an air filled parallel plate capacitor has a capacitor of 1.3 pico farad. the separation of the plates is doubled and wax is inserted between them, the new capacitance is 2.6 pico farad. find the dielectric constant of the wax​

Answer:

(a) 0.829 m/s

(b) 3.27 m/s

(c) 0.000153 m²

55.8%

Explanation:

(a) Flow rate equals velocity times cross-sectional area. (1 L = 0.001 m³)

Q = vA

(0.001 m³ / 2.00 s) = v (48 × π (0.002 m)²)

v = 0.829 m/s

(b) Use Bernoulli equation.  Choose point 1 to be the exit of the pump, and point 2 to be exit of the shower head.  Choose 0 elevation to be at point 1.

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

(1.50 atm × 1.0×10⁵ Pa/atm) + ½ (1000 kg/m³) v² + 0 = (1 atm × 1.0×10⁵ Pa/atm) + ½ (1000 kg/m³) (0.829 m/s)² + (1000 kg/m³) (10 m/s²) (5.50 m)

1.50×10⁵ Pa + (500 kg/m³) v² = 1×10⁵ Pa + 414.5 Pa + 55000 Pa

v = 3.27 m/s

(c) Flow rate is constant.

Q = vA

(0.001 m³ / 2.00 s) = (3.27 m/s) A

A = 0.000153 m²

Flow rate is proportional to the pressure difference and the radius raised to the fourth power.

Q ∝ ΔP r⁴

Q₂/Q₁ = (ΔP₂/ΔP₁) (r₂/r₁)⁴

Q₂/Q₁ = (1.120) (0.840)⁴

Q₂/Q₁ = 0.558

The flow decreases to 55.8% of the original value.

Answer:

Explanation:

Regarding the point of "Flow rate is proportional to the pressure difference and the radius raised to the fourth power", flow rate depends on pressure, cross-section area and speed.  As speed also depends on cross-section area, flow rate becomes dependent on pressure and cross-section area squared.

In a round pipe like blood vessel, the cross-section area is equal to pi*radius squared. So flow rate is proportional to the pressure difference and (radius squared) squared; i.e. the radius raised to the fourth power.

The new flow rate = (1.12)*(0.84)^4

=0.5576 or 55.76% of the original flow rate

Answer:speeding up constantly

Explanation:

The graph between the time and the speed of the car shows that the speed is increasing constantly, so, option C is correct.

A moving object's rate of change in distance traveled is measured as speed. Speed is a scalar, which implies it is a measurement with a magnitude but no direction.

A thing that moves quickly and with high speed, covering a lot of ground in a short time. On the other hand, a slow-moving object traveling at a low speed covers a comparatively small distance in the same amount of time. An object with zero speed does not move at all.

Given:

The graph shows the speed of a car traveling east over a 12-second period,

As you can see from the graph, at time t = 0 sec the speed is 10 m/s,

At t = 3 sec, the speed = 15.3 m/s

At t = 6 sec, the speed = 20.3 m/s

Thus, speed is increasing constantly.

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Answer: Calculate the energy required in joules to raise the temperature of 450 grams of water from 15°C to 85°C? (The specific heat capacity of water is 4.18 J/g/°C)

Explanation:

The quantity of heat is required to raise the temperature of of water is 94050 joule.

Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.

The law of energy conservation is adhered to by all energy forms. The law of conservation of energy essentially says that the total energy of the system is conserved in a closed system, also known as an isolated system.

Mass of water: m = 450 grams

Initial temperature of water = 35° C

Final temperature of water = 85° C

Capacity of water is 4.18 J/g °C.

Hence, the  quantity of heat is required to raise = mass × Capacity  × raise in temperature

= 450 × 4.18 × (85 - 35) joule

= 94050 joule.

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Answer:

Acceleration(a) = 0.75 m/s²

Explanation:

Given:

Force(F) = 3 N

Mass of thing(m) = 4 kg

Find:

Acceleration(a)

Computation:

Force(F) = ma

3 = (4)(a)

Acceleration(a) = 3/4

Acceleration(a) = 0.75 m/s²

The complete question is;

A person with body resistance between his hands of 10 kΩ accidentally grasps the terminals of a 16-kV power supply. What is the power dissipated in his body?

A) If the internal resistance of the power supply is 1600 Ω , what is the current through the person's body?

B) What is the power dissipated in his body?

C) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be I_max = 1.00mA or less?

Answer:

A) I = 1.379 A

B) P = 19016.41 W

C) r = 15990000 Ω

Explanation:

A) We are given;

Internal resistance of the power supply; r = 1600 Ω

Body resistance between hands; R = 10kΩ = 10000 Ω

Power supply voltage; E =16 kV = 16000 V

Formula for the current through the person's body with internal resistance is given by;

I = E/(R + r)

Thus;

I = 16000/(10000 + 1600)

I = 1.379 A

B) Formula for power dissipated is;

P = I²R

P = 1.379² × 10000

P = 19016.41 W

C) Now, we are told that the maximum current should be I_max = 1.00mA or less. So, I_max = 0.001 A

Thus, from I = E/(R + r) and making r the subject, we have;

r = (E/I) - R

r = (16000/0.001) - 10000

r = 15990000 Ω

Answer:

Explanation:

wavelength λ = 580 x 10⁻⁹ m

slit width d = 3.7 x 10⁻⁶ m

distance of screen D = 10 m

distance of first bright fringe =  1.5 x λ D / d

= 1.5 x 580 x 10⁻⁹  x 10 / 3.7 x 10⁻⁶

= 2351.34  x 10⁻³ m

= 2351.34 mm .

x = 1/2 at²

where x = length of runway, a = acceleration, and t = time.

600 m = 1/2 (12 m/s²) t²

t² = (1200 m) / (12 m/s²)

t² = 100 s²

t = 10 s

Answer:

a = 2 [m/s^2]

Explanation:

To solve this problem we must use the expressions of kinematics, we must bear in mind that when a body is at rest its velocity is zero.

[tex]v_{f} = v_{i} - (a*t)[/tex]

where:

Vf = final velocity = 0

Vi = initial velocity = 60 [m/s]

a = desacceleration [m/s^2]

t = time = 30 [s]

Note: the negative sign of the above equation means that the car is slowing down, i.e. its speed decreases.

0 = 60 - (a*30)

a = 2 [m/s^2]

Answer:

Explanation:

For a body on a ramp with mass m, the forces acting on the body along the vertical component are the weight and the normal reaction.

The weight of the body acts in the negative y direction while the normal reaction acts in the positive y direction

Taking the sum of forces along the y component

Sum Fy = -W+R = ma

Since acceleration is zero

-W+R = m(0)

-W+R = 0

-W = -R

W = R

Hence the Normal reaction force acting on the on the body is equal to normal force

Explanation:

Let east = E, and, west = opposite to east = - E.

Here, displacement:

=> 2m east + 4m west + 1m east

=> 2E + 4(-E) + 1E

=> 2E - 4E + 1E

=> - 1E

=> 1(-E)

=> 1m west

And, distance,

=> 2m + 4m + 1m = 7m

The distance of a person is 7 m and the displacement of the person is 1m west.

To find the distance and displacement, the given values are,

A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east.

Displacement:

Distance:

Let us consider East = E and west = opposite to east = - E.

Calculating the displacement:

= 2m east + 4m west + 1m east

= 2E + 4(-E) + 1E

= 2E - 4E + 1E

= - 1E

= 1(-E)

= 1m west.

The displacement is 1m west.

Now calculating the distance,

= 2m + 4m + 1m

= 7m

The distance is 7m.

Thus, the displacement and the distance is found as 1 m west and 7m.

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Answer:

c. initial (x and y)

Explanation:

When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.

Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"

Thus, this method resolves the initial x and y velocities.

Answer:

The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.

Explanation:

The spring constant is 2000 newtons per meter. Let consider the spring-block system, from Principle of Energy Conservation we can represent it by the following model:

[tex]U_{k,1}+K_{1} = U_{k,2}+K_{2}[/tex]

[tex]K_{2} = K_{1}+(U_{k,1}-U_{k,2})[/tex] (Eq. 1)

Where:

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies of the block, measured in joules.

[tex]U_{k,1}[/tex], [tex]U_{k,2}[/tex] - Initial and final elastic potential energy, measured in joules.

And we expand the equation above by definitions of elastic potential energy and kinetic energy:

[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = \frac{1}{2}\cdot m\cdot v_{1}^{2} + \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})[/tex]

[tex]v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }[/tex] (Eq. 1b)

Where:

[tex]m[/tex] - Mass of the block, measured in kilograms.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final velocities of the block, measured in meters per second.

[tex]x_{1}[/tex], [tex]x_{2}[/tex] - Initial and final positions of spring, measured in meters.

If we know that [tex]v_{1} = 6\,\frac{m}{s}[/tex], [tex]k = 2000\,\frac{N}{m}[/tex], [tex]m = 2\,kg[/tex], [tex]x_{1} = 0\,m[/tex] and [tex]x_{2} = 0.15\,m[/tex], the final speed of the block moving at the instant the spring has been compressed is:

[tex]v_{2} = \sqrt{\left(6\,\frac{m}{s} \right)^{2}+\left(\frac{2000\,\frac{N}{m} }{2\,kg} \right)\cdot [(0\,m)^{2}-(0.15\,m)^{2}]}[/tex]

[tex]v_{2}\approx 3.674\,\frac{m}{s}[/tex]

The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.

Answer: the earth

Explanation: Earth exerts a gravitational pull on the moon 80 times stronger than the moon's pull on the Earth. Over a very long time, the moon's rotations created fiction with the Earth's tugging back, until the moon's orbit and rotational locked with Earth.

and that's why the earth pulls the moon

Answer:

The value is    [tex]t = 56.68 \  s  [/tex]

Explanation:

From the question we are told that

   The rating of the hoist motor is  [tex]k  =  159hp = 159 *746 =118614 \ W[/tex]

    The  percentage of it power used is  [tex]z = 0.72 * 118614=85402.08 \ W[/tex]

      The  mass of the load is m  = 5550 kg

      The distance is  h = 89.0 m

The potential  energy required to lift the load through that distance is

     [tex]E =  m *  g * h[/tex]

=>    [tex]E =  5550 *  9.8 *  89.0[/tex]

=>   [tex]E =  4840710 \ J[/tex]

Generally the time taken is mathematically represented as

       [tex]t = \frac{E}{ z}[/tex]

=>    [tex]t = \frac{4840710}{ 85402.08}[/tex]

=>    [tex]t = 56.68 \  s  [/tex]

Answer:

2156 N

Explanation:

Data obtained from the question include:

Mass of satellite (m) = 220 Kg

Force (F) of gravity =?

The force of gravity exerted on the satellite on the surface of the earth can be obtained by using the following formula:

Force (F) of gravity = mass (m) × acceleration due to gravity (g)

F = mg

Mass of satellite (m) = 220 Kg

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

Force (F) of gravity =?

F = mg

F = 220 × 9.8

F = 2156 N

Thus, the force of gravity exerted on the satellite on the surface of the earth is 2156 N

Answer:

The electric potential energy of the system is 7.87x10⁻²⁰ J.

Explanation:

The electric potential energy is given by:

[tex]E = \int{Fdr} = \frac{Kq_{1}q_{2}}{r}[/tex]

Where:

q₁ = q₂ is the charge of the protons = 1.62x10⁻¹⁹ C

r is the distance = 3x10⁻⁹ m

K: is the electrostatic constant = 9x10⁹ Nm²/C²

[tex] E = \frac{Kq_{1}q_{2}}{r} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*(1.62 \cdot 10^{-19} C)^{2}}{3\cdot 10^{-9} m} = 7.87 \cdot 10^{-20} J [/tex]

Therefore, the electric potential energy of the system is 7.87x10⁻²⁰ J.

I hope it helps you!

The electric potential energy of the system should be 7.87x10⁻²⁰ J.

SInce We know that

fdr = kq1q2/r

Here

q₁ = q₂ i.e. is the charge of the protons = 1.62x10⁻¹⁹ C

r should be the distance = 3x10⁻⁹ m

K should be the electrostatic constant = 9x10⁹ Nm²/C²

Now electric potential energy should be

= (9x10⁹ Nm²/C² * 1.62x10⁻¹⁹ C) /  3x10⁻⁹ m

=  7.87x10⁻²⁰ J.

hence, The electric potential energy of the system should be 7.87x10⁻²⁰ J.

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Answer:

A ) E/8

Explanation:

If the sphere of radius R  carries charge Q,  then the volumetric charge density is:

ρ₁ = [Q/ (4/3)*π*R³]

Therefore the net charge inside r  ( r < R ) is:

q₁ = ρ * (4/3)*π*r³

And E = K * q₁/r                  K = 9,98 *10⁹ [N*m²/C²]

E = K *  ρ * (4/3)*π*r³/r

E = K *  ρ * (4/3)*π*r²

If now the charge is distributed over a sphere of radius 2R

ρ₂ =  [Q/ (4/3)*π*(2R)³]

ρ₂ =  [Q/ (4/3)*π*8*R³]

Then  ρ₂ < ρ₁    in fact     ρ₂ = (1/8)*ρ₁

The electric field depends on the net charge enclosed by a gaussian surface, and the distance between the net charge and the considered point, ( considering the net charge as being at the center of the gaussian surface) In this case, there was no distance change then

E₂ = E₁/8

The right answer is lyrics  A ) E/8

Answer:

It must be 4 times high.

Explanation:

⇒      [tex]m*g*h = \frac{1}{2} * m*v^{2}[/tex]

       So, at the bottom of the ramp, all the gravitational potential energy

      must be equal to the kinetic energy of the car (Defining the bottom of

      the ramp as our zero reference for the gravitational potential energy):

       [tex]m*g*h_{1} = \frac{1}{2} * m*v_{1} ^{2}[/tex]  (1)

        [tex]m*g*h_{2} = \frac{1}{2} * m*(2*v_{1}) ^{2}[/tex] (2)

Answer:

The average force exerted on the car during this time is 5,943 N

Explanation:

Given;

mass of the car, m = 849 kg

initial velocity of the car, u = 0

time of motion of the car, t = 5.00 s

final velocity of the car, v = 35 m/s

The average force exerted on the car during this time is given by;

[tex]F = ma \\\\F = \frac{m(v-u)}{t}\\\\F = \frac{849(35-0)}{5}\\\\F = \frac{849(35)}{5}\\\\ F = 849*7\\\\F = 5,943 \ N[/tex]

Therefore, the average force exerted on the car during this time is 5,943 N

Answer:

5943N

Let's say (+x) = eastward

Average horizontal acceleration

ax = vx -v0x/5.00s

= 35.0m/s-0/5.00s

= +7.09m/s

From here we apply the second law of newton

During this period average horizontal force acting on car

Summation x = max = (849kg)(+7.09m/s²)

= 5943N

+5.943x10³N

= 5.94kN east ward.

Answer:

Your parrot, from earth, that weighs 1.33kg is 1.33kg on earth. as far as i'm aware there is only one earth and everything always weighs the same on one planet as it did on that same planet.

Explanation:

Answer: the long day in neptune would be .18383562 years!

Explanation:also for every day is 16 hours

Answer:

Isaac walked a distance of 5 blocks, and his displacement was 3 blocks north.

Explanation:

Distance is what he covered from the beginning, while displacement was what he covered in a specific direction

Answer:

kenietic and potential i guess

Explanation:

Answer:

The effective value of g at 6700 m above the Earth's surface is 9.79 m/s².

Explanation:

The value of g can be found using the following equation:

[tex] F = \frac{GmM}{r^{2}} [/tex]

[tex] ma = \frac{GmM}{r^{2}} [/tex]

[tex] a = \frac{GM}{r^{2}} [/tex]

Where:

a is the acceleration of gravity = g

G: is the gravitational constant = 6.67x10⁻¹¹ m³/(kg.s²)

M: is the Earth's mass = 5.97x10²⁴ kg

r: is the Earth's radius = 6371 km      

Since we need to find g at 6700 m, the total distance is:

[tex] r_{T} = 6371000 m + 6700 m = 6377700 m [/tex]

Now, the value of g is:

[tex] a = \frac{GM}{r_{T}^{2}} = \frac{6.67\cdot 10^{-11} m^{3}/(kg*s^{2})*5.97 \cdot 10^{24} kg}{(6377700 m)^{2}} = 9.79 m/s^{2} [/tex]

Therefore, the effective value of g at 6700 m above the Earth's surface is 9.79 m/s².

I hope it helps you!

Answer:

light reflects when it hits the surface

Explanation:

Youngs double slit is a evidence for wave nature,

The properties refraction are attributed as properties of waves. The phenomena of interference and diffraction also fall in this category.

So,

     the answer must be B

Answer: Light does not need a medium to travel.

Explanation: I took the test and got it right :]

Answer: 11400 m

Explanation:

Given:

t = 3.8 x 10^-5 s

v = 3 x 10^8 m/s

d = ?

Formula:

d = vt

  = (3.8 x 10^-5 s)(3 x 10^8 m/s)

  = 11400 m

hope this helps :)

The uniform movement allows to find the results for the distance from the radar to the plane is: 5.7 10³ m or 5.7 km

Kinematics studies the motion of objects looking for relationships between position, velocity and acceleration, in the special case that the acceleration is zero is called uniform motion and is described by the expression

                [tex]v = \frac{d}{t}[/tex]

                d = v t

Where v is the velocity, d the displacement and t the time.

Radar waves are electromagnetic waves with constant velocity

            v = 3 10⁸ m/s

They indicate that the time of the waves to go to the plane and return is 3.8 10⁻⁵ s, therefore if the speed is constant, the time to reach the plane is half of the total time.

          t = [tex]\frac{t_{total} }{ 2}[/tex]

          t = [tex]\frac{3.8 \ 10^{-5}}{2}[/tex]  

          t = 1.9 10⁻⁵ s

Let's calculate

        d = 3 10⁸ 1.9 10⁻⁵

        d = 5.7 10³

In conclusion with the uniform movement we can find the results for the distance from the radar to the plane is: 5.7 10³ m or 5.7 km

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Density ρ is mass m per unit volume v, or

ρ = m / v

Solving for v gives

v = m / ρ

So the given object has a volume of

v = (130 g) / (65 g/cm³) = 2 cm³

Answer:

1.25 g/cm^3

Explanation:

mass-30.0g

volume- 4cm×2cm×3cm=24cm^3

density?

*to find density

Density=Mass/Volume

=30÷24

=1.25g/cm^3