The acceleration due to gravity on the Moon's surface is
one-sixth that on Earth. An astronaut's life support backpack
weighs300lb on Earth. What does it weigh on the moon?

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:

11.306 nm/s

or

113.06 atomic layers/sec

Explanation:

Hello!

First we need to know how much an inch equals in nanometers and a day in seconds:

Since 1inch = 2.54cm and 1cm=10^7nm

     1 inch = 2.54 * 10^7 nm

Also 1day = 24hours = 24*60minutes = 24*60*60seconds

   1 day = 86.4 * 10^3 s

Therefore the rate at which the hair grows in nanometers per seconds is:

    1/26 in/day = (1/26) * (2.54*10^7)/(86.4*10^3) = 11.306 nm/s

Now, if 1 atomic layer = 0.1 nm this means that 1 nm = 10 atomic layers.

Therefore:

The rate in atomic layers is

11.306 nm/s = 11.306 (10 atomic layers)/s = 113.06 atomic layers/sec

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:

Time taken by motorboat to reach [tex]17.0m/s[/tex] equals 13 seconds.

Explanation:

From the first equation of kinematics we have

[tex]v=u+at[/tex]

where,

'v' is the final speed of the accelerating object

'u' is the initial speed of the object

'a' is the accleration of the object

't' is the time for which the object accelerates

Applying the given values in the equation above we get

[tex]17=4+1.0\times t\\\\\\\therefore t=17-4=13seconds[/tex]

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:

(a) v =  399.3 m/s, (b) v =  1742.4 m/s, (c) v =  399.3 m/s, (d) v = 1742.4 m/s

Explanation:

The velocity of a wave can be defined as:

[tex]v = \lambda f[/tex]   (1)

Where [tex]\lambda[/tex] and f are the wavelength and frequency of the sound wave.

The values for each case will be replaced in equation (1).

(a) f = 36.3 Hz, λ = 11.0 m

[tex]v = (11.0 m)(36.3 Hz)[/tex]

But 1 Hz = s⁻¹, therefore:

[tex]v = (11.0 m)(36.3 s^{-1})[/tex]

[tex]v = 399.3 m.s^{-1}[/tex]

[tex]v = 399.3 m/s[/tex]

So the sound wave has a velocity of 399.3 m/s.

(b) f = 363.0 Hz, λ = 4.80 m

[tex]v = (4.80 m)(363.0 Hz)[/tex]

[tex]v = (4.80 m)(363.0 s^{-1})[/tex]

[tex]v = 1742.4 m.s^{-1}[/tex]

[tex]v = 1742.4 m/s[/tex]

So the sound wave has a velocity of 1742.4 m/s.

(c) f = 3,630.0 Hz, λ = 11.0 cm

Before using equation (1) it is necessary to express [tex]\lambda[/tex] in meters.

[tex]11.0 cm . \frac{1 m}{100 cm}[/tex] ⇒ [tex]0.11 m[/tex]

[tex]v = (0.11 m)(3630.0 Hz)[/tex]

[tex]v = (0.11 m)(3630.0 s^{-1})[/tex]

[tex]v = 399.3 m.s^{-1}[/tex]

[tex]v = 399.3 m/s[/tex]

So the sound wave has a velocity of 399.3 m/s.

(d) f = 36,300.0 Hz, λ = 4.80 cm

[tex]4.80 cm . \frac{1 m}{100 cm}[/tex] ⇒ [tex]0.048 m[/tex]

[tex]v = (0.048 m)(36300.0 Hz)[/tex]

[tex]v = (0.048 m)(36300.0 s^{-1})[/tex]

[tex]v = 1742.4 m.s^{-1}[/tex]

[tex]v = 1742.4 m/s[/tex]

So the sound wave has a velocity of 1742.4 m/s.

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:

0.93 m/s

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 90 m

a = Acceleration = 9.81 m/s²

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 90=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{90\times 2}{9.81}}\\\Rightarrow t=4.3\ s[/tex]

So, the raft covered 6-2 = 4 m in 4.3 seconds

Speed = Distance / Time

[tex]\text{Speed}=\frac{4}{4.3}=0.93\ m/s[/tex]

Speed of the raft is 0.93 m/s

To find the speed of the raft, we can use the principle of conservation of energy. When the woman drops the stone, it starts with potential energy due to its height and then converts to kinetic energy as it falls.

To find the speed of the raft, we can use the principle of conservation of energy. When the woman drops the stone, it starts with potential energy due to its height and then converts to kinetic energy as it falls. The kinetic energy of the stone when it hits the water is equal to the potential energy it had initially. We can use the equation:

mgh = 0.5mv^2

Where m is the mass of the stone, g is the acceleration due to gravity, h is the height of the bridge, and v is the speed of the stone when it hits the water. Rearranging the equation, we can solve for v:

v = √(2gh)

Substituting the given values h = 90.0 m and g = 9.8 m/s^2, we can calculate the speed of the stone when it hits the water. This speed is equal to the speed of the raft.

Answer:

(C) 3.78 seconds

Explanation:

At the highest point, the velocity is equal to 0m/s

[tex]v_{f}=v_{o}-gt[/tex]

[tex]t=\frac{v_{o}}{g}[/tex]  ; t is the time to reach the highest point

The  time the ball takes to return to its starting point after the ball  reach its maximum height is the same:

[tex]T_{descent}=t=\frac{v_{o}}{g}=\frac{37}{9.81}=3.78s[/tex]

Explanation:

An electron is released from rest, u = 0

We know that charge per unit area is called the surface charge density i.e. [tex]\sigma=\dfrac{q}{A}=2.1\times 10^{-7}\ C/m^2[/tex]

Distance between the plates, [tex]d=1.2\times 10^{-2}\ m[/tex]

Let E is the electric field,

[tex]E=\dfrac{\sigma}{\epsilon_o}[/tex]

[tex]E=\dfrac{2.1\times 10^{-7}}{8.85\times 10^{-12}}[/tex]

E = 23728.81 N/C

Now, [tex]ma=qE[/tex]

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

[tex]a=\dfrac{1.6\times 10^{-19}\times 23728.81}{9.1\times 10^{-31}}[/tex]

[tex]a=4.17\times 10^{15}\ m/s^2[/tex]

Let v is the speed of the electron just before it reaches the positive plate. So, third equation of motion becomes :

[tex]v^2=2ad[/tex]

[tex]v^2=2\times 4.17\times 10^{15}\times 1.2\times 10^{-2}[/tex]

[tex]v=10.003\times 10^6\ m/s[/tex]

Hence, this is the required solution.

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;

https://brainly.com/question/506926

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]

Explanation:

We know that if

e = 1  then collision is called perfectly elastic collision.

0<e <1  then collision is called partial elastic collision.

e =0  then collision is called inelastic collision.

So when two automobile collide then they usually do not stick together then this collision is called as elastic collision.

When object collide head to head it become more dangerous than other type of collision because when object come toward each other and due to suddenly velocity of object become zero due to this it produce large amount of force.Usually this force produce two time more as compare to when object moving in same direction.

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:

The maximum height of the ball is 20 m. The ball needs 2 s to reach that height.

Explanation:

The equation that describes the height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration

v = velocity at time t

When the ball is at its maximum height, its velocity is 0, then, using the equation of the velocity, we can calculate the time at which the ball is at its max-height.

v = v0 + g · t

0 = 20 m/s - 9.8 m/s² · t

-20 m/s / -9.8 m/s² = t

t = 2.0 s

Then, the ball reaches its maximum height in 2 s.

Now,  we can calculate the max-height obtaining the position at time t = 2.0 s:

y = y0 + v0 · t + 1/2 · g · t²

y = 0 m + 20 m/s · 2 s - 1/2 · 9,8 m/s² · (2 s)²

y = 20 m

The maximum height reached by the ball is 20.4 meters, and it takes approximately 2.04 seconds to reach this height.

When a ball is thrown vertically into the air with an initial velocity of 20 m/s, we can calculate the maximum height using the kinematic equation:

[tex]v^2 = u^2 + 2gh,[/tex]

where v is the final velocity (0 m/s at the highest point), u is the initial velocity (20 m/s), g is the acceleration due to gravity (9.81 m/s2), and h is the maximum height. Solving for h gives us:

[tex]h = u^2 / (2g).[/tex]

By substituting the values we get:

[tex]h = (20 m/s)^2 / (2 * 9.81 m/s^2) = 20.4 m.[/tex]

To find the time needed to reach the maximum height, we use the equation:

v = u + gt,

Solving for t when v is 0 m/s, we get:

t = u / g = 20 m/s / 9.81 m/s2 = approx. 2.04 seconds.

Thus, the maximum height of the ball is 20.4 meters and the time needed to reach the maximum height is approximately 2.04 seconds.

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:

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.

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.

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:

(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.

Answer:d- Away from it

Explanation:

For a positive point charge, the electric field vectors point away from the charge. Electric field line radiates out of positive charge and could terminate to a  negative charge if it is placed in its vicinity.

Similarly for negative charge electric field lines seems to come inside of negative charge. It is basically opposite of positive charge.

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:

6.86 cm

Explanation:

Given:

Assume:

When the first bead rests on the table, then electrostatic force due to the second bead acts on it in the upward direction, Normal force acts in the upward direction and its weight in the downward direction.

So, using Newton's second law on the first bead resting on the table, we have

[tex]F+N-W=0\\[/tex]

At the closest distance of the second bead to the first bead, it just lifts off the table and the normal force becomes zero.

[tex]\therefore F-W=0\\\Rightarrow F=W\\\Rightarrow \dfrac{kqQ}{r^2}=mg\\\Rightarrow r^2=\dfrac{kqQ}{mg}\\\Rightarrow r^2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}\times 5.3\times 10^{-9}}{3\times 10^{-6}\times 9.8}\\\Rightarrow r^2=4.70\times 10^{-3}\\\textrm{Taking square root on both the sides}\\r = \pm 0.0686\ m\\\textrm{Since the distance is never negative}\\\therefore r = 0.0686\ m\\\Rightarrow r = 6.86\ cm[/tex]

Hence, the centers of the two beads must be brought closest to 6.86 cm before the lower bead is lifted off the table.

Answer:

a) 30.84m/s

b) 348.32Hz

c) 32.34m/s

d) 289.69Hz

Explanation:

a) If 1 mile=1609,34m, and 1 hour=3600 seconds, then 69mph=69*1609.34m/3600s=30.84m/s

b) Based on Doppler effect:

/*I will take as positive direction the vector [tex]\vec r_{observer}-\vec r_{emiter}[/tex] */

[tex]f_{observed}=(\frac{v_{sound}-v_{observed}}{v_{sound}-v_{emited}})f_{emited}[/tex]

[tex]f_{observed}=(\frac{343m/s-0m/s}{343m/s-30.84m/s})317Hz=348.32Hz[/tex]

c) [tex]350Hz=(\frac{343m/s-0m/s}{343m/s-v_{ambulance}})317Hz, V_{ambulance}=343m/s-\frac{317Hz}{350Hz}.343m/s=32.34m/s[/tex]

d) [tex]f_{observed}=(\frac{343m/s-0m/s}{343m/s+32.34m/s})317Hz=289.69Hz[/tex]

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:

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:decreases by 30.55%

Explanation:

Given

length of wire is increased by 20 % keeping volume constant

Let the length of wire be L and its area of cross section be A

Thus new length=1.2 L

Volume is constant

[tex]AL=1.2 L\times A'[/tex]

A'=0.833 A

and resistance is given by

[tex]R=\frac{\rho L}{A}[/tex]

where [tex]\rho [/tex]=resistivity

New resistance [tex]R'=\frac{\rho\times 1.2L}{0.833A}[/tex]

R'=1.44 R

heat produced for same potential

[tex]H_1=\frac{V^2t}{R}[/tex]

[tex]H_2=\frac{V^2t}{1.44R}=0.694H_1[/tex]

% change in heat

[tex]\frac{H_2-H_1}{H_1}\times 100[/tex]

[tex]=\frac{0.694-1}{1}[/tex]

=30.55 decreases

Answer:

30.55 %

Explanation:

Assumptions:

According to the question, we have

[tex]L = l + 20\ \% l = \dfrac{120l}{100}\\V=v\\\Rightarrow LA=la\\\Rightarrow A= \dfrac{la}{L}\\\Rightarrow A= \dfrac{la}{\dfrac{120l}{100}}\\\Rightarrow A= \dfrac{100a}{120}[/tex]

Using the formula of resistance of a wire in terms of its length, cross sectional area and the resistivity of the material, we have

[tex]r =  \dfrac{\rho l}{a}\\R=\dfrac{\rho L}{A}=\dfrac{\rho\times \dfrac{120l}{100} }{\dfrac{100a}{120}}=(\dfrac{120}{100})^2\dfrac{\rho l}{a}= 1.44r\\[/tex]

Using the formula of heat generated by the wire for potential diofference created across its end for time t, we have

[tex]h = \dfrac{P^2}{r}t\\H = \dfrac{P^2}{R}t= \dfrac{P^2}{1.44r}t\\\therefore \Delta H = h-H\\\Rightarrow \Delta H = \dfrac{P^2}{r}t-\dfrac{P^2}{1.44r}t\\\Rightarrow \Delta H = \dfrac{P^2t}{r}(-\dfrac{1}{1.44})\\\Rightarrow \Delta H = \dfrac{P^2t}{r}(\dfrac{0.44}{1.44})\\\therefore \textrm{Percentage change in the heat produced}= \dfrac{\Delta H}{h}\times 100\ \%= \left (\dfrac{\dfrac{P^2t}{r}(\dfrac{0.44}{1.44})}{\dfrac{P^2}{r}t}  \right )\times 100\ \% = 30.55\ \%[/tex]

Hence, the percentage change in the heat produced in the wire is 30.55 %.

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration ([tex]g=9.8 m/s^2[/tex]) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

[tex]u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s[/tex]

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

[tex]v=18.9 m/s[/tex]

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

[tex]v_y^2 - u_y^2 = 2ah[/tex] (1)

where

[tex]u_y = 16.4 m/s[/tex] is the initial vertical velocity

[tex]v_y[/tex] is the vertical velocity at height h

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

At the top of the parabolic path, [tex]v_y = 0[/tex], so we can use the equation to find the maximum height

[tex]h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m[/tex]

So, at half of the maximum height,

[tex]h = \frac{13.7}{2}=6.9 m[/tex]

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

[tex]v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s[/tex]

And so, the speed of the stone at half of the maximum height is

[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s[/tex]

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

[tex]v_1 = 18.9 m/s[/tex]

while the speed at half of the maximum height (part b) is

[tex]v_2 = 22.2 m/s[/tex]

So the difference is

[tex]\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s[/tex]

And so, in percentage,

[tex]\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%[/tex]

So, the stone in part (b) is moving 17.4% faster than in part (a).

Answer:

The correct option is 'D': 9

Explanation:

We know that the magnitude of the centripetal acceleration of  a body moving in circular orbit of radius 'r' with speed 'v' is given by

[tex]a_{c}=\frac{v^{2}}{r}[/tex]

Now when the speed of the body is tripled the speed becomes [tex]3v[/tex]

Hence the new centripetal acceleration is obtained as

[tex]a'_{c}=\frac{(3v)^{2}}{r}\\\\a'_{c}=\frac{9v^{2}}{r}=9a_{c}[/tex]

Thus we can see that the new centripetal acceleration becomes 9 times the oroginal value.

The acceleration of a body traveling in a circular route is known as centripetal acceleration. The magnitude of the centripetal acceleration increases by a factor of 9.

The acceleration of a body traveling in a circular route is known as centripetal acceleration. Because velocity is a vector quantity. It has both a magnitude and a direction.

When a body moves on a circular route, its direction changes constantly, causing its velocity to vary, resulting in acceleration.

Mathematically it is given as,

[tex]\rma_c=\frac{v^2}{r} \\\\ a_c'=\frac{(3v)^2}{r} \\\\ \rm v=9\frac{v^2}{r}\\\\ a_c'=9a_c[/tex]

Hence the magnitude of the centripetal acceleration increases by a factor of 9. Option c is correct.

To learn more about centripetal acceleration refer to the link;

https://brainly.com/question/17689540