Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperature of 2.81 K.

Answer:

This is because motion is intended to occur but at zero acceleration. It means at a constant velocity, henceFor that to happen the pulling force F must exactly equal the frictional force Fk .

Answer:

a)v = 476.28 m / s , b) T = 6.69 10⁵ N , c)  λ = 0.486 m , d)     λ = 0.35 m

Explanation:

a) The speed of a wave on a string is

          v = √T /μ

also all the waves fulfill the relationship

          v = λ f

they indicate that the fundamental frequency is f = 980 Hz.

The wavelength that is fixed at its ends and has a maximum in the center

          L = λ / 2

          λ = 2L

we substitute

           v = 2 L f

let's calculate

           v = 2  0.243  980

           v = 476.28 m / s

b) The tension of the rope

             T = v² μ

the density of the string is

            μ = m / L

            T = v² m / L

            T = 476.28²   0.717 / 0.243

            T = 6.69 10⁵ N

c)          λ = 2L

            λ = 2  0.243

            λ = 0.486 m

d) The violin has a resonance process with the air therefore the frequency of the wave in the air is the same as the wave in the string. Let's find the wavelength in the air

          v = λ f

          λ= v / f

          λ = 343/980

          λ = 0.35 m

Answer:

Explanation:

Given data

angle of incidence, i=40°

angle of refraction,r = ?

index of refraction u=  1.33

applying the formula

[tex]u=\frac{ sin i}{ sin r}[/tex]

According to Snell's law, the incident, normal and refracted rays all act on the same point.

Substituting and solving for r we have.

[tex]1.33= \frac{sin (40)}{sin (r)} \\\\sin(40)= 1.33* sin(r)\\\0.642= 1.33* sin(r)[/tex]

divide both sides by 1.33 we have

[tex]sin(r)= \frac{0.642}{1.33} \\\sin(r)= 0.48\\\r= sin^-^10.48\\\r= 28.68[/tex]

r= 28.68°

Answer:

 y <8 10⁻⁶ m

Explanation:

For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.

 Therefore the diffraction equation for slits with m = 1 remains

             a sin θ = λ

in general these experiments occur for oblique angles so

             sin θ = θ

             θ = λ / a

in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant

           θ = 1.22 λ / a

The angles in these measurements are taken in radians, therefore

          θ = s / R

as the angle is small the arc approaches the distance s = y

          y / R = 1.22 λ / s

          y = 1.22 λ R / a

let's calculate

            y = 1.22 500 10⁻⁹ 0.42 / 0.032

            y = 8 10⁻⁶ m

with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)

                 y <8 10⁻⁶ m

Answer:

(a) aₓ = 1.33 m/s² and aᵧ = -0.770 m/s²

(b) x = 0.665 m and y = 4.62 m

(c) 3.61 s

Explanation:

(a) There are two ways we can solve this.  The first way is to sum the forces in the x and y direction, then use the relation tan 30° = -aᵧ/aₓ, where aᵧ is the acceleration in the +y direction (up) and aₓ is the acceleration in the +x direction (right).

The second way is to sum the forces in the parallel and perpendicular directions to find the acceleration parallel to the incline, a.  Then, use the relations aᵧ = -a sin 30° and aₓ = a cos 30°.

Let's try the first method.  Sum of forces in the +y direction:

∑F = ma

N cos 30° + Nμ sin 30° − mg = maᵧ

N cos 30° + Nμ sin 30° − mg = -maₓ tan 30°

Sum of forces in the +x direction:

∑F = ma

N sin 30° − Nμ cos 30° = maₓ

Substituting:

N cos 30° + Nμ sin 30° − mg = -(N sin 30° − Nμ cos 30°) tan 30°

N cos 30° + Nμ sin 30° − mg = -N sin 30° tan 30° + Nμ sin 30°

N cos 30° − mg = -N sin 30° tan 30°

N (cos 30° + sin 30° tan 30°) = mg

N = mg / (cos 30° + sin 30° tan 30°)

N = (10 kg) (10 m/s²) / (cos 30° + sin 30° tan 30°)

N = 86.6 N

Now, solving for the accelerations:

N sin 30° − Nμ cos 30° = maₓ

aₓ = N (sin 30° − μ cos 30°) / m

aₓ = (86.6 N) (sin 30° − 0.4 cos 30°) / 10 kg

aₓ = 1.33 m/s²

N cos 30° + Nμ sin 30° − mg = maᵧ

aᵧ = N (cos 30° + μ sin 30°) / m − g

aᵧ = (86.6 N) (cos 30° + 0.4 sin 30°) / 10 kg − 10 m/s²

aᵧ = -0.770 m/s²

Now let's try the second method.

Sum of forces in the perpendicular direction:

∑F = ma

N − mg cos 30° = 0

N = mg cos 30°

Sum of forces in the parallel direction:

∑F = ma

mg sin 30° − Nμ = ma

mg sin 30° − mgμ cos 30° = ma

a = g (sin 30° − μ cos 30°)

a = (10 m/s²) (sin 30° − 0.4 cos 30°)

a = 1.536 m/s²

Solving for the accelerations:

aₓ = a cos 30°

aₓ = 1.33 m/s²

aᵧ = -a sin 30°

aᵧ = -0.770 m/s²

As you can see, the second method is faster and easier, but both methods will give you the same answer.

(b) In the x direction:

Given:

x₀ = 0 m

v₀ = 0 m/s

aₓ = 1.33 m/s²

t = 1 s

Find: x

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (1 s) + ½ (1.33 m/s²) (1 s)²

x = 0.665 m

In the y direction:

Given:

y₀ = 5 m

v₀ = 0 m/s

aᵧ = -0.770 m/s²

t = 1 s

Find: y

y = y₀ + v₀ t + ½ at²

y = 5 m + (0 m/s) (1 s) + ½ (-0.770 m/s²) (1 s)²

y = 4.62 m

(c) In the y direction:

Given:

y₀ = 5 m

y = 0 m

v₀ = 0 m/s

aᵧ = -0.770 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 m = 5 m + (0 m/s) t + ½ (-0.770 m/s²) t²

t = 3.61 s

Answer:

Explanation:

Before we determine the amount of current in each bulb, we must first know that the same current flows in a series connected resistors. Since the Two 15-Ω and three 25-Ω light bulbs are connected in series, same current will flow in all of them.

According to Ohm's law, E = IRt where;

E is the supply voltage = 24V

I is the total current flowing in the circuit

Rt is the total equivalent resistance.

First, we need to calculate Rt.

Rt = 15Ω+15Ω+25Ω+25Ω+25Ω (Two 15-Ω and three 25-Ω light bulb in series)

Rt = 105Ω

From ohms law formula, I = E/Rt

I = 24/105

I = 0.229Amp

Since the total current in the circuit is 0.229A, therefore the amount of current that passes through each bulb is the same as the total current i.e  0.229A

The current that passes via each bulb is 0.229A

According to the above law,

E = IRt

Here

E should be the supply voltage = 24V

I should be the total current flowing in the circuit

Rt should be the total equivalent resistance.

Now Rt should be

Rt = 15Ω+15Ω+25Ω+25Ω+25Ω

Rt = 105Ω

Now the current is

I = E/Rt

I = 24/105

I = 0.229Amp

Therefore, The current that passes via each bulb is 0.229A.

Learn more about current here: https://brainly.com/question/21075693

Answer:

a) before immersion

C = εA/d = (8.85e-12)(25e-4)/(1.31e-2) = 1.68e-12 F

q = CV = (1.68e-12)(255) = 4.28e-10 C

b) after immersion

q = 4.28e-10 C

Because the capacitor was disconnected before it was immersed, the charge remains the same.

c)*at 20° C

C = κεA/d = (80.4*)(8.85e-12)(25e-4)/(1.31e-2) = 5.62e-10 F

V = q/C = 4.28e-10 C/5.62e-10 C = 0.76 V

e)

U(i) = (1/2)CV^2 = (1/2)(1.68e-12)(255)^2 = 5.46e-8 J

U(f) = (1/2)(5.62e-10)(0.76)^2 = 1.62e-10 J

ΔU = 1.62e-10 J - 5.46e-8 J = -3.84e-8 J

Answer:

Explanation:

Let the velocity of astronaut be u and the velocity of flat sheet of metal plate be v . They will move in opposite direction ,  so their relative velocity

= u + v = 2.3 m /s ( given )

We shall apply conservation of momentum law for the movement of astronaut and metal plate

mu  = M v where m is mass of astronaut , M is mass of metal plate

71 u = 230 x v

71 ( 2.3 - v ) = 230 v

163.3 = 301 v

v = .54 m / s

u = 1.76 m / s

honeycomb will be at rest  because honeycomb surface  is frictionless . Plate will slip over it . Over plate astronaut is walking .

a ) velocity of metal sheet relative to honeycomb will be - 1.76 m /s

b ) velocity of astronaut relative to honeycomb will be + .54 m /s

Here + ve direction is assumed to be the direction of astronaut .  

Answer:

F= 3.56e22N

Explanation:

Using the force of radiation acting on the earth which is

force = radiation pressure x area = (intensity/c)xpi R^2

force = 1370W/m^2 x pi x( 6.37x10^6m)^2/3x10^8m/s

force = 5.82x10^8 N

But the sun's gravitational attraction means the magnitude of the solar gravitational force on earth: If that's the case, the answer is approx 10^22 N:

F=GMm/r^2

G=6.67x10^(-11)=6.67e-11

M=mass sun = 2x10^30kg=2e30

m=mass earth = 6x10^24kg

r=earth sun distance = 1.5x10^11m

F=(6.6e-11)(2e30)(6e24)/(1.5e11)^2 =

F= 3.56e22N

Answer:

Explanation:

The formula for hydrogen atomic  spectrum is as follows

energy of photon due to transition from higher orbit n₂ to n₁

[tex]E=13.6 (\frac{1}{n_1^2 } - \frac{1}{n_2^2})eV[/tex]

For layman series n₁ = 1 and n₂ = 2 , 3 , 4 ,   ...   etc

energy of first line

[tex]E_1=13.6 (\frac{1}{1^2 } - \frac{1}{2 ^2})[/tex]

10.2 eV

wavelength of photon = 12375 / 10.2 = 1213.2 A

energy of 2 nd line

[tex]E_2=13.6 (\frac{1}{1^2 } - \frac{1}{3 ^2})[/tex]

= 12.08 eV

wavelength of photon = 12375 / 12.08 = 1024.4 A

energy of third line

[tex]E_3=13.6 (\frac{1}{1^2 } - \frac{1}{4 ^2})[/tex]

12.75 e V

wavelength of photon = 12375 / 12.75 = 970.6 A

energy of fourth line

[tex]E_4=13.6 (\frac{1}{1^2 } - \frac{1}{5 ^2})[/tex]

= 13.056 eV

wavelength of photon = 12375 / 13.05 = 948.3 A

energy of fifth line

[tex]E_5=13.6 (\frac{1}{1^2 } - \frac{1}{6 ^2})[/tex]

13.22 eV

wavelength of photon = 12375 / 13.22 = 936.1 A

Answer:

Amplitude, A = 5 m

Explanation:

Let a progressive wave is given by equation :

[tex]y=5\sin (100\pi t-0.4\pi x)[/tex] .....(1)

The general equation of a progressive wave is given by :

[tex]y=A\sin (\omega t-kx)[/tex] ....(2)

Here,

A is the amplitude of the wave

[tex]\omega[/tex] is the angular frequency

k is propagation constant

We need to find the amplitude of the wave.

If we compare equations (1) and (2), we find that the amplitude of the given plane progressive wave is 5 m.

Answer:

A) W = 103.55 N

B) mass of displaced water = 4186 g

C) W_displaced water = 41.06 N

D) Buoyant force = 41.06 N.

E) ZERO

F) 62.54 N

Explanation:

We are given;

mass of statuette;m = 10,566 g = 10.566 kg

volume = 4,064 cm³

Density of seawater;ρ = 1.03 g/mL = 1.03 g/cm³

A) The dry weight of the statuette can be calculated as;

W = mg

So;

W = 10.556 × 9.81

W = 103.55 N

B) Mass of displaced water is calculated from;

Density = mass/volume

So, mass = Density × Volume

m = 1.03 × 4,064 = 4186 g

C) Weight of displaced water is given by;

W_displaced water = (m_displaced water) × g

W_displaced water = 4.186 kg × 9.81 m/s^2 = 41.06 N

D) The buoyant force is the same as the weight of the displaced water.

Thus, Buoyant force = 41.06 N.

E) The apparent weight of the statuette is calculated from;

Apparent weight = Dry weight - Weight of displaced water

Apparent weight = 103.6 N - 41.06 N = 62.54 N. It is sitting on the bottom of the sea, so the sea floor is providing an opposite force that is equal but opposite the weight so that the net force on the statuette is zero. Since It has zero acceleration, in any direction, hence the net force on it is zero.

F. From E above, The Force required to lift the statuette = 62.54 N

Answer:

Explanation:

a )

Moment of inertial of four masses about axis that coincides with one side :

Out of four masses . location of two masses will lie on the axis so their moment of inertia will be zero .

Moment of inertia of the two remaining masses

= m L² + m L²

= 2 mL²

b )

Axis that bisects two opposite sides

Each of the four masses will lie at a distance of L / 2 from this axis so moment of inertia of the four masses

= 4 x m x ( L/2 )²

= 4 x  mL² / 4

= m L² .

Answer:

[tex]1.357\times 10^{-12}[/tex]

Explanation:

Relevant Data provided

Area which indicates A = 2.3 cm^2 = 2.3 x 10^-4 m^2

Distance which indicates d = 1.50 x 10^-3 m

Voltage which indicates V = 12 V

According to the requirement, the computation of value of its capacitance is shown below:-

[tex]Capacitance, C = \frac{\epsilon oA}{D}[/tex]

[tex]= \frac{= 8.854\times 10^{-12}\times 2.3\times 10^{-4}}{(1.5 \times 10^{-3})}[/tex]

= [tex]1.357\times 10^{-12}[/tex]

Therefore for computing the capacitance we simply applied the above formula.

Answer:

F= 175.5N

Explanation:

Given:

Torque which can also be called moment is defined as rotational equivalent of linear force. It is the product of the external force and perpendicular distance

torque of 38N⋅m

angle of 120∘

Torque(τ): 38Nm

position r relative to its axis of rotation: 25cm , if we convert to metre for consistency we have 0.25m

Angle: 120°

To find the Force, the torque equation will be required which is expressed below

τ = Frsinθ

We need to solve for F, if we rearrange the equation, we have the expression below

F= τ/rsinθ

Note: the torque is maximum when the angle is 90 degrees

But θ= 180-120=60

F= 38/0.25( sin(60) )

F= 175.5N

Answer:

its supposed to be (a) 1W

Answer:

0.0665 days

Explanation:

We are given;

The mean distance from the Earth's center to the moon;a1 = 385000 km

The mean distance from the Earth's center to the space craft;a2 = 6965 km

Formula for kepplers third law is;

T² = 4π²a³/GM

However, the proportion of both distances would be;

(T1)²/(T2)² = (a1)³/(a2)³

Where;

T1 is the period of orbit of the moon around the earth. T1 has a standard value of 27.322 days

T2 is the period of the space craft orbit.

Making T2 the subject, we have;

T2 = √((T1)²×(a2)³)/(a1)³)

Thus, plugging in the relevant values;

T2 = √(27.322² × 6965³)/(385000)³

T2 = 0.0665 days

Answer:

f = 500 x 10^12Hz

Explanation:

E=hc/wavelength

E=hf

hc/wavelength =hf

c/wavelength =f

f = 3 x 10^8 / 600 x 10^-9 = 500 x 10^12Hz

Answer:

Explanation:

induced emf = ?

induced emf = induced current x resistance of the circuit

= .54 x 3.6

= 1.944 V

induced emf required = 1.944 V .

Answer:

moment of inertia I ≈ 4.0 x 10⁻³ kg.m²

Explanation:

given

point masses = 50g = 0.050kg

note: m₁=m₂=m₃=m₄=50g = 0.050kg

distance, r, from masses to eachother = 20cm = 0.20m

the distance, d, of each mass point from the centre of the mass, using pythagoras theorem is given by

= (20√2)/ 2 = 10√2 cm =14.12 x 10⁻² m  

moment of inertia is a proportion of the opposition of a body to angular acceleration about a given pivot that is equivalent to the entirety of the products of every component of mass in the body and the square of the component's distance from the center

mathematically,

I = ∑m×d²

remember, a square will have 4 equal points

I = ∑m×d² = 4(m×d²)

I = 4 × 0.050 × (14.12 x 10⁻² m)²

I = 0.20 × 1.96 × 10⁻²

I =  3.92 x 10⁻³ kg.m²

I ≈ 4.0 x 10⁻³ kg.m²

attached is the diagram of the equation

Answer:

5.4x10^4km

Explanation:

See attached file

Answer:

The speed of m2 just before it hits the ground is 2.1 m/s

Explanation:

mass on the ground m1 = 30 kg

mass oat rest at the above the ground m2 = 35 kg

height of m2 above the ground =2.9 m

Let the tension on the string be taken as T

for the mass m2 to reach the ground, its force equation is given as

[tex]m_{2} g - T = m_{2}a[/tex]    ....equ 1

where g is acceleration due to gravity = 9.81 m/s^2

and a is the acceleration with which it moves down

For mass m1 to move up, its force equation is

[tex]T - m_{1} g = m_{1} a[/tex]

[tex]T = m_{1}a + m_{1}g[/tex]

[tex]T = m_{1}(a + g)[/tex]    ....equ 2

substituting T in equ 1, we have

[tex]m_{2} g - m_{1}(a+g) = m_{2}a[/tex]

imputing values, we have

 [tex](35*9.81) - 30(a+9.81) = 35a[/tex]

 [tex]343.35 - 30a-294.3 = 35a[/tex]

[tex]343.35 -294.3 = 35a+ 30a[/tex]

[tex]49.05 = 65a[/tex]

a = 49.05/65 = 0.755 m/s^2

The initial velocity of mass m2 = u = 0

acceleration of mass m2 = a = 0.755 m/s^2

distance to the ground = d = 2.9 m

final velocity = v = ?

using Newton's equation of motion

[tex]v^{2}= u^{2} + 2ad[/tex]

substituting values, we have

[tex]v^{2}= 0^{2} + 2*0.755*2.9[/tex]

[tex]v^{2}= 2*0.755*2.9 = 4.379\\v = \sqrt{4.379}[/tex]

v = 2.1 m/s

Answer:

The horizontal acceleration of the block is 4.05 m/s².

Explanation:

The horizontal acceleration can be found as follows:

[tex] F = m \cdot a [/tex]

[tex] Fcos(\theta) - \mu_{k}N = m\cdot a [/tex]

[tex] Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)] = m\cdot a [/tex]  

[tex] a = \frac{Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)]}{m} [/tex]

Where:

a: is the acceleration

F: is the force exerted by the rope = 28.2 N

θ: is the angle = 30°

[tex]\mu_{k}[/tex]: is the kinetic coefficient = 0.12

m: is the mass = 5 kg

g: is the gravity = 9.81 m/s²

[tex] a = \frac{28.2 N*cos(30) - 0.12[5 kg*9.81 m/s^{2} - 28.2 N*sen(30)]}{5 kg} = 4.05 m/s^{2} [/tex]

Therefore, the horizontal acceleration of the block is 4.05 m/s².

I hope it helps you!

Answer:

2.9tons

Explanation:

Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:

parallel ("tangential"): F_t = F sin a

perpendicular ("normal"): F_n = F cos a

At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:

F sin a ~~ (46 tons)*0.064 ~~ 2.9tons

Also see attached file

The required force parallel to the incline to hold the monolith on this​ causeway will be "2.9 tons".

According to the question,

Angle, a = 3.7 degrees or,

Sin a = 0.064

Force, F = 46 tons

We know the relation,

Parallel (tangential), [tex]F_t[/tex] = F Sin a

By substituting the values,

                                       = 46 × 0.064

                                       = 2.9 tons

Thus the response above is appropriate answer.

Find out more information about Force here:

https://brainly.com/question/25239010

Answer:

F = 44.22 N

Explanation:

Let force 1, [tex]F_1=19\ N[/tex]

Force 2, [tex]F_2=32\ N[/tex]

The angle between forces, [tex]\theta=118^{\circ}[/tex]

We need to find the magnitude of the resultant force. It is based on the law of cosines. The formula is given by :

[tex]F^2=F_1^2+F_2^2-2AB\cos\theta\\\\F^2=(19)^2+(32)^2-2\times 19\times 32\times \cos(118)\\\\F=\sqrt{(19)^{2}+(32)^{2}-2\times19\times32\times\cos(118)}\\\\F=44.22\ N[/tex]

So, the magnitude of resultant force is 44.22 N.

Correct question: Two forces act on a point on the plane. The angle between the two forces is given. Find the magnitude of the resultant force. forces of 19 and 32 newtons, forming an angle of 118 degrees.

the magnitude of the resultant force is 28.53 N.

The resultant of the two vectors can be calculated using parallelogram theorem.

parallelogram theorem states that if two vectors are represented by the adjacent side of a parallelogram, the resultant of the vectors is the diagonal of the parallelogram drawn from the point of intersection of the vectors.

This can be expressed mathematically as

R² = P²+Q²-2PQcos(180-∅).............. Equation 1

Where R = resultant of the vectors, P and Q = the two vectors respectively, ∅ = angle between the vectors.

From the question,

Given: P = 19 N, Q = 32 N, ∅ = 118°

Substitute these values into equation 2

R² = 19²+32²-2×19×32cos(180-118)

R² = 361+1024-1216cos62°

R² = 1385-1216(0.4695)

R² = 1385-570.878

R² = 814.122

R = √(814.122)

R = 28.53 N

Hence, the magnitude of the resultant force is 28.53 N

Learn more about resultant force   here:https://brainly.com/question/21852571

Answer:

a

    [tex]\alpha = 2327.7 \ rev/s^2[/tex]

b

   [tex]\theta = 9124.5 \ rev[/tex]

Explanation:

From the question we are told that

    The maximum  angular   speed is  [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]

     The  time  taken is  [tex]t = 2.8 \ s[/tex]

     The  minimum angular speed is  [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest

Apply the first equation of motion to solve for acceleration we have that

       [tex]w_{max} = w_{mini} + \alpha * t[/tex]

=>     [tex]\alpha = \frac{ w_{max}}{t}[/tex]

substituting values

       [tex]\alpha = \frac{40950.73}{2.8}[/tex]

       [tex]\alpha = 14625 .3 \ rad/s^2[/tex]

converting to [tex]rev/s^2[/tex]

  We have

           [tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]

           [tex]\alpha = 2327.7 \ rev/s^2[/tex]

According to the first equation of motion the angular displacement is  mathematically represented as

       [tex]\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2[/tex]

substituting values

      [tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]

      [tex]\theta = 57331.2 \ radian[/tex]

converting to revolutions  

        [tex]revolution = 57331.2 * 0.159155[/tex]

        [tex]\theta = 9124.5 \ rev[/tex]

Answer:

The object with the twice the area of the other object, will have the larger drag coefficient.

Explanation:

The equation for drag force is given as

[tex]F_{D} = \frac{1}{2}pu^{2} C_{D} A[/tex]

where [tex]F_{D}[/tex] IS the drag force on the object

p = density of the fluid through which the object moves

u = relative velocity of the object through the fluid

p = density of the fluid

[tex]C_{D}[/tex] = coefficient of drag

A = area of the object

Note that [tex]C_{D}[/tex] is a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. The most interesting things is that it is dependent on the linear dimension, which means that it will vary directly with the change in diameter of the fluid

The above equation can also be broken down as

[tex]F_{D}[/tex] ∝ [tex]P_{D}[/tex] A

where [tex]P_{D}[/tex] is the pressure exerted by the fluid on the area A

Also note that [tex]P_{D}[/tex] = [tex]\frac{1}{2}pu^{2}[/tex]

which also clarifies that the drag force is approximately proportional to the abject's area.

In this case, the object with the twice the area of the other object, will have the larger drag coefficient.

Answer:

`1. charge Q, on the capacitor increases, while the current will decrease

2. τ = t = secs

Explanation:

1. consider RC  of a circuit to be am external source

voltage across the circuit is given as

v =v₀(1 - [tex]e^{\frac{t}{τ} }[/tex])

where v = voltage

v₀ = peak voltage

t = time taken

τ= time constant

as the charge across the capacitor increases, current decreases

the charge across the circuit is given as

Q= Q₀(1 - [tex]e^{\frac{t}{τ} }[/tex])

charge Q is inversely proportional to the current I

hence the charge across the circuit increases

2. τ = RC

unit of time constant, τ,

= Ω × F

=[tex]\frac{V}{I}[/tex] ˣ [tex]\frac{C}{V}[/tex]

=[tex]\frac{C}{A}[/tex]

=[tex]\frac{C}{C/t}[/tex]

τ = t = secs