K

Explanation:

Given the data in the question;

a)

Maximum probable speed of hydrogen molecule (H₂)

[tex]V_{H_2[/tex] = √( 2RT / [tex]M_{H_2[/tex] )

where R = 8.314 m³.Pa.K⁻¹.mol⁻¹ and given that T = 20.3 K

molar mass of H₂; [tex]M_{H_2[/tex] = 2.01588 g/mol

we substitute

[tex]V_{H_2[/tex] = √( (2 × 8.314 × 20.3 ) / 2.01588 × 10⁻³  )

[tex]V_{H_2[/tex] = √( 337.5484 / 2.01588 × 10⁻³  )

[tex]V_{H_2[/tex] = 409.2 m/s

Therefore, the most probable speed of the molecules is 409.2 m/s

b)

Temperature of xenon  = ?

Temperature of hydrogen = 20.3 K

we know that;

T = (Vxe² × Mxe) / 2R

molar mass of xenon; Mxe = 131.292 g/mol

so we substitute

T = ( (409.2)² × 131.292 × 10⁻³) / 2( 8.314  )

T = 21984.14167 / 16.628

T = 1322 K

Therefore, required temperature of xenon is 1322 K

Answer:

v = 6491.94 m/s

Explanation:

We are given;

Frequency; f = 87 Hz

Wavelength;λ = 74.62 m

Formula for velocity(v) of waves from the wave equation is;

v = fλ

Thus;

v = 87 × 74.62

v = 6491.94 m/s

Answer:

F' = 73.7 N

F = 8.749×10⁻³ N

a' = a =  9.83 m/s²

Explanation:

(a)

For the rock

Applying

F' = Gm'm/r²................... Equation 1

Where F = magnitude of the gravitational force on the rock, G = Gravitational constant, m' = mass of the rock, m = mass of the earth, r = radius of the earth.

From the question,

Given: m' = 7.5 kg

Constant: m = 5.98×10²⁴ kg, G = 6.67×10⁻¹¹ Nm²/kg², r = 6.37×10⁶ m

Substitute these values into equation 1

F' = 6.67×10⁻¹¹ (7.5)(5.98×10²⁴)/(6.37×10⁶)²

F' = 7.37×10¹ N

F' = 73.7 N

Also, For the pebble,

F = GMm/r².............. Equation 2

Where M = mass of the pebble, F = Gravitational force exerted on the pebble by the earth

Given: M = 8.9×10⁻⁴ kg,

Substitute into equation 2

F = 6.67×10⁻¹¹(8.9×10⁻⁴)(5.98×10²⁴)/(6.37×10⁶)²

F = 8.749×10⁻³ N

(b)

For the rock,

a' = F'/m'

Where a' = magnitude of the acceleration of the rock

a' = 73.7/7.5

a' = 9.83 m/s²

For the pebble,

a = F/M

Where a = acceleration of the pebble

a = (8.749×10⁻³)/(8.9×10⁻⁴)

a = 9.83 m/s²

Answer:

Explanation:

There are two main ways to de-starch leaves of a plant - the 'Light Exclusion' Method and the 'Carbon Dioxide Deprivation' Method. The 'Light Exclusion' method is a simpler procedure and is used often. Leaves can be destarched by depriving them of light for an extended period of time, usually 24-48 hours.

By the work-energy theorem, the total work done on the car is equal to the change in its kinetic energy:

W = ∆K

W = 1/2 (0.34 kg) (22.9 m/s)² - 1/2 (0.34 kg) (6.5 m/s)²

W ≈ 82 J

Answer:

time of fall is 1.75 s and the velocity with which it strikes the water is 17.15 m/s.

Explanation:

Height, h =  15 m

Newton's second law

Force = mass x acceleration

The unit of gravitational force is Newton and the value is m x g.

where, m is the mas and g is the acceleration due to gravity.  

Let the time of fall is t.

Use second equation of motion

[tex]s= u t +0.5 at^2\\\\15 = 0 +0.5\times 9.8\times t^{2}\\\\t = 1.75 s[/tex]

Let the final speed is v.

Use third equation of motion

[tex]v^2 = u^2 + 2 a s\\\\v^2 = 0 + 2 \times 9.8\times 15\\\\v =17.15 m/s[/tex]

Answer:

The angle is 4.1 rad.

Explanation:

The centripetal acceleration (α) is given by:

[tex] \alpha = \omega^{2} r [/tex]    (1)                  

Where:

ω: is the angular velocity  

r: is the radius

And the tangential acceleration (a) is:                      

[tex] a = \alpha r [/tex]      (2)

Since the magnitude of "α" is 8.2 times the magnitude of "a" (equating (2) and (1)) we have:

[tex] \omega^{2} r = 8.2\alpha r   [/tex]

[tex] \omega^{2} = 8.2\alpha [/tex]    (3)      

Now, we can find the angle with the following equation:

[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \Delta \theta [/tex]

Where:

[tex] \omega_{f}[/tex]: is the final angular velocity                                                                              [tex] \omega_{0}[/tex]: is the initial angular velocity = 0 (it starts from rest)

[tex]\Delta \theta[/tex]: is the angle

[tex] \omega^{2} = 2\alpha \Delta \theta [/tex]     (4)    

By entering equation (3) into (4) we can calculate the angle:

[tex] 8.2\alpha = 2\alpha \Delta \theta [/tex]

[tex] \Delta \theta = 4.1 rad [/tex]

Therefore, the angle is 4.1 rad.

I hope it helps you!                  

Answer:

We can also prove the conservation of mechanical energy of a freely falling body by the work-energy theorem, which states that change in kinetic energy of a body is equal to work done on it. i.e. W=ΔK. And ΔE=ΔK+ΔU. Hence the mechanical energy of the body is conserved

Explanation:

Answer:

No of proton is 13 and nucleus is 13

If a person wants to calculate the length of time it takes for a ball to fall from a height of 20m, the correct equation that they should use is:

D. Δt= √2Δd/a

The free fall formula should be used to obtain the length of time that it takes for a ball to fall from a given height. This formula also factors the height or distance from which the fall occurred and this is denoted by the letter d. The small letter 'a' is denotative of acceleration due to gravity and this is a constant pegged at -9.98 m/s².

So, the change in height is obtained and multiplied by two. This is further divided by the acceleration and the square root of the derived answer translates to the time taken for the ball to fall from the height of 20m. Of all the options listed, option D represents the correct equation.

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Answer:

Fr = 1.91 * 9*10⁹*q²/L²

Explanation:

Let´s say that the corners of the square are  A B C and D

We are going to find out the force on the charge placed on B  ( the charge placed in the upper right corner.

As all the charges are positive (the same sign), then all the three forces on the charge in B are of rejection.

Force due to charge placed in A

module   Fₓ =  K* q² / L²   in the direction of x

Force due to charge placed in C

module  Fy = K* q²/L²   in the direction of y

Force due to  the charge placed in D

That force will have the direction of the diagonal of the square, and the distance between charges placed in D and A is the length of the diagonal.

d²  =  L²  +  L²  =  2*L²

d  =  √2 * L

The module of the force due to charge place in D

F₄₅ = K*q²/ 2*L²

To get the force we need to add first  Fₓ  and  Fy  

Fx + Fy  =  F₁

module of  F₁ = √ Fx² + Fy²    the direction will be the same as the diagonal of the square then:

F₁   =   √  ( K* q²/L² )²  +   ( K* q²/L² )²

F₁  =  √ 2  *  K*q²/L²

And now we add forces F₁   and F₄₅   to get the net force Fr on charge in point B.

The direction of Fr is the direction of the diagonal and is of rejection

the module is

Fr  =  F₁  *  F₄₅

Fr  =  √ 2  *  K*q²/L²  +   K*q²/ 2*L²

Fr  = ( √ 2 + 0,5 ) * K*q² /L²

K  =  9*10⁹  Nm²C²

Fr = 1.91 * 9*10⁹*q²/L²

We don´t know units of L and q

Answer:

[tex]\triangle P=1.95*10^{-4}[/tex]

Explanation:

Mass [tex]m=0.001[/tex]

Diameter [tex]d=1.2m[/tex]

Length [tex]l=10m[/tex]

Generally the equation for Volume flow rate is mathematically given by

 [tex]Q=AV[/tex]

 [tex]V=\frac{Q}{\pi/4D^2}[/tex]

 [tex]V=\frac{0.001}{\pi/4(1.2)^2}[/tex]

 [tex]V=8.84*10^{-4}[/tex]

Generally the equation for Friction factor is mathematically given by

 [tex]F=\frac{64}{Re}[/tex]

Where Re

Re=Reynolds Number

 [tex]Re=\frac{pVD}{\mu}[/tex]

 [tex]Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}[/tex]

 [tex]Re=1040[/tex]

Therefore

 [tex]F=\frac{64}{Re}[/tex]

 [tex]F=\frac{64}{1040}[/tex]

 [tex]F=0.06[/tex]

Generally the equation for Friction factor is mathematically given by

 [tex]Head loss=\frac{fLv^2}{2dg}[/tex]

 [tex]H=\frac{0.06*10*(8.9*10^-4)^2}{2*1.2*9.81}[/tex]

 [tex]H=19.9*10^{-9}[/tex]

Where

[tex]H=\frac{\triangle P}{\rho g}[/tex]

[tex]\triangle P=\frac{19.9*10^{-9}}{10^3*(9.81)}[/tex]

[tex]\triangle P=H*\rho g[/tex]

[tex]\triangle P=1.95*10^{-4}[/tex]

Answer:

D) The bicycle is not in motion.

Explanation:

Study the position-time graph for a bicycle.

Which statement is supported by the graph?

A) The bicycle has speed but not velocity.

B) The bicycle is moving at a constant velocity.

C) The bicycle has a displacement of 3 m.

D) The bicycle is not in motion.

Solution:

Velocity is the time rate of change of displacement. It is the ratio of displacement to time taken.

Speed is the time rate of change of distance. It is the ratio of distance to time taken.

From the position-time graph, we can see that the bicycle has a constant positon of 3 m for the whole of the time. That is the position remains 3 m even as the time changes. Therefore, we can conclude that the bicycle is not in motion.

From the position-time data provided, it can concluded that the bicycle is not in motion.

Motion of a body involves a change in the position of that body with time.

A body in motion is constantly changing position or orientation as time passes.

The body may move with constant velocity/speed or changes in its velocity.

A position-time graph provides information about the motion of a body.

From the data provided:

The position of the bicycle remains the same for all time intervals.

Therefore, from the position-time data provided, it can concluded that the bicycle is not in motion.

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Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a)  at 4.75 cm

b)  at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula [tex]E = \dfrac{kq }{d}[/tex]

[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}[/tex]

E = 1461.95 N/C

c) The electric field E is calculated as:

[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}[/tex]

E = 239.76 N/C

Answer: 31.89seconds

Explanation:

Based on the information given, we are meant to calculate deceleration which will be:

t = V/a

where, a = mg

Therefore, t = V/mg

t = 25/0.08 × 9.8

t = 25/0.784

t = 31.89seconds

Therefore, the time that it will take before the hockey puck slides to a stop is 31.89seconds.

Answer:

the person  is sitting 1.5 m from the left end of the board

Explanation:

Given the data in the question;

Wb = 125 N

Wm = 500 N

T₂ = 250 N

Now, we know that;

T₁ + T₂ = Wb + Wm

T₁ + 250 = 125 + 500

T₁ = 125 + 500 - 250

T₁ = 375 N

so tension of the left chain is 375 N.

Now, taking torque about the left end

500 × d + 125 × 2 = 250 × 4

500d + 250 = 1000

500d = 1000 - 250

500d = 750

d = 750 / 500

d = 1.5 m

Therefore, the person  is sitting 1.5 m from the left end of the board.