A research team developed a robot named Ellie. Ellie ran 1,000 meters for 200 seconds from the research building, rested for 100 seconds, and walked back to the research building for 1000 seconds. To find out Ellie’s average velocity for each case while running, resting, and walking, begin by plotting a graph between position and time. 1.List the velocity from greatest to least among running, resting, and walking
2.List the speed from greatest to least among running, resting, and walking.

Answer:

20.88 mN

Explanation:

given data:

mass of object = 7.2 gm

acceleration of object = 2.9 m/s2

we know that force is given as

F = ma

where m is mass of object and a is acceleration of moving object.

putting all value to get required force

    = 7.2*10^{-3}\ kg *2.9 m/s2

   = 20.88*10^{-3} N

force in milli newton is

  = 20.88*10^{-3} * 1000 = 20.88 mN

Answer:

[tex]E = \frac{(f^2c^2 - v_o^2)M}{2QD}[/tex]

Part c)

[tex]E = \frac{2.07 \times 10^5}{D}[/tex]

Explanation:

Part a)

As per Coulomb's law we know that force on a charge placed in electrostatic field is given as

[tex]F = QE[/tex]

now acceleration of charge is given as

[tex]a = \frac{QE}{M}[/tex]

now if charge moved through the distance D in electric field and its speed changes from vo to fraction f of speed of light c

then we will have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex](fc)^2 - v_o^2 = 2(\frac{QE}{M})D[/tex]

so we have

[tex]E = \frac{(f^2c^2 - v_o^2)M}{2QD}[/tex]

Part b)

Now using work energy theorem we can say that total work done by electric force on moving charge will convert into kinetic energy

So we will have

[tex]QED = \frac{1}{2}M(cf)^2 - \frac{1}{2}Mv_o^2[/tex]

so we have

[tex]E = \frac{M(c^2f^2 - v_o^2)}{2QD}[/tex]

Part c)

Now if an electron is accelerated using this field

then we have

[tex]M = 9.11 \times 10^{-31} kg[/tex]

[tex]Q = 1.6 \times 10^{-19} C[/tex]

[tex]c = 3\times 10^8 m/s[/tex]

so we have

[tex]E = \frac{(9.1 \times 10^{-31})(0.9^2 - 0.001^2)\times 9 \times 10^{16}}{2(1.6 \times 10^{-19})D}[/tex]

[tex]E = \frac{2.07 \times 10^5}{D}[/tex]

Answer:14.72 m/s

Explanation:

Given

Initial velocity (u)=16.6 m/s

[tex]\theta =40.9^{\circ}[/tex]

Horizontal velocity component ([tex]u_x[/tex])=16.6cos40.9=12.54 m/s

As the ball comes down so its vertical displacement is zero except 3 m elevation

Thus [tex]v_y=\sqrt{\left ( 16.6sin40.9\right )^2+2\left ( -9.81\right )\left ( 3\right )}[/tex]

[tex]v_y=\sqrt{10.868^2-58.86}[/tex]

[tex]v_y=\sqrt{59.253}[/tex]

[tex]v_y=7.69 m/s[/tex]

there will be no change is horizontal velocity as there is no acceleration

Therefore Final Velocity

[tex]v=\sqrt{u_x^2+v_y^2}[/tex]

[tex]v=\sqrt{12.54^2+7.69^2}[/tex]

v=14.72 m/s

Answer:

It means at a diverging boundary when magma spreads to both sides it is almost identical in both the side.

Explanation:

At a diverging boundary when magma spreads to both sides it is almost identical in both the side.

If you take a picture of one side of the boundry and the spreading of sea floor and place it before a mirror you can see the image is identical to the picture of other side.

Therefore, the meaning of saying, each side of the sea floor away from the mid ocean ridge is a mirror image of the other side is explained above.

The statement that each side of the seafloor away from the mid ocean ridge is a mirror image of the other side refers to the symmetric nature of seafloor spreading. Molten material rises from the Earth's mantle at the mid-ocean ridge, cools, and forms new oceanic crust, creating symmetric magnetic striping patterns. This is akin to the bilaterally symmetric structure observed in certain organisms.

When we say that each side of the seafloor away from the mid ocean ridge is a mirror image of the other side, we are referencing the symmetric nature of seafloor spreading. Just as a mirror image reflects an object exactly, so does the seafloor on one side of the mid-ocean ridge reflect the seafloor on the opposite side. This is because molten material rises from the Earth's mantle at the mid-ocean ridge, cools, and forms new oceanic crust. This new crust then moves away from the ridge due to tectonic forces, and the process repeats, creating a pattern of symmetrical magnetic striping on the seafloor.

This 'mirroring' effect is similar to the bilaterally symmetric structure seen in certain organisms, where a plane cut from the front to back of the organism produces distinct left and right sides that are mirror images of each other. You can see this symmetry in the images of the Moon provided – two different sides, yet mirroring similar physical traits.

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Answer: 6.45 s

Explanation:

We have the following equation:

[tex]y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}[/tex] (1)

Where:

[tex]y=0[/tex] is the height when the rock hits the ground

[tex]y_{o}=75 m[/tex] the height at the edge of the cilff

[tex]V_{o}=20 m/s[/tex] the initial velocity

[tex]g=9.8 m/s^{2}[/tex] acceleration due gravity

[tex]t[/tex] time

[tex]0=75 m+(20 m/s)t-(4.9 m/s^{2})t^{2}[/tex]  (2)

Rearranging the equation:

[tex]-(4.9 m/s^{2})t^{2} + (20 m/s)t + 75 m=0[/tex] (3)

At this point we have a quadratic equation of the form [tex]at^{2}+bt+c=0[/tex], and we have to use the quadratic formula if we want to find  [tex]t[/tex]:

[tex]t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]  (4)

Where [tex]a=-4.9[/tex], [tex]b=20[/tex], [tex]c=75[/tex]

Substituting the known values and choosing the positive result of the equation:

[tex]t=\frac{-20\pm\sqrt{20^{2}-4(-4.9)(75)}}{2(-4.9)}[/tex]  (5)

[tex]t=6.453 s[/tex]  This is the time it takes to the rock to hit the ground

Answer : The molar mass of unknown compound is 128.22 g/mole

Explanation :

Mass of unknown compound = 9.72 g

Mass of solvent = 50.0 g

Formula used :

[tex]\Delta T_b=i\times K_b\times m\\\\T_2-T_1=i\times K_b\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of solvent in Kg}}[/tex]

where,

[tex]\Delta T_b[/tex] = elevation in boiling point

[tex]T_1[/tex] = temperature of solvent = [tex]80.7^oC=273+80.7=353.7K[/tex]

[tex]T_2[/tex] = temperature of solution = [tex]84.93^oC=273+84.93=357.93K[/tex]

i = Van't Hoff factor = 1 (for non-electrolyte)

[tex]K_f[/tex] = boiling point constant for solvent = 2.79 K/m

m = molality

Now put all the given values in this formula, we get:

[tex](357.93-353.7)K=1\times (2.79K/m)\times \frac{9.72g\times 1000}{\text{Molar mass of unknown compound}\times 50.0g}[/tex]

[tex]\text{Molar mass of unknown compound}=128.22g/mole[/tex]

Therefore, the molar mass of unknown compound is 128.22 g/mole

Answer:

a) 0.063 m/s

b)  243.83 m/s

Explanation:

given,

distance = 7900 km

time = 4 years

a) average velocity for the trip = [tex]\dfrac{7900}{4\times 365}[/tex]

                                                   = 5.41 km/day

                                                   = [tex]\dfrac{5.41\times 1000}{24\times 60\times 60}[/tex]

        average velocity for the trip = 0.063 m/s

b) average velocity  = [tex]\dfrac{7900\times 1000}{9\times 60\times 60}[/tex]

average velocity for the trip = 243.83 m/s

Answer:

if Y is the position and X the time: in the first one you will see a crescent function that starts sharp and starts to curve down as the time pases. as the cart is slowing down, you will need more time to move the same as before.

Y (position)

I

sensor-------------------------------------------------------------------

I                                                    o

I                                     o

I                           o

I                   o

I            o

I       o

I   o

I------------------------------------------------------------------------------------- X (time)

in the second case the cart starts close to the sensor and starts getting sharper and sharper as the time pases. This is because the velocity is increasing, so for each second that pases, you will travel more distance that the second before it.

Y (position)

I

sensor ----------------------

I       o

I                 o

I                          o

I                                 o

I                                       o

I                                            o

I                                              o

I------------------------------------------------------------------------------------- X (time)

i hope you can understand it, kinda hard to do graphs here.

Answer:

H = 171.90 m

Explanation:

given data

distance = 53.2 m

height = H

to find out

height H

solution

we know height is here H = [tex]\frac{1}{2} gt^2[/tex]    ......................1

here t is time and a is acceleration

so

we find t first

we know during time (t -1) s , it fall distance (H - 53.2) m

so equation of distance

[tex]H - 53.2 = \frac{1}{2} g (t-1)^2[/tex]

[tex]H - 53.2 = \frac{1}{2} g (t^2-2t+1)[/tex]

[tex]H - 53.2 = \frac{1}{2} gt^2-gt+\frac{1}{2} g[/tex]     ................2

now subtract equation 2 from equation 1 so we get

[tex]H - (H - 53.2) =\frac{1}{2} gt^2- (\frac{1}{2} gt^2-gt+\frac{1}{2} g)[/tex]

53.2 = gt - [tex]\frac{1}{2} g [/tex]

53.2 = 9.81 t - [tex]\frac{1}{2} 9.8 [/tex]

t = 5.92 s

so from equation 1

H = [tex]\frac{1}{2} (9.81)5.92^2[/tex]

H = 171.90 m

Answer:

Sound Intensity at microphone's position is [tex]9.417\times 10^{- 4} W/m^{2}[/tex]

The amount of energy impinging on the microphone is [tex]9.417\times 10^{- 8} W/m^{2}[/tex]

Solution:

As per the question:

Emitted Sound Power, [tex]P_{E} = 32.0 W[/tex]

Area of the microphone, [tex]A_{m} = 1.00 cm^{2} = 1.00\times 10^{- 4} m^{2}[/tex]

Distance of microphone from the speaker, d = 52.0 m

Now, the intensity of sound, [tex]I_{s}[/tex] at a distance away from the souce of sound follows law of inverse square and is given as:

[tex]I_{s} = \frac{P_{E}}{Area} = \frac{P_{E}}{4\pi d^{2}}[/tex]

[tex]I_{s} = \frac{32.0}{4\pi (52.0)^{2}} = 9.417\times 10^{- 4} W/m^{2}[/tex]

Now, the amount of sound energy impinging on the microphone is calculated as:

If [tex]I_{s}[/tex] be the Incident Energy/[tex]m^{2}/s[/tex]

Then

The amount of energy incident per 1.00 [tex]cm^{2} = 1.00\times 10^{- 4} m^{2}[/tex] is:

[tex]I_{s}(1.00\times 10^{- 4}) = 9.417\times 10^{- 4}\times 1.00\times 10^{- 4} = 9.417\times 10^{- 8} J[/tex]

Final answer:

The sound intensity at the position of the microphone is calculated using the formula Intensity = Power/Area. The amount of sound energy impinging on the microphone each second is found using the formula Energy = Power * Time.

Explanation:

To find the sound intensity at the position of the microphone, we can use the formula:

Intensity = Power/Area

Given that the sound power is 32.0W and the microphone has an area of 1.00cm^2 (converted to m^2 by dividing by 10000), we can calculate:

Intensity = 32.0W / (1.00cm^2 / 10000)

Next, to find the amount of sound energy impinging on the microphone each second, we can use the formula:

Energy = Power * Time

Since the time is 1 second, we have:

Energy = 32.0W * 1s

Therefore, the sound intensity at the position of the microphone is the calculated value, and the amount of sound energy impinging on the microphone each second is 32.0 joules.

Answer:

4 %

2 ) 3.42 %

Explanation:

Sensitivity requirement of 4 milligram means it is not sensitive below 4 milligram or can not measure below 4 milligram .

Given , 4 milligram is the maximum error possible .

Measured weight = 100 milligram

So percentage maximum potential error

= (4 / 100)  x 100

4 %

2 )

As per measurement

weight of 6 milliliters of water

= 48.540 - 42.745 = 5.795 gram

6 milliliters of water should measure 6 grams

Deviation = 6 - 5.795 = - 0.205 gram.

Percentage of error =(.205 / 6 )x 100

= 3.42 %

Answer:

t=89.44 sec

Explanation:

Given that

d= 100 m

Mass of electron

[tex]m=9.11\times 10^{-31}kg[/tex]

Force acting on electron

[tex]F=\dfrac{Kq^2}{d^2}[/tex]

Now by putting the values of K and charge on electron

[tex]F=\dfrac{Kq^2}{d^2}[/tex]

[tex]F=\dfrac{9\times 10^9(1.6\times 10^{-19})^2}{100^2}[/tex]

[tex]F=2.3\times 10^{-32}[/tex]

As we know that

F= m a

So acceleration of electron

a=F/m

[tex]a=\dfrac{2.3\times 10^{-32}}{9.11\times 10^{-31}}\ m/s^2[/tex]

[tex]a=0.025\ m/s^2[/tex]

We know that

[tex]S=ut+\dfrac{1}{2}at^2[/tex]

here electron move from rest so u= 0

[tex]100=\dfrac{1}{2}\times 0.025^2\times t^2[/tex]

t=89.44 sec

So time taken by electron is 89.44 sec.

Force on electron, released from rest at a distance from conducting plane indirectly proportional to this distance coulombs law. Time required for electron to strike the plane is 89.44 seconds.

Coulombs law states the the force of attraction or repulsion between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of distance between them.

It can be given as,

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

Here, [tex]k[/tex] is coulombs constant.

Given information-

The electron is released from rest at a distance 100 m from an infinite conducting plane.

The acceleration of a object is the ratio of force applied on it to the mass of the object. Thus the acceleration of the electron is,

[tex]a=\dfrac{F}{m}[/tex]

Force is the ratio of charge on particles and the square of distance between them, multiplied by coulomb's constant. Thus,

[tex]a=\dfrac{k\dfrac{q_1q_2}{r^2}}{m}\\a=\dfrac{k{q_1q_2}}{m{r^2}}[/tex]

As the mass of the electron is [tex]9.11\times10^{-31}kg[/tex] and the charge on a electron is [tex]1.6\times10^{-19} C[/tex]. Thus put the value in above expression as,

[tex]a=\dfrac{(9\times10^9)(1.6\times10^{-19})(1.6\times10^{-19})}{9.11\times10^{-31}\times100^2}\\a=0.025\rm m/s^2[/tex]

As the value of acceleration is 0.025 meter per second squared and initial velocity is zero. Thus by the distance formula of equation of motion,

[tex]100=0+\dfrac{1}{2}\times0.025\times t^2\\t=89.44 \rm sec[/tex]

Therefore, the time required for the electron to strike the plane is 89.44 seconds.

Learn more about the coulombs law here;

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Answer:

f = 40 cm

image formed at a distance of 40 cm from lens is magnified and virtual.

when lens is turned around focal length is f = 40 cm

Explanation:

given data:

[tex]R_1 = 20 cm[/tex]

[tex]R_2 = \infty[/tex]

Refraction index of lens = 1.5

focal length of lens is given as

[tex]\frac{1}{f} = (n-1) (\frac{1}{R_1} - \frac{1}{R_1})[/tex]

Putting all value to get focal length value

[tex]\frac{1}{f} = (1.5 -1) \frac{1}{20}[/tex]

[tex]\frac{1}{f} =\frac{0.5}{20}[/tex]

f = 40 cm

image formed at a distance of 40 cm from lens is magnified and virtual.

when lens is turned around

[tex]R_1 =  \infty [/tex]

[tex]R_2 = -20cm[/tex]

Refraction index of lens = 1.5

focal length of lens is given as

[tex]\frac{1}{f} = (n-1) (\frac{1}{R_1} - \frac{1}{R_1})[/tex]

Putting all value to get focal length value

[tex]\frac{1}{f} = (1.5 -1) \frac{1}{20}[/tex]

[tex]\frac{1}{f} =\frac{0.5}{20}[/tex]

f = 40 cm

there is no change can be seen between two condition. image will form at 40 cm from lens

Answer:

[tex]4.50*10^8\frac{N}{C}[/tex]

Explanation:

The electric field is generated by a charge and represents the force exerted on a test charge, that is, the electric force per unit of charge. Therefore the equation for the electric field can be obtained from Coulomb's law.

[tex]E=\frac{F}{q}\\E=\frac{kq}{r^2}\\E=\frac{5C*8.99*10^9\frac{Nm^2}{C^2}}{(10 m)^2}=4.50*10^8\frac{N}{C}[/tex]

The magnitude of the electric field created by a larger charge Q at the location of a test charge is calculated by the formula E = k*Q/r² where k is Coulomb's constant, Q is the charge creating the field, and r is the distance between the charges. Substituting the given values into the formula, we find the electric field magnitude to be 4.495 x 10^8 N/C.

The magnitude of the electric field created by a point charge Q can be calculated using Coulomb's law as follows:

The equation for calculating the electric field E in relation to the force F imparted on a small test charge q is defined as E = F/q.

However, the Coulomb’s law gives the force F between two charges as F = k*Q*q/r², where, k is the Coulomb's constant (8.99 × 10^9 N.m²/C²), Q is the charge creating the field, q is the test charge, and r is the distance between them.

In the case where we want to find the electric field created by a larger charge Q at the location of a small charge q, we consider the force on the small charge exerted by Q, and therefore, we rewrite the equation as E = k*Q/r².

Therefore for your question, the magnitude of the electric field (E) at the location of the test charge is calculated as E = (8.99 x 10^9 N.m²/C² * 5.0 C) / (10m)² = 4.495 x 10^8 N/C.

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Answer: A) we can write a wave packet like Y = ∑aₙ*cos(w*n*t)

So you have lots of different wavelengths here, and you think that a wave packet has a unified frequency, but not wavelength.

B) mathematically speaking you will write a wave packet like i did up there, has a sum of many waves, where for each one you can have an intensity aₙ and a different frequency, the only thing you must see is that all the waves you are suming are moving in the same direction.

Answer:

(a): [tex]F_e = 8.202\times 10^{-8}\ \rm N.[/tex]

(b): [tex]F_g = 3.6125\times 10^{-47}\ \rm N.[/tex]

(c): [tex]\dfrac{F_e}{F_g}=2.27\times 10^{39}.[/tex]

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053[tex]\times 10^{-9}[/tex] m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges [tex]q_1[/tex] and [tex]q_2[/tex] respectively is given by

[tex]F_e = \dfrac{k|q_1||q_2|}{r^2}[/tex]

where,

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, [tex]q_1 = +1.6\times 10^{-19}\ C.[/tex]

The charge on the electron, [tex]q_2 = -1.6\times 10^{-19}\ C.[/tex]

These two are separated by the distance, [tex]r = 0.053\times 10^{-9}\ m.[/tex]

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

[tex]F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.[/tex]

Part (b):

The gravitational force of attraction between two objects of masses [tex]m_1[/tex] and [tex]m_1[/tex] respectively is given by

[tex]F_g = \dfrac{Gm_1m_2}{r^2}.[/tex]

where,

For the given system,

The mass of proton, [tex]m_1 = 1.67\times 10^{-27}\ kg.[/tex]

The mass of the electron, [tex]m_2 = 9.11\times 10^{-31}\ kg.[/tex]

Distance between the two, [tex]r = 0.053\times 10^{-9}\ m.[/tex]

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

[tex]F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.[/tex]

The ratio [tex]\dfrac{F_e}{F_g}[/tex]:

[tex]\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.[/tex]

Answer:

1556.906J/K=ΔS

Explanation:

Entropy is a thermodynamic property that measures the level of molecular disorder in a substance.

This property is already calculated for all pressure and temperature values.

Therefore, to solve this problem we must use thermodynamic tables for water and calculate the specific entropy in the two states, finally multiply by the mass to find the entropy change.

the entropy change is given by the following equation

ΔS=m(s2-s1)

where

ΔS=change in the entropy

s= especific entropy

m=mass=257g=0.257Kg

for the state 1: entropy for liquid water at 99°C

s1=1296J/kgK

for the state 2: entropy for steam  at 100°C

s2=7354J/kgK

solving

ΔS=0.257(7354-1296)=1556.906J/K

Answer:

T =3227 °C

Explanation:

Given data:

P1 = 3.0 atm

T1 = 27 degree celcius

P2 = 35 atm

from ideal gas equation

PV/T=const

[tex]\frac{P_1 V_1}{T_1} =\frac{P_2 V_2}{T_2}[/tex]

[tex]\frac{3*V}{(27+273)}=\frac{35*V*10}{T}[/tex]

solving for T WE GET

[/tex]0.01V = \frac{350V}{T}[/tex]

T=3500K

T=3500-273=3227 °C

T =3227 °C

Answer:

39.8 m

Explanation:

Let h be the required height and t be the time to reach the egg which was thrown later on. The egg thrown earlier took ( t +2.2) s , with initial velocity zero to cover distance of h.,

So

h = 1/2 g( t +2.2)²

For the egg thrown with velocity of 58 m/s

h = 58 t + 1/2 g t²

Equating these two equations

1/2 g( t +2.2)²  =  58 t + 1/2 g t²

1/2 g ( t² + 4.4 t + 4.84 ) = 58 t + 1/2 g t²

58 t = 21.56 t + 23.716

t = .65 s

h = 1/2 x 9.8 x (2.2+.65)²

h = 39.8 m

Answer:

a) The coin´s maximum height is 9.79 m above the ground.

b) The coin is 2.17 s in the air.

c) The speed is 13.82 m/s when the coin hits the ground

Explanation:

The equations for the position and velocity of the coin are the following:

y = y0 + v0 · t + 1/2 · g · t²

v =  v0 + g · t

Where

y = height at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity

v = velocity at time t

a) At its max-height, the velocity of the coin is 0. Using the equation of velocity, we can obtain the time at which the velocity is 0.

v =  v0 + g · t

0 = 7.4 m/s - 9.8 m/s² · t

- 7.4 m/s / - 9.8 m/s² = t

t = 0.76 s

Now calculating the height of the coin at t = 0.76 s, we will obtain the maximum height:

y = y0 + v0 · t + 1/2 · g · t²

y = 0 m + 7.4 m/s · 0.76 s - 1/2 · 9.8 m/s² · (0.76 s)²

y = 2.79 m

The coin´s maximum height above the ground is 7 m + 2.79 m = 9.79 m

b) After the coin reaches its maximum height, it falls to the ground. The initial position will be the max-height (2.8 m) and the final position is the ground (-7 m). The initial velocity, v0, will be 0, because the coin is at the max-height. Then, using the equation of position we can calculate the time the coin is falling:

y = y0 + v0 · t + 1/2 · g · t²

-7 m = 2.79 m - 1/2 · 9.8 m/s² · t²

-2 ·(-7 m - 2.79 m)/ 9.8 m/s² = t²

t = 1.41 s

The coin is (1.41 s + 0.76 s) 2.17 s in the air

c) Using the equation of velocity, we can calculate the speed at time 1.41 s, when the coin hits the ground.

v =  v0 + g · t

v = 0 m/s - 9.8 m/s² · (1.41 s)

v = -13.82 m/s

The speed is 13.82 m/s when the coin hits the ground.

Answer:

[tex]n_l = 1.97[/tex]

Explanation:

given data:

refractive index of lens 1.50

focal length in air is 30 cm

focal length in water is -188 cm

Focal length of lens is given as

[tex]\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ][/tex]

[tex]\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ][/tex]

[tex]\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ][/tex]

focal length of lens in liquid is

[tex]\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ][/tex]

                [tex]=\frac{n_{g} -n_{l}}{n_{l}}  [\frac{1}{(n_{g} - 1) f}[/tex]

rearrange fro[tex] n_l[/tex]

[tex]n_l = \frac{n_g f_l}{f_l+f(n_g-1)}[/tex]

[tex]n_l = \frac{1.50*(-188)}{-188 + 30(1.50 -1)}[/tex]

[tex]n_l = 1.97[/tex]

Final answer:

To calculate the refractive index of a liquid in which a convex lens acts as a diverging lens, the lensmaker's formula and given focal lengths are used. The formula is manipulated to solve for the refractive index of the liquid, providing an approximation of 0.2394 for the unknown index.

Explanation:

To find the refractive index of the liquid in which a convex lens becomes a diverging lens, we need to apply the lensmaker's formula, considering refractive indices of the lens material and the surrounding medium. The lens has a focal length of 30 cm in air, and its refractive index is 1.50. When the lens is immersed in the liquid, its focal length changes to -188 cm, indicating that it now diverges light rays.

We use the formula for the focal length of a lens in a medium:

1 / f = (n_lens / n_medium - 1) * (1 / R1 - 1 / R2)

which is derived from the lensmaker's formula, where f is the focal length, n_lens is the refractive index of the lens, n_medium is the refractive index of the surrounding medium, and R1 and R2 are the radii of curvature for the lens surfaces. For a thin lens, this can be simplified to:

f in medium = (n_medium / n_lens) * f in air.

Rearranging for n_medium, we get:

n_medium = (f in air / f in medium) * n_lens.

Plugging in the values, we calculate:

n_medium = (30 cm / -188 cm) * 1.50 = -0.2394 (approximately).

The negative sign indicates that we need to take the absolute value, so the refractive index of the liquid is approximately 0.2394.

Answer:

D. Equalize voltage

Explanation:

Of the following lead-acid  battery the battery with voltage value is Equalize voltage. EQUALIZING lead acid battery is process of de-sulphating the battery by carrying out a controlled overcharge. When battery plates acquire sulphate coating over time, their efficiency reduces, by overcharging this sulpahte coating is blown-off and the battery is rejuvenated.

Answer:

Explanation:

length L = 5±.1

percentage error

= 0.1/5 x 100 = 2%

breadth b = 3.8 ±.1

percentage error

= .1 / 3.8 x 100

= 2.6 %

Maximum reading of length = 5.1 m

maximum reading of breadth = 3.9m

maximum area possible = 5.1 x 3.9

= 19.9 m².

maximum possible error in the measurement of area in percentage terms

= 2+2.6 = 4.6 %

error in maximum area

19.9 x 4.6 /100

= .9 m²

Minimum possible measurement of area

4.9 x 3.7 = 18.1 m²

possible error

18.1 x 4.6 / 100

= .8 m²

Answer:

Explanation:

Distance = 50 m

1. To find the average speed, first start the stop watch as the sprinter starts running and then stop it when he reaches the finish line.

Now note the time taken by the sprinter to run for 50 m.

The average speed of the sprinter is defined as the ratio of total distance covered to the total time taken.

Average speed = total distance / total time

2. To find the instantaneous speed, check the seed of the sprinter as he is at the finish line.

Explanation:

Given that,

Separation between the plates, d = 3.1 mm = 0.0031 m

Capacitance of the capacitor, C = 23 pF

We need to find the area of the plate. The capacitance of a parallel plate capacitor is given by :

[tex]C=\dfrac{A\epsilon_0}{d}[/tex]

[tex]A=\dfrac{Cd}{\epsilon_0}[/tex]

[tex]A=\dfrac{23\times 10^{-12}\times 0.0031}{8.85\times 10^{-12}}[/tex]

[tex]A=0.0080\ m^2[/tex]

or

[tex]A=80\ cm^2[/tex]

So, the area of the plate is 80 square centimetres. Hence, this is the required solution.

Answer:

3.25 m

Explanation:

t = Time taken = 0.166 seconds

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s²

s = 1 because meter stick is 1 meter in length

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1-\frac{1}{2}\times 9.81\times 0.166^2}{0.166}\\\Rightarrow u=5.21\ m/s[/tex]

Here, the initial velocity of point B is calculated from the time which is given. This velocity will be the final velocity of the acorn which falling from point A.

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.21^2-0^2}{2\times 9.81}\\\Rightarrow s=1.38\ m[/tex]

The distance of the acorn from the ground is 1.87+1.38 = 3.25 m

Answer:

news will reach to the listener who are 42 km apart

Explanation:

given,

distance of the radios from the station = 42 Km

speed of the sound in the air = 343 m/s

distance of the people = 9.26 m

[tex]time =\dfrac{distance}{speed}[/tex]

time taken by the signal to reach to the radio

speed of the electromagnetic wave to reach to the people

speed of electromagnetic wave = 3 × 10⁸ m/s

[tex]t =\dfrac{42000}{3 \times 10^8}[/tex]

t = 1.4 μs

time taken to reach to the people

[tex]time =\dfrac{distance}{speed}[/tex]

[tex]t =\dfrac{9.26}{343}[/tex]

t = 27 ms

time taken by the station to radio is less.

hence, news will reach to the listener who are 42 km apart

Answer:

"Magnitude of a vector can be zero only if all components of a vector are zero."

Explanation:

"The magnitude of a vector can be smaller than length of one of its components."

Wrong, the magnitude of a vector is at least equal to the length of a component. This is because of the Pythagoras theorem. It can never be smaller.

"Magnitude of a vector is positive if it is directed in +x and negative if is is directed in -X direction."

False. Magnitude of a vector is always positive.

"Magnitude of a vector can be zero if only one of components is zero."

Wrong. For the magnitude of a vector to be zero, all components must be zero.

"If vector A has bigger component along x direction than vector B, it immediately means, the vector A has bigger magnitude than vector B."

Wrong. The magnitude of a vector depends on all components, not only the X component.

"Magnitude of a vector can be zero only if all components of a vector are zero."

True.

The correct statements are A and E.

A) True. The magnitude of a vector can indeed be smaller than the length of one of its components, especially when the vector has components in multiple directions.

B) False. The magnitude of a vector is always positive or zero, regardless of its direction. It is not negative if directed in the -x direction.

C) False. The magnitude of a vector can be zero only if all of its components are zero. Having only one component as zero does not guarantee a zero magnitude.

D) False. The magnitude of a vector depends on the vector's components in all directions, not just the component along the x direction. It cannot be immediately concluded that vector A has a bigger magnitude than vector B solely based on the x-component.

E) True. The magnitude of a vector can be zero only if all components of the vector are zero. This is a fundamental property of vectors; they have zero magnitude only if they have no components in any direction.

Complete question:

Which of the following statements about vectors is definitely true? (Section 3.3)

A) The magnitude of a vector can be smaller than the length of one of its components.

B) The magnitude of a vector is positive if it is directed in the +x direction and negative if it is directed in the -x direction.

C) The magnitude of a vector can be zero if only one of its components is zero.

D) If vector A has a bigger component along the x direction than vector B, it immediately means that vector A has a bigger magnitude than vector B.

E) The magnitude of a vector can be zero only if all components of the vector are zero.

Answer:

p = 727.273 kg/m3,  y = 7134.545 N/m^3, SG = 0.7273

Explanation:

Density is simply the amount of mass of a substance per unit of volume. It can be found by dividing the mass in kg by the volume im m^3:

[tex]p = \frac{m}{v}  = \frac{40kg}{55lt*\frac{1 m^3}{1000 lt}} = 727.273 kg/m^3[/tex]

Specific weight is the weight of the substance per unit of volume. The weight is the mass of the material times the gravity, and it represents the force that the earth exerts on an object. Another way of calculate this value, its multiplying the density of the fuel times the gravity. Then:

[tex]y =  p*g = 727.273 kg/m^3 * 9.81 m/s^2 = 7134.545 N/m^3[/tex]

Specific Gravity is the ratio of the density of the substance to the density of a reference substance. For liquids, the reference substance is water at 4°C, which has a density of about 1000 kg/m^3.

[tex]SG =\frac{ p_{fuel}}{p_{water}}  = \frac{727.273 kg/m^3}{1000 kg/m^3} = 0.7273[/tex]

Answer:

(a) 1852259 [tex]ft^3[/tex] (b) 489085.47 pound

Explanation:

We have given auditorium measures 35 m×30 m×5 m

We know that 1 meter = 3.28 feet

So the measure of auditorium = 35×3.28 feet ×30×3.28 feet× 5×3.28 feet

(a) So the volume of the auditorium [tex]=35\times 3.28\times 30\times 3.28\times 5\times 3.28=185259.648ft^3[/tex]

Density is given as [tex]d=1.20kg/m^3[/tex]  

(b) weight of air  = volume × density [tex]=185259.648\times 1.2=222311.577kg[/tex]

We know that 1 kg = 2.20 pound

So 222311.577 kg =222311.577×2.20=489085.47 pound

The volume of the room is 185,197 cubic feet and the weight of air in the room is 13,889 pounds.

To convert the volume of the auditorium from cubic meters to cubic feet, we can use the conversion factor 1 cubic meter = 35.3147 cubic feet. With dimensions of 35.0 m x 30.0 m x 5.0 m, the volume of the auditorium is 5250 cubic meters. Multiplying this by the conversion factor, we find that the volume of the room is approximately 185,197 cubic feet.

To calculate the weight of air in the room, we can multiply the volume of air by its density. The density of air is given as 1.20 kg/m³. Using the volume of the room in cubic meters (5250 m³), we can multiply it by the density to find the mass of air, which is 6300 kg. To convert this to pounds, we can multiply by the conversion factor 1 kg = 2.20462 pounds. The weight of air in the room is therefore approximately 13,889 pounds.