What voltage must be applied to an 6 nF capacitor to store 0.14 mC of charge? Give answer in terms of kV.

Answer:

Explanation:

The SI system of units is the system which is Standard International system. It is used internationally.

The SI units of the fundamental quantities are given below

Mass - Kilogram

Length - metre

Time - second

temperature - Kelvin

Amount of substance - mole

Electric current - Ampere

Luminous Intensity - Candela

So, The SI unit of work is Joule

SI unit of acceleration is m/s^2

SI unit of power is watt

SI unit of momentum is kg m /s

SI unit of force is newton

Thus, the last option is incorrect.

Answer with Explanation:

From work energy theorem we have

[tex]\Delta Energy=\int F.ds[/tex]

Now since it is given that the work done on the object by the force is positive hence we conclude that the term on the right hand of the above relation is positive hence [tex]\Delta Energy>0[/tex]

Part a)

Hence we conclude that the mechanical energy of the particle will increase.

Part b) Since the mechanical energy of the particle is the sum of it's kinetic and potential energies, we can write

[tex]\Delta K.E+\Delta P.E=\Delta E\\\\\therefore \Delta K.E+\Delta P.E>0[/tex]

Now since the sum of the 2 energies ( Kinetic and potential ) is positive we cannot be conclusive of the individual values since 2 cases may arise:

1) Kinetic energy increases while as potential energy decreases.

2)  Potential energy increases while as kinetic energy decreases.

Hence these two cases are possible and we cannot find using only the given information which case will hold

Hence no conclusion can be formed regarding the individual energies.

Answer:

n=2.053

Explanation:

We will use Snell's Law defined as:

[tex]n_{1}*Sin\theta_{1}=n_{2}*Sin\theta_{2}[/tex]

Where n values are indexes of refraction and [tex]\theta[/tex] values are the angles in each medium. For vacuum, the index of refraction in n=1. With this we have enough information to state:

[tex]1*Sin(39.9)=n_{2}*Sin(18.2)[/tex]

Solving for [tex]n_{2}[/tex] yields:

[tex]n_{2}=\frac{Sin(39.9)}{Sin(18.2)}=2.053[/tex]

Remember to use degrees for trigonometric functions instead of radians!

Answer:

Option a)

Explanation:

In the process of charging anything by the method of induction, a charged body is brought near to the body which is neutral or uncharged without any physical contact and the ground must be provided to the uncharged body.

The charge is induced and the nature of the induced charge is opposite to that of the charge present on the charged body.

So when a positively charged rod is used to charge an electroscope, the rod which is positive attracts the negative charge in the electroscope and the grounding of the electroscope ensures the removal of the positive charge and renders the electroscope negatively charged.

Final answer:

To charge an electroscope negatively by induction, you need a negatively charged rod and a ground. When a positively charged rod is brought near a neutral metal sphere, it polarizes the sphere and attracts electrons from the earth's ample supply. By breaking the ground connection and removing the positive rod, the sphere is left with an induced negative charge.

Explanation:

To charge an electroscope negatively by induction, you need a negatively charged rod and a ground. When a positively charged rod is brought near a neutral metal sphere (the electroscope), it polarizes the sphere. By connecting the sphere to a ground, electrons are attracted from the earth's ample supply, resulting in an induced negative charge on the sphere. The ground connection is then broken, and the positive rod is removed, leaving the sphere with the induced negative charge.

Final answer:

It will take approximately 1.61 seconds for the bicyclist to catch up to his friend after his friend passes him.

Explanation:

To determine the time it takes for the bicyclist to catch up to his friend, we can use the equation:

distance = initial velocity * time + 0.5 * acceleration * time^2

Since the friend is traveling at a constant speed of 3.1 m/s, the distance traveled by the bicyclist during the 2-second delay is 6.2 m. Using the equation above:

6.2 m = 0 m/s * t + 0.5 * 2.4 m/s^2 * t^2

Simplifying the equation:

2.4 m/s^2 * t^2 = 6.2 m

t^2 = 6.2 m / 2.4 m/s^2

t^2 = 2.5833 s^2

t ∼ 1.61 s

Therefore, it will take approximately 1.61 seconds for the bicyclist to catch up to his friend.

Answer:

The charge on sphere A and C is [tex]\frac{3q}{8} C[/tex]

Solution:

As per the question:

Charge on sphere A = q C

Now,

When sphere A with 'q' charge is brought in contact with B then conduction takes place and equal charges are distributed on both the spheres as:

[tex]\frac{q + q}{2} = \frac{q}{2}[/tex]

Charge on each A and B is [tex]\frac{q}{2}[/tex]

Now, after separating both the spheres, when sphere B with charge [tex]\frac{q}{2}[/tex] touches uncharged sphere C, again conduction occurs and the charge is distributed as:

[tex]\frac{\frac{q}{4} + \frac{q}{4}}{2} = \frac{q}{4}[/tex]

Now, after the charges are separated with charge [tex]\frac{q}{4}[/tex] on each sphere B and C

Now,

When sphere A and Sphere C with charge [tex]\frac{q}{2}[/tex] and [tex]\frac{q}{4}[/tex] comes in contact with each other, conduction takes place and charge distribution is given as:

[tex]\frac{\frac{q}{2} + \frac{q}{4}}{2} = \frac{3q}{8}[/tex]

Answer:

(a). The tangential acceleration is 1.107 m/s².

(b). The radial acceleration is 102.08 m/s².

Explanation:

Given that,

Initial angular velocity = 7.85 rad/s

Final angular velocity = 10.65 rad/s

Time = 2.3 s

Acceleration = 1.23 rad/s²

Radius = 90 cm

(a). We need to calculate the tangential acceleration

Using formula of radial acceleration

[tex]a_{t}=\alpha R[/tex]

Put the value into the formula

[tex]a_{t}=1.23\times90\times10^{-2}[/tex]

[tex]a_{t}=1.107\ m/s^2[/tex]

The tangential acceleration is 1.107 m/s².

(b). We need to calculate the radial acceleration

Using formula of radial acceleration

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

Put the value into the formula

[tex]a_{c}=(10.65)^2\times90\times10^{-2}[/tex]

[tex]a_{c}=102.08\ m/s^2[/tex]

Hence, (a). The tangential acceleration is 1.107 m/s².

(b). The radial acceleration is 102.08 m/s².

Answer:

q = -7.691 × [tex]10^{-13}[/tex] C

so magnitude of charge is 7.691 × [tex]10^{-13}[/tex] C

and negative sign mean charge is negative potential

Explanation:

given data

electric potential = −2.36 V

distance = 2.93 mm = 2.93 × [tex]10^{-3}[/tex] m

to find out

What are the sign and magnitude of a point charge

solution

we know here that electric potential due to charge is

V = [tex]k\frac{q}{r}[/tex]       ..............................1

here k is coulomb force that is 8.99 ×[tex]10^{9}[/tex] Nm²/C² and r is distance and q is charge  and V is electric potential

put here all value we get q  in equation 1

V = [tex]k\frac{q}{r}[/tex]

-2.36 = [tex]8.99*10^{9} \frac{q}{2.93*10^{-3}}[/tex]

q = -7.691 × [tex]10^{-13}[/tex] C

so magnitude of charge is 7.691 × [tex]10^{-13}[/tex] C

and negative sign mean charge is negative potential

The electric potential is negative, indicating that the charge is negative. So, the point charge is a negative charge with a magnitude of [tex]\( 7.68 \times 10^{-12} \) C.[/tex]

To find the sign and magnitude of the point charge ( q ) that produces an electric potential of ( -2.36 ) V at a distance of [tex]\( 2.93 \) mm (\( 2.93 \times 10^{-3} \) m)[/tex], we can use the formula for the electric potential created by a point charge:

[tex]\[ V = \frac{k \cdot |q|}{r} \][/tex]

where:

- ( V ) is the electric potential (given as ( -2.36 ) V),

- ( k ) is Coulomb's constant [tex](\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),[/tex]

- |q| is the magnitude of the point charge we want to find, and

- ( r ) is the distance from the point charge to the point where the electric potential is being measured (given as [tex]\( 2.93 \times 10^{-3} \) m[/tex]).

Rearranging the equation to solve for |q|, we get:

[tex]\[ |q| = \frac{V \cdot r}{k} \[/tex]]

Substituting the given values, we have:

[tex]\[ |q| = \frac{-2.36 \times 2.93 \times 10^{-3}}{8.99 \times 10^9} \][/tex]

[tex]\[ |q| = \frac{-2.36 \times 2.93 \times 10^{-3}}{8.99 \times 10^9} \][/tex]

[tex]\[ |q| \approx 7.68 \times 10^{-12} \, \text{C} \][/tex]

Now, the electric potential is negative, indicating that the charge is negative. So, the point charge is a negative charge with a magnitude of [tex]\( 7.68 \times 10^{-12} \) C.[/tex]

Answer:

Explanation:

We know that 3.20 grams of salt in 9.10 grams of water gives us a saturated solution at 25°C. To find how many grams of salt will gives us a saturated solution in 100 grams of water at the same temperature, we can use the rule of three.

[tex]\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{100 \ g \ water}[/tex]

Working it a little this gives us :

[tex] x = 100 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water} [/tex]

[tex] x = 35.16 \ g \ salt [/tex]

So, the solubility of the salt is 35.16 (g/100 g of water).

Using the rule of three, we got:

[tex]\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{25 \ g \ salt}{x \ g \ water}[/tex]

Working it a little this gives us :

[tex] x = \frac{25 \ g \ salt}{ \frac{3.20 \g \ salt}{9.10 \ g \ water}} [/tex]

[tex] x = 71.09 g \ water [/tex]

So, it would take 71.09 grams of water to dissolve 25 grams of salt.

Using the rule of three, we got that for 15.0 grams of water the salt dissolved will be:

[tex]\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{15.0 \ g \ water}[/tex]

Working it a little this gives us :

[tex] x = 15.0 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water} [/tex]

[tex] x = 5.27\ g \ salt [/tex]

This is the salt dissolved

The percentage of salt dissolved is:

[tex]percentage \ salt \ dissolved = 100 \% * \frac{g \ salt \ dissolved}{g \ salt}[/tex]

[tex]percentage \ salt \ dissolved = 100 \% * \frac{ 5.27\ g \ salt }{ 10.0 \ g \ salt}[/tex]

[tex]percentage \ salt \ dissolved = 52.7 \% [/tex]

a. The solubility (in g salt/100 g of water) of the salt is 35.16 g/100 g water. b. Amount of water it would take to dissolve 25 g of this salt is 71.10 g. c. If 10.0 g of the salt is mixed with 15.0 g of water, 52.7% of the salt will dissolve.

To solve the problem, we need to determine the solubility of the salt at 25°C using the given data:

Part a: Solubility

Part b: Amount of Water Needed to Dissolve 25 g of Salt

Part c: Percentage of Salt Dissolved

Therefore, 52.7% of the 10.0 g of salt will dissolve in 15.0 g of water.

Answer:

Explanation:

a ) Force = a charge q in an electric field E = q x E

Acceleration = Force / mass

Acceleration of proton

= q [tex]\frac{1.6\times10^{-19}\times700}{1.67\times10^{-27}}[/tex]E / m

= 6.7 x 10¹⁰ m /s²

b ) Initial speed u = 0 , Final speed v = 1.10 x 10⁶ m/s , acceleration a = 6.7 x 10¹⁰ m /s²

v = u + at

1.1 x 10⁶ = 0 +6.7 x 10¹⁰ t

t = 16.4 x 10⁻⁶ s

c ) s = ut + 1/2 at²

s = 0 +.5 x 6.7 x 10¹⁰ x (16.4x 10⁻⁶ )²

= 9.01 m .

d ) K E = 1/2 m v²

= .5 x 1.67 x 10⁻²⁷x (1.1 x 10⁶ )²

= 10⁻¹⁵ J

Answer:

the height of the balcony from where the ball is thrown is 27.874 m.

Explanation:

given,

initial velocity (u) = 5.1 m/s

time (t) = 1.92 s

height of balcony = ?

using equation;          

[tex]s = u t + \dfrac{1}{2} at^2[/tex]

[tex]h =5.1 \times 1.92 + \dfrac{1}{2}\times 9.81\times 1.92^2[/tex]

h= 9.792 + 18.082                      

h = 27.874 m

hence, the height of the balcony from where the ball is thrown is 27.874 m.

Final answer:

Using the kinematic equations of motion, we find that the ball was thrown from a height of approximately 8.211 meters.

Explanation:

To calculate the height from which the ball was thrown, we use the kinematic equations of motion for an object under constant acceleration due to gravity. The specific equation that relates the initial velocity (Vi), time (t), acceleration due to gravity (g), and the displacement (height h in this case) is:

h = Vi * t + 0.5 * g * t2

Where:

Plugging in these values, we get:

h = -5.10 m/s * 1.92 s + 0.5 * 9.81 m/s2 * (1.92 s)2

h = -9.792 m + 18.003 m

h = 8.211 m

Therefore, the height of the balcony is approximately 8.211 meters above the ground.

Explanation:

Let [tex]m_p[/tex] is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.

The magnetic force is balanced by the centripetal force acting on the proton as :

[tex]\dfrac{mv^2}{r}=qvB[/tex]

r is the radius of path,

[tex]r=\dfrac{mv}{qB}[/tex]

Time period is given by :

[tex]T=\dfrac{2\pi r}{v}[/tex]

[tex]T=\dfrac{2\pi m_p}{qB}[/tex]

Frequency of proton is given by :

[tex]f=\dfrac{1}{T}=\dfrac{qB}{2\pi m_p}[/tex]

The wavelength of radiation is given by :

[tex]\lambda=\dfrac{c}{f}[/tex]

[tex]\lambda=\dfrac{2\pi m_pc}{qB}[/tex]

So, the wavelength of radiation produced by a proton is [tex]\dfrac{2\pi m_pc}{qB}[/tex]. Hence, this is the required solution.

Answer:

Magnitude of electric field on Earth = Q = 6.8 × 10⁵ C

Explanation:

Electric field = E = 150 N/C

Distance from the center of the earth to the surface = Radius of the earth

Radius of the earth = R = 6.38× 10⁶ m

E = k Q / R²  is the basic formula for the electric field. k = 9 × 10⁹ N m²/C²

150 = (9 × 10⁹)(Q) / (6.38× 10⁶ )²

⇒ Charge = Q = (150)(6.38× 10⁶ )²/(9 × 10⁹)

                        = 6.8 × 10⁵ C(2 significant figures).

Answer:

The charge of earth is[tex]-6.8\times 10^{5}Columbs[/tex]

Explanation:

Assuming earth as a spherical body we have

For a sphere of radius 'r' and charge 'q' the electric field generated at a distance 'r' form the center of sphere is given by the equation

[tex]E=\frac{1}{4\pi \epsilon _o }\cdot \frac{Q}{r^{2}}[/tex]

where

'Q' is the total charge on sphere

Now at a distance 'r' equal to radius of earth(6371 km)  we have the electric field strength is 150N/C

Using the given values we obtain

[tex]150=\frac{1}{4\pi \epsilon _o}\frac{Q}{(6371\times 10^{3})^2}\\\\\therefore Q=150\times (6371\times 10^{3})^{2}\times 4\pi \epsilon _o\\\\\therefore Q=6.8\times 10^{5}Columbs[/tex]

Now since the electric field is inwards thus we conclude that this charge is negative in magnitude.

Answer:259.80 N

Explanation:

Given

mass of ball=3 kg

ball velocity =10 m/s

angle made by ball with plane of wall [tex]\theta [/tex]=60

Momentum change in Y direction remains same and there is change only in x direction

therefore

initial momentum[tex]=mvsin\theta [/tex]

=30sin60

Final momentum=-30sin60

Change in momentum is =30sin60+30sin60

=60sin60

and Impulse = change in momentum

Fdt=dP

where F=force applied

dP=change in momentum

[tex]F\times 0.2=60sin60[/tex]

[tex]F\times 0.2=51.96[/tex]

F=259.80 N

The average force exerted by the wall on the 3.00 kg steel ball after it bounces off is 150 N. We calculated this using the principle of conservation of momentum and Newton's second law.

The subject of this question is Physics, specifically, the topic of motion and force. To answer this question, we should apply the law of conservation of momentum and Newton's second law (Force = change in Momentum / change in Time).

Firstly, the ball is projected against the wall at an angle of 60 degrees, but we are concerned with the component of velocity perpendicular to the wall. This means we will consider the velocity component of the ball towards the wall, which is 10 cos(60), giving us 5.0 m/s. The incoming momentum of the ball can then be calculated as the mass times the velocity (3.0 kg * 5.0 m/s = 15 kg*m/s).

Since the ball bounces off with the same speed at the same angle, its outgoing momentum is -15 kg*m/s. The change in momentum is therefore Outgoing momentum - Incoming momentum = -15 kg*m/s - 15 kg*m/s = -30 kg*m/s. The force exerted by the wall on the ball equals the Change in momentum divided by the time it takes for the change to occur (-30 kg*m/s / 0.200 s = -150 N). Given that force is a vector and we are asked for the magnitude of the force, the answer is 150 N.

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Answer:

53.36 m.

Explanation:

Initial velocity of the toolbox u = 0

acceleration due to gravity g = 9.8 m s⁻²

time t = 3.3 s

height of roof h = ?

h = ut + 1/2 g t²

= 0 + .5 x 9.8 x 3.3²

= 53.36 m.

Explanation:

Given that,

Charge 1, [tex]q_1=-23\ C[/tex]

Charge 2, [tex]q_1=+67\ C[/tex]

Distance between charges, r = 3.18 cm = 0.0318 m

(a) Let F is the magnitude of force that each sphere experiences. The force is given by :

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

[tex]F=9\times 10^9\times \dfrac{-23\times 67}{(0.0318)^2}[/tex]

[tex]F=-1.37\times 10^{16}\ N[/tex]

(b) The spheres are brought into contact and then separated to a distance of 3.18 cm. When they are in contact, both possess equal charges. Net charge is :

[tex]q=\dfrac{q_1+q_2}{2}[/tex]

[tex]q=\dfrac{-23+67}{2}=22\ C[/tex]

Electric force is given by :

[tex]F=9\times 10^9\times \dfrac{22^2}{(0.0318)^2}[/tex]

[tex]F=4.307\times 10^{15}\ N[/tex]

Hence, this is the required solution.    

Answer:

The value of 'c' is 4.

Explanation:

We know from the basic relation between work and force is

[tex]W_{2}-W_{1}=\int_{x_{1}}^{x_{2}}F\cdot dx[/tex]

Now since the energy of the particle changes from 20 Joules to 11 Joules  as position changes from  x=0 to x=3.00 thus applying the given vales in above relation we get

[tex]11-20=\int_{0}^{3}(cx-3x^{2})\cdot dx\\\\-9=[\frac{cx^2}{2}]_0^3-[\frac{3x^3}{3}]_0^3\\\\-9=\frac{9c}{2}-27\\\\4.5c=18\\\\\therefore c=4[/tex]

Answer:

a)Total distance = 399. 5 m

b)Total time =51.51 sec

c)Average speed = 7.75 m/s

Explanation:

For A to B:

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

[tex]v^2=u^2+2as[/tex]

v= u + at

8.33 = 0.42 x t

t=19.83 sec

1 Km/h=0.27 m/s

30 Km/h=8.33 m/s

[tex]8.33^2=2\times 0.42\times s[/tex]

s=82.6 m

For B to C

V= 8.33 m/s

s= V x t

s=8.33 x 30

s=249.9 m

For C to D

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

v= u - at

Final speed v=0

So

s=v x t/2

7= 8.33 x t/2

t=1.68 sec

Total distance = 82.6 + 249 .9 +7

Total distance = 399. 5 m

Total time = 19.83 + 30 + 1.68

Total time =51.51 sec

Average speed =Total distance/Total time

Average speed = 399.5/51.5

Average speed = 7.75 m/s

Answer:

(177.94 ± 3.81) cm^2

Explanation:

l + Δl = 21.7 ± 0.2 cm

b + Δb = 8.2 ± 0.1 cm

Area, A = l x b = 21.7 x 8.2 = 177.94 cm^2

Now use error propagation

[tex]\frac{\Delta A}{A}=\frac{\Delta l}{l}+\frac{\Delta b}{b}[/tex]

[tex]\frac{\Delta A}{A}=\frac{0.2}{21.7}+\frac{0.1}{8.2}[/tex]

[tex]\Delta A=177.94 \times \left ( 0.0092 + 0.0122 \right )=3.81[/tex]

So, the area with the error limits is written as

A + ΔA = (177.94 ± 3.81) cm^2

Answer:

acceleration of person = 9.77 m/s²

Explanation:

given data

latitude = 40 degree

to find out

Calculate the acceleration of a person

solution

we know that here 40 degree = 0.698 rad

so

acceleration of person = g - ω²R    ...............1

and 1 rotation complete in 24 hours = 360 degree

here g is 9.81

so we know Earth angular speed ω = 7.27 × [tex]10^{-5}[/tex] rad/s and R is earth radius that is 6.37 × [tex]10^{6}[/tex] m

so

put here value in equation 1 we get

acceleration of person = g - ω²R

acceleration of person = 9.81 - (7.27 × [tex]10^{-5}[/tex])² × 6.37 × [tex]10^{6}[/tex]

acceleration of person = 9.77 m/s²

Final answer:

To calculate the centripetal acceleration at a latitude of 40°, use the formula ω2 r, considering Earth's angular velocity and the adjusted radius at that latitude. The resulting acceleration is less than that at the equator and significantly weaker than gravity, ensuring no risk of 'flying off' due to Earth's rotation.

Explanation:

To calculate the acceleration due to Earth's rotation for a person at a latitude of 40 degrees, we can use the formula for centripetal acceleration, ac = ω2 r, where ω is the angular velocity of Earth and r is the distance from the axis of rotation. The angular velocity of Earth is approximately 7.29 × 10-5 radians per second (based on a sidereal day). The distance from the axis of rotation at latitude λ can be expressed as r = REcos λ, where RE is the Earth's radius.

The formula for a person at latitude 40° becomes ac = (ω2)(RE)(cos 40°). By using the Earth's radius RE = 6380 km and the given values, we can calculate the centripetal acceleration and compare it with the acceleration due to gravity, g.

At latitude 40°, the velocity due to Earth's rotation is reduced as a function of the cosine of the latitude, hence, the centripetal acceleration will also be less than what it is at the equator. To find out if there is any danger of 'flying off', we must understand that the gravitational force at Earth's surface is significantly stronger than the centripetal force due to Earth's rotation, ensuring that we remain safely on the ground.

Answer:

ΔV = ±0.175 cm

Explanation:

The equation for volume is

V = π/4 * d^2 * h

All the measurements are multiplied. To propagate uncertainties in multiplication we add the relative uncertainties together.

The relative uncertainty of the diameter is:

εd = Δd/d

εd = 0.0005/5.1 = 0.000098

The relative uncertainty of the height is:

εh = Δh/h

εh = 0.005/37.6 = 0.00013

Then, the relative uncertainty of the volume is:

εV = 2 * εd + εh

εV = 2 * 0.000098 + 0.00013 = 0.000228

Then we get the absolute uncertainty of the volume, for that we need the volume:

V = π/4 * 5.1^2 * 37.6 = 768.1 cm^3

So:

ΔV = ±εV * V

ΔV = ±0.000228 * 768.1 = ±0.175 cm

Answer:

1. 9.8 x 10^-7 cm/s

2. 50970 L

Explanation:

1.

time, t = 30 days = 30 x 24 x 60 x 60 seconds = 2592000 seconds

length of hair, d = 1 inch = 2.54 cm

rate of growth = length of hair grows per second = 2.54 / 2592000

                        = 9.8 x 10^-7 cm/s

Thus, the grown rate of hair is 9.8 x 10^-7 cm/s.

2. length of the pool, l = 15 feet

width of the pool, w = 15 feet

height of pool, h = 8 feet

Volume of the pool, V = length x width x height

V  = 15 x 15 x 8 = 1800 ft^3

To convert ft^3 into m^3 , we use

1 ft = 0.3048 m

so, 1 ft^3 = (0.3048)^3 m^3 = 0.0283 m^3

So, 1800 ft^3 = 1800 x 0.0283 = 50.97 m^3

now, 1 m^3 = 1000 L

So, 50.97 m^3 = 50.97 x 1000 L = 50970 L

Hair grows at approximately 9.8 x 10^-9 m/s based on the 30-day/1-inch growth rate. The swimming pool's volume is calculated by converting its dimensions into meters, then multiplying to find cubic meters and converting that volume to liters.

To estimate how fast your hair grows in meters per second, consider that 1 inch is equivalent to 2.54 cm, and there are 30 days in the given period. Hair growing 1 inch in 30 days is roughly 0.0254 meters in 2592000 seconds (30 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute), which results in a speed of approximately 9.8 x 10-9 meters per second.

For the swimming pool volume, the dimensions in feet need to be converted to meters: 15.0 feet is about 4.572 meters (since 1 foot = 0.3048 meters). Therefore, the pool volume in cubic meters would be 4.572 m x 4.572 m x 2.4384 m (since 8.0 feet is 2.4384 meters). To convert this volume to liters, note that 1 m3 is 1000 liters. So, we find the pool's volume in liters by multiplying the volume in cubic meters by 1000.

To find the ratio of the magnitude of E2 to the magnitude of E1, we analyze the motion of the proton in the electric fields. By equating the work done in the first and second electric fields and using the equations of motion, we can determine the ratio of E2 to E1. The ratio is -1.

To find the ratio of the magnitude of E2 to the magnitude of E1, we need to analyze the motion of the proton in the electric fields. Since the proton returns to its starting point, its net displacement is zero. This means that the net work done on the proton in the combined electric fields is also zero. Using the work-energy theorem, we can equate the work done in the first electric field (E1) to the work done in the second electric field (E2).

The work done by a constant electric field is given by the formula W = μEd, where W is the work done, μ is the magnitude of the electric charge of the proton, E is the magnitude of the electric field, and d is the displacement. Since the displacement is zero, the work done in both fields is zero. Therefore, we have the equation: μE1d₁ + μE2d₂ = 0.

Since the proton is released from rest, its initial velocity is zero. Using the equation of motion, v = u + at, where v is the final velocity (which is also zero), u is the initial velocity, a is the acceleration, and t is the time, we can solve for the times taken in each field: t₁ = 32.3 s and t₂ = 3.45 s.

We can rewrite the equation as: E₁d₁ = -E₂d₂.

Using the equation of motion, d = ut + 0.5at², we can substitute the values of t₁ and t₂ to obtain: d₁ = 0.5a₁t₁² and d₂ = 0.5a₂t₂².

Substituting these values into the equation, we get: E₁(0.5a₁t₁²) = -E₂(0.5a₂t₂²).

The accelerations a₁ and a₂ can be determined using the formula a = μE/m, where μ is the magnitude of the charge of the proton, E is the magnitude of the electric field, and m is the mass of the proton. Since the proton is released from rest, its initial velocity is zero, so the acceleration is the same in both fields. Using this formula, we can express the accelerations in terms of the electric fields: a₁ = (μE₁)/m and a₂ = (μE₂)/m.

Substituting these values into the equation, we get: E₁(0.5((μE₁)/m)t₁²) = -E₂(0.5((μE₂)/m)t₂²). Simplifying the equation, we have: E₁²t₁² = -E₂²t₂².

Dividing both sides of the equation by E₁²t₁², we obtain: 1 = -(E₂/E₁)²(t₂/t₁)².

Taking the square root of both sides of the equation, we get: 1 = -(E₂/E₁)(t₂/t₁).

Dividing both sides of the equation by (t₂/t₁), we finally obtain the ratio of the magnitude of E₂ to the magnitude of E₁: E₂/E₁ = -1.

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Since the proton returned to its starting point, it must have decelerated and then accelerated in the opposite direction due to the electric field E2. Upon calculation, we found that the electric field E2 must be zero. Therefore, the ratio of the magnitude of E2 to E1 is 0.

The problem given can be resolved using the principles of Electromagnetism and Classical mechanics. The proton under the influence of the electric field E1 pointing due North will gain a certain amount of velocity during the first 32.3 seconds. When the direction of the electric field changes to E2 pointing due South, the proton will experience a force in the opposite direction, causing it to decelerate, stop, and then accelerate back to its starting point.

We know that the proton's velocity upon reaching the starting point is zero, and the total time it takes to return to the starting point is 3.45 seconds. Using the equations of motion, we can calculate the deceleration rate caused by the second electric field E2. The acceleration or deceleration experienced by a charged particle in an electric field is given by equation F=qE, where F is force, q is the charge, and E is the electric field.

Considering the proton is experiencing uniform deceleration, we have a = 2s/t². We know s=0 (as it returns to the same point) and the time t = 3.45s. From this, we can find a = 2*0/(3.45)² = 0 m/s². The deceleration caused by the second electric field E2 is therefore 0 m/s². The force acting on the proton is given by F = m*a, where m is the mass of the proton, and a is the deceleration. Since a = 0, F = 0. We know from the equation F = qE, the electric field E2 must be zero. Therefore, the ratio of the magnitude of E2 to E1 is 0.

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Answer:

2899.24 N

Explanation:

W = Weight of pilot

r = Radius

v = Velocity

g = Acceleration due to gravity = 9.81 m/s²

[tex]mg=m\frac{v^2}{r}\\\Rightarrow v=\sqrt{gr}\\\Rightarrow v=\sqrt{9.81\times 200}\\\Rightarrow v=44.3\ m/s[/tex]

Speed of the aeroplane at the top of the loop is 44.3 m/s

Now, v = 280 km/h = 280/3.6 = 77.78 m/s

Apparent weight

[tex]A=W+\frac{W}{g}\frac{v^2}{r}\\\Rightarrow A=710+\frac{710}{9.81}\times \frac{77.78^2}{200}\\\Rightarrow A=2899.24\ N[/tex]

Apparent weight at the bottom of the loop is 2899.24 N

Answer:

ΔT = 258°C

Explanation:

mass of bullet, m = 0.07 kg

velocity of bullet, v = 257 m/s

According to the energy conservation law, the kinetic energy of bullet is totally converted into form of heat energy.

let ΔT be the rise in temperature of the bullet, c be the specific heat of lead.

c = 0.128 J / g°C = 128 J/kg°C

[tex]\frac{1}{2}mv^{2}=mc\Delta T[/tex]

By substituting the values

0.5 x 0.07 x 257 x 257 = 0.07 x 128 x ΔT

ΔT = 258°C

Answer:

The diver is in the air for [tex]1.65s[/tex].

Explanation:

Hi

Finally, we need to find the time in free fall, so we use [tex]y_{f}=y_{i}+v_{i}t-\frac{1}{2}gt^{2}[/tex], at this stage [tex]v_{i}=0m/s, y_{i}=5.03m[/tex] and [tex]y_{f}=0m[/tex], therefore we have [tex]0m=5.03-\frac{1}{2}(9.8m/s^{2})t^{2}[/tex], and solving for [tex]t_{2}=\sqrt{\frac{5.03m}{4.9m/s^{2}}} =\sqrt{1.02s^{2}}=1.01s[/tex].

Last steep is to sum [tex]t_{1}[/tex] and [tex]t_{2}[/tex], so [tex]t_{T}=t_{1}+t_{2}=0.64s+1.01s=1.65s[/tex].

The total time spent by the diver in the air from the moment he leaves the board until he gets to the water is 1.65 s

We'll begin by calculating the time taken to get to the maximum height from the board.

Initial velocity (u) = 6.3 m/s

Final velocity (v) = 0 (at maximum height)

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

Time to reach maximum height (t₁) =?

v = u – gt (since the diver is going against gravity)

0 = 6.3 – 9.8t₁

Collect like terms

0 – 6.3 = –9.8t₁

–6.3 = –9.8t₁

Divide both side by –9.8

t₁ = –6.3 / –9.8

Next, we shall determine the maximum height reached by the diver from the board

Initial velocity (u) = 6.3 m/s

Final velocity (v) = 0 (at maximum height)

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

Maximum Height from the board (h₁) =?

v² = u² – 2gh (since the diver is going against gravity)

0² = 6.3² – (2 × 9.8 × h₁)

0 = 39.69 – 19.6h₁

Collect like terms

0 – 39.69 = –19.6h₁

–39.69 = –19.6h₁

Divide both side by –19.6

h₁ = –39.69 / –19.6

Next, we shall determine the height from the maximum height reached by the diver to the water.

Maximum height from the board (h₁) = 2.03 m

Height of board from water (h₂) = 3 m

Height of diver from maximum height to water (H) =?

H = h₁ + h₂

H = 2.03 + 3

Next, we shall determine the time taken by the diver to fall from the maximum height to the water.

Height (H) = 5.03 m

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

Time to fall from maximum height to water (t₂) =?

H = ½gt²

5.03 = ½ × 9.8 × t₂²

5.03 = 4.9 × t₂²

Divide both side by 4.9

t₂² = 5.03 / 4.9

Take the square root of both side

t₂ = √(5.03 / 4.9)

Finally, we shall determine the total time spent by the diver in the air.

Time to reach maximum height (t₁) = 0.64 s

Time to fall from maximum height to water (t₂) = 1.01 s

T = t₁ + t₂

T = 0.64 + 1.01

Therefore, the total time spent by the diver in the air is 1.65 s

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Answer:

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i:  unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R=  131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

[tex]D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2}  }[/tex]

[tex]D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2}  }[/tex]

D₂= 167.21 cm

Direction of the second displacement

[tex]\beta = tan^{-1} \frac{D_{y}}{D_{x} }[/tex]

[tex]\beta = tan^{-1} \frac{62.18}{155.22 }[/tex]

β= 21.8°

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

Answer:

[tex]a=5.66*10^{23} \frac{m}{s^2}[/tex]

Explanation:

In this case we will use the Bohr Atomic model.

We have that: [tex]F=m*a[/tex]

We can calculate the centripetal force using the coulomb formula that states:

[tex]F=k*\frac{q*q'}{r^2}[/tex]

Where K=[tex]9*10^9 \frac{Nm^2}{C}[/tex]

and r is the distance.

Now we can say:

[tex]m*a=k*\frac{q*q'}{r^2}[/tex]

The mass of the electron is = [tex]9.1*10^{-31}[/tex] Kg

The charge magnitud of an electron and proton are= [tex]1.6*10^{-19}C[/tex]

Substituting what we have:

[tex][tex]a=\frac{9*10^{9}*(1.6*10^{-19} )*(2(1.6*10^{-19} ))}{9.1*10^{-31}*(2.99*10^{-11})^2 }[/tex][/tex]

so:

[tex]a=5.66*10^{23} \frac{m}{s^2}[/tex]

Answer:

Explanation:

We shall convert the movement of grasshopper in vector form. Suppose the grass hopper is initially sitting at the origin or (00) position .

It went 40 cm due west so

D₁ = -40 i

It then moves 26 cm 32 ° south of west so

D₂ = - 26 Cos32i - 26 Sin32 j = - 22 i -13.77 j

Then it moves 19 cm 50° south of east

D₃ = 19 Cos 50 i - 19 Sin 50 j = 12.2 i - 14.55 j

Then it moves 18 cm 60° north of east

D₄ = 18 Cos 60 i + 18 Sin 60 j = 9 i + 15.58 j

Total displacement = D₁ +D₂+D₃+D₄

= - 40i -22 i - 13.77 j + 12.2 i - 14.55 j + 9 i + 15.58 j

= - 40.8 i - 12.74 j

Magnitude of displacement D

D² = ( 40.8 )² + ( 12.74)²

D = 42 .74 cm

If ∅ be the required angle

Tan∅ = 12.74 / 40.80 = .31

∅ = 17 ° positive angle with respect to due west.

Answer: the right answer is C

Explanation: