Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawing, the time constant to charge up this circuit is 1.48 s. What is the time constant when they are connected with the same resistor, as in part b

Answer:

3.6 arcsec

Explanation:

 angular diameter = diameter / distance

diameter is constant

so angular diameter ∝ 1 / distance

angular diameter = k / distance

For first case ,

18 = k / .5

for second case let angular diameter be D .

D = k / 2.5

dividing ,

D / 18 = .5 / 2.5 = 1 / 5

D = 18 / 5 = 3.6 arcsec

3.6 arcsec is the answer .

Answer:

the average force exerted on the ball by the bat is 11,613.27 N

Explanation:

Given;

mass of the baseball, m = 151 g = 0.151 kg

initial velocity of the baseball, u = 39.5 m/s

final velocity of the baseball, v = 45.1 m/s

time of action, t = 1.10 ms = 1.10 x 10⁻³ s

The average force exerted on the ball by the bat is calculate as;

[tex]F = ma = \frac{m(v-u)}{t} \\\\F = \frac{0.151(45.1-(-39.5))}{1.10\times 10^{-3}} \\\\F = \frac{0.151(45.1\ +\ 39.5)}{1.10\times 10^{-3}} \\\\F = 11,613.27 \ N[/tex]

Therefore, the average force exerted on the ball by the bat is 11,613.27 N

Force excerted is defined as the product of mass and rate of change of velocity .The average force exerted on the ball by the bat will be 11,613.27 N.

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

m is the mass of the baseball = 151 g = 0.151 kg

u is the  initial velocity of the baseball =39.5 m/s

v is the final velocity of the baseball = 45.1 m/s

t is time of contact = 1.10 ms = 1.10 x 10⁻³ s

The average force mathematically given as

[tex]\rm F=\frac{m(v-u)}{t} \\\\\rm F=\frac{0.151(45.1-39.85)}{1.10\times10^-3} \\\\\rm F=11,613.27 N[/tex]

Hence the average force exerted on the ball by the bat will be 11,613.27 N.

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Answer:

A: 1,212 kg

Explanation:

Answer:

she should write about how big is it and what the sun looks and how far away is it from earth.

Answer:

If it accelerates at 20 m/s2 for a period of 22 seconds, what is its final velocity? ... How fast is the ball falling after 5 seconds? v = v0 + gt v = 0 + 10(5) v = 50 m/s. 4. ... + ½ 2.5(15)2 x = 281 m. 5. What is the total displacement of the car in question 2? ... 8. A base jumper falls until he reaches a speed of 200 m/s

Explanation:

Answer:

the power generation potential is 2.705 x 10⁶ J/s.

Explanation:

Given;

height below the free surface of a large water reservoir, h = 120 m

mass flow rate of the water, m' = 2300 kg/s

The power generation potential is calculated as;

[tex]Power = \frac{Energy}{time} = \frac{F\times h}{t} = \frac{(mg) \times h}{t} = \frac{m}{t}\times gh = m' \times gh\\\\Power = m' \times gh\\\\Power = 2300 \ kg/s \ \times \ 9.8 \ m/s^2 \ \times \ 120 \ m\\\\Power = 2.705 \times 10^6 \ J/s[/tex]

Therefore, the power generation potential is 2.705 x 10⁶ J/s.

Answer: 3m

Explanation: If he is already 2m away from the mirror then if he walks away 1m then it would equal out to 3. You could also add 1 to 2 so you could get the same results.

Answer:

a.432

B. 562

C.402

D. 316

A

Explanation:

Work is defined as the product of applied force and the distance through which the body is displaced on which the force is applied. Work was done by the workman will be 432 Nm.

What is work?

Work is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.

Work may be zero, positive and negative.it depend on the diection of body displaced . if the body is displaced in thw same direction of force it will be positive .

while if the displacement is in the opposite direction of force applied the work will be negative work . if their is no displacement of the body the work done will b e zero.

As given in question wall is stationary diplacement will be zero so in that work will also z

The given data in the problem is;

x is the displacement in the horizontal direection=22m

y is the displacement in the verical direection=9.6m

W is the weight of lumber = 45 N

[tex]\rm W=F\times y[/tex]

[tex]\rm W=F\times y \\\\\rm W=45\times 9.6\\\\ \rm W=432\;Nm[/tex]

Hence Work was done by the workman will be 432 Nm.

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Answer:

#_galon = 8,873 gallons

Explanation:

For this exercise let's use a direct rule of proportions (rule of three). If 1 gallon is 142 MJ, how many gallons is 1260 MJ

          #_galon = 1260 MJ (1 gallon / 142 MJ)

          #_galon = 8,873 gallons

therefore air travel uses less fuel than car travel

Answer:

I think its 71.4 if u get it wrong u can slap me

Answer:

The answer is below

Explanation:

The resistance of a wire is directly proportional to the length of the wire and inversely proportional to its area. The resistance (R) is given by:

[tex]R=\frac{\rho L}{A}\\\\where\ L=length \ of\ wire,A=cross\ sectional\ area, \rho=resistivity\ of\ wire.[/tex]

Let us assume that all the wires have the same resistivity.

a) Wire of Length L and area A

[tex]R_1=\frac{\rho L}{A}[/tex]

b) Wire of Length 2L and area A

[tex]R_2=\frac{\rho *2L}{A}=2R_1[/tex]

C) Wire of Length L and area 2A

[tex]R_3=\frac{\rho L}{2A}=\frac{1}{2}R_1[/tex]

Therefore the wire of least resistance is R3 and R2 has the highest resistivity.

R₃ < R₁ < R₂

Therefore, the ranking of the wires from most current (least resistance) to least current (most resistance) is:

R₃ < R₁ < R₂

Answer:

c

Explanation:

Answer: C

Explanation:

Answer:

d=20m/sx60s=1200m=1200/1000Km=1.2km

Explanation:

Answer:

Explanation:

velocity of first projectile after 3 s

v = u - gt

v = 49.4 - 9.8 x 3

= 20 m /s

Velocity of second projectile after 3 s after being dropped from rest

v = u + gt

= 0 + 9.8 x 3

= 29.4 m /s

They will be moving in opposite direction at the time of meeting , so their relative velocity

= 20 + 29.4 = 49.4 m /s

From the frame of reference of the first projectile, the velocity of the second projectile will be 49.4 m /s .

Complete Question:

Gauss's law:

Group of answer choices

A. can always be used to calculate the electric field.

B. relates the electric field throughout space to the charges distributed through that space.

C. only applies to point charges.

D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.

E. relates the surface charge density to the electric field.

Answer:

D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.

Explanation:

Gauss's law states that the total (net) flux of an electric field at points on a closed surface is directly proportional to the electric charge enclosed by that surface.

This ultimately implies that, Gauss's law relates the electric field at points on a closed surface to the net charge enclosed by that surface.

This electromagnetism law was formulated in 1835 by famous scientists known as Carl Friedrich Gauss.

Mathematically, Gauss's law is given by this formula;

ϕ = (Q/ϵ0)

Where;

ϕ is the electric flux.

Q represents the total charge in an enclosed surface.

ε0 is the electric constant.

Answer:

q = 3.6 10⁵  C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

          r = 6 , 37 106 m

          E = k q / r²

          q = E r² / k

          q = [tex]\frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}[/tex]

          q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

           r = 0.6 r_ Earth

           r = 0.6 6.37 10⁶ = 3.822 10⁶ m

           E = k q / r²

            q = E r² / k

            q = [tex]\frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}[/tex]

            q = 3.6 10⁵  C

Answer:

With or without.

Explanation:

According to the National Electrical Code (NEC), here the air conditioner disconnecting means is not within sight from the equipment, the provision for locking or adding a lock to the disconnecting means shall be on the switch or circuit breaker and remain in place with or without the lock installed. Thus, this is in accordance with section 110.25 of the National Electrical Code (NEC).

Answer:

A. [tex]2\cdot t[/tex]

Explanation:

The SUV accelerates uniformly. In this case, speed ([tex]v[/tex]) is directly proportional to time ([tex]t[/tex]). That is:

[tex]v \propto t[/tex] (1)

[tex]v = k\cdot t[/tex]

Where [tex]k[/tex] is the proportionality constant.

Then, we eliminate this constant by creating the following relationship:

[tex]\frac{v_{1}}{t_{1}} = \frac{v_{2}}{t_{2}}[/tex] (2)

If we know that [tex]v_{1} = v[/tex], [tex]t_{1} = t[/tex] and [tex]v_{2} = 2\cdot v[/tex], then the time interval for the SUV to accelerate from rest to a speed [tex]2\cdot v[/tex] is:

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

[tex]t_{2} = 2\cdot t[/tex]

Hence, correct answer is A.

Answer:

The net acceleration of the SUV is 0.429 meters per square second due west.

Explanation:

Statement is incomplete. Description is presented below:

A SUV along with 5 passengers has a mass of 3500 kg. It has a driving force of 2500 N directed along west on a perfectly horizontal road. The surface of the road exerts a resistance force of 500 N due east. At the same time a high wind is blowing a force of 500 N due east in the opposite direction of the car's drive force. Does the car has any acceleration? If yes, then what are the magnitude and direction of the car's acceleration?

According to Newton's Laws of Motion, the SUV will accelerate if and only if net acceleration is different of zero. Let suppose as positive the direction of driving force ([tex]F[/tex]), measured in newtons:

[tex]\Sigma F = F - R -f = F_{net}[/tex] (1)

Where:

[tex]R[/tex] - Resistance force, measured in newtons.

[tex]f[/tex] - Wind force, measured in newtons.

[tex]F_{net}[/tex] - SUV net force, measured in newtons.

If we know that [tex]F = 2500\,N[/tex], [tex]R = 500\,N[/tex] and [tex]f = 500\,N[/tex], then net force experimented by the SUV is:

[tex]F_{net} = 2500\,N-500\,N-500\,N[/tex]

[tex]F_{net} = 1500\,N[/tex]

The car has acceleration.

By definition of force for systems with constant mass, we calculate the acceleration of the vehicle below:

[tex]a_{net} = \frac{F_{net}}{m}[/tex] (2)

Where [tex]m[/tex] is the mass of the SUV, measured in kilograms.

If we know that [tex]F_{net} = 1500\,N[/tex] and [tex]m = 3500\,kg[/tex], then the net acceleration of the car is:

[tex]a_{net} = \frac{1500\,N}{3500\,kg}[/tex]

[tex]a_{net} = 0.429\,\frac{m}{s^{2}}[/tex]

The net acceleration of the SUV is 0.429 meters per square second due west.

Answer:

a) the rate at which 235U is fissioned is 2895.9 grams per day

b) the rate at which 235U is consumed is 3385.3071 gram/day

Explanation:

Given the data in the question;

a)

designed operation thermal power = 2758 MW.

we know that, the burn up rate fission rate of 235U is 1.05 grams per MW-day.

so, the rate at which 235U is fissioned will be;

⇒ 2758 × 1.05 = 2895.9 grams per day

Therefore, the rate at which 235U is fissioned is 2895.9 grams per day

b)

Consumption rate is given as;

Cr = 1.05 × ( 1 + ∝ )P gram/day

where ∝ is 0.169 for U-235

so,

Cr = 1.05 × ( 1 + 0.169 )2758 gram/day

Cr = 1.05 × 1.169 × 2758 gram/day

Cr = 3385.3071 gram/day

Therefore, the rate at which 235U is consumed is 3385.3071 gram/day

Answer:

$84

Explanation:

The coefficient of performance (COP) show the relationship between the power (kW) output of the heat pump and the power (kW) input to the compressor.

The heater consumed by the heater is 1200 kWh.

For a heat pump with a COP of 2.4, the electric input needed to produce an output of 1200 kWh is:

Electric input to heat pump = 1200 / 2.4 = 500 kWh

That means that supplying a heat pump with 500 kWh produces an output of 1200 kWh

The amount of power saved = power consumed by heater - power consumed by heat pump = 1200 - 500 = 700 kWh

Money saved = $0.12/kWh * 700 kWh = $84

The amount of money the homeowner would have saved during the month is $84.

Given the following data:

To determine the amount of money the homeowner would have saved during the month:

In Science, the coefficient of performance (COP) is a mathematical expression that is typically used to show the relationship between the power output (kilowatts) of a heater and the power input (kilowatts) to the compressor.

Mathematically, the coefficient of performance (COP) is given by the formula:

[tex]COP = \frac{E_o}{E_i}[/tex]

Where:

First of all, we would determine the energy input of the heaters.

[tex]E_i = \frac{E_o}{COP} \\\\E_i=\frac{1200}{2.4}[/tex]

Energy input = 500 kWh

For power saved:

[tex]Power = 1200 -500[/tex]

Power = 700 kWh

For the amount of money saved while using these heaters:

[tex]Money = 0.12 \times 700[/tex]

Amount of money = $84

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Answer:

beacuse it isnt on

Explanation:

Explanation:

The wires isn't in a circulation, the negative side must be connected to the wire

Answer:

[tex]c=0.45\ J/g^{\circ} C[/tex]

Explanation:

Given that,

A 25 g sample of iron (initially at 800.00°C) is dropped into 200 g of water (initially at  30.00°C). The final temperature of the system is 40.22°C.

We need to find the specific heat of iron.

It can be calculated as:

Cooler water gains = hot metal loses

mc∆T = - mc∆T

Put all the values,

[tex]200g(4.184\ J/g^{\circ} C)(T_f-T_i) = -25g(c)(T_f-T_i) \\\\200g(4.184 )( 40.22-30.00) = -25\times (c)\times (40.22-800.00)\\\\8552.096 = 18994.5c\\\\c=\dfrac{8552.096 }{18994.5}\\\\c=0.45\ J/g^{\circ} C[/tex]

So, the specific heat of iron is [tex]0.45\ J/g^{\circ} C[/tex]

Answer:

1) elastic shock, the velocity of the center of mass does not change

2) inelastic shock, he velocity of the mass center   change

Explanation:

The position of the center of mass of your system is defined by

          [tex]x_{cm}[/tex] = [tex]\frac{1}{M} \sum x_i m_i[/tex]

in this case we have two bodies

          x_{cm} = [tex]\frac{1}{M}[/tex] (x₁m₁ + x₂ m₂)

the velocity of the center of mass is

          x_{cm} = dx_{cm} / dt = [tex]\frac{1}{M} ( m_1 \frac{dx_1}{dt} \ + m_2 \frac{dx_2}{dt} )[/tex]

          x_{cm} = [tex]\frac{1}{M} ( m_1 v_1 + m_2 v_2 )[/tex]

where M is the total mass of the system.

Therefore to answer this question we have to find the velocity of the body after the collision.

Let's use momentum conservation, where the system is formed by the two bodies, so that the forces have been internal during the collision.

Let's solve each case separately.

2) inelastic shock

initial instant. Before the crash

         p₀ = m₁ v₀ + 0

final instant. After the collision with the cars together

        p_f = (m₁ + m₂) v

         p₀ = p_f

         m₁ v₀ = (m₁ + m₂) v

         v = [tex]\frac{m_1}{m_1+m_2}[/tex]  v₀

let's find the velocity of the center of mass

         M = m₁ + m₂

initial.

         [tex]v_{cm o}[/tex] = [tex]\frac{1}{m_1 +m_2}[/tex] (m₁ vo)

final

         [tex]v_{cm f}[/tex] = [tex]\frac{1}{M} ( \frac{m_1}{m_1 + m_2} v_o )[/tex] ( v) = v

         v_{cm f} =  [tex]\frac{m_1}{M^2} v_o[/tex]

Let's find the ratio of the velocities of the center of mass

          vcmf / vcmo = [tex]\frac{1}{M} = \frac{1}{m_1 +m_2}[/tex]

therefore the velocity of the mass center   change

1) elastic shock

initial instant.

           p₀ = m₁ v₀

final moment

           p_f = m₁ v_{1f} + m₂ v_{2f}

           p₀ = p_f

           m₁ v₀ = m₁ v_{1f} + m₂ v_{2f}

           m₁ (v₀ - v_{2f}) = m₂ v_{2f}

in this case the kinetic energy is conserved

           K₀ = K_f

          ½ m₁ v₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²

           m₁ (v₀² - v_{1f}²) = m₂ v_{2f}²

           m₁ (v₀ + v_{1f}) (v₀ - v_{1f}) = m₂ v_{2f}

we write our system of equations

           m₁ (v₀ - v_{1f}) = m₂ v_{2f}             (1)

           m₁ (v₀ - v_{1f}) (v₀ + v_{1f}) = m₂ v_{2f}²

we solve the system

             v₀ + v_{1f} = v_{2f}

we substitute and look for the final speeds

             v_{1f} = [tex]\frac{m_1 -m_2}{m1 +m2 } v_o[/tex]

             v_{2f} = [tex]\frac{2 m_1}{m-1+m_2} vo[/tex]

now let's find the velocity of the center of mass

initial

          [tex]v_{cm o}[/tex] = [tex]\frac{1}{M}[/tex] m₁ v₀

final

          [tex]v_{cm f}[/tex] = [tex]\frac{1}{M}[/tex]  (m₁ v_{1f} + m₂ v_{2f} )

          v_{cm f} = [tex]\frac{1}{M}[/tex] [  [tex]m_1 \frac{m_2}{M}[/tex] + [tex]m_2 \frac{2 m_1}{M}[/tex] ] v₀

          v_{cm f} = [tex]\frac{1}{M^2}[/tex] ( m₁² - m₁m₂ +2 m₁m₂) v₂

          v_{cm f} = [tex]\frac{1}{M^2}[/tex] (m₁² + m₁ m₂) v₀

let's look for the relationship

         v_{cm f} / v_{cm o} = [tex]\frac{1}{M}[/tex] M

         v_{cm f} / v_{cm o} = 1

therefore the velocity of the center of mass does not change

we see in either case the velocity of the center of mass does not change.

Answer:

a

Explanation:

Answer:

a= 0

Explanation:

       [tex]F_{net} = m*a = 0 (1)[/tex]

      ⇒ a = 0

Answer:

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