The circuit diagram below shows the locations of four switches. Which switch configuration creates a short circuit?

Answer:

summer June 21 is the answer

Answer:

49 N

Explanation:

F = m a

 = 5 * 9.81 =~ 49 N

If the process is smooth and only takes the given time, the magnitude of the emf induced in the loop is 3.33 × 10⁻⁴Volts.

Given the data in the question

Magnitude of the emf induced in the loop; \(|e| =\ ?\)

From Faraday's law, formula for Induced emf is given as:

\(|e| = -\frac{\delta\theta}{\delta t} \\\\|e| = -\frac{\delta (BAcos\theta)}{\delta t} \\\\|e| = -BA\frac{\delta (cos\theta)}{\delta t}\)

Where \(\delta\theta\) is the change in flux, \(\delta t\) is time taken to change the flux, B is the Magnetic field and A is the area.

We know that; \(\theta = 0^o\ to\ 90^o\), so

\(|e| = -BA\frac{(cos90^o -cos0^o)}{\delta t}\)

Now, we substitute in our given valueslux

\(|e| = [-(20*10^{-3}T)*0.02m^2]\frac{(cos90^o -cos0^o)}{1.2s}\\\\|e| = (-0.0004Tm^2)\frac{(0-1)}{1.2s}\\\\|e| = \frac{0.0004Tm^2}{1.2s}\\\\|e| = 3.33*10^{-4}Volts\)

Therefore, if the process is smooth and only takes the given time, the magnitude of the emf induced in the loop is 3.33 × 10⁻⁴Volts

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Answer:

letter B.

Explanation:

Answer: A

0.1

Explanation: I did this

Answer:

t = T / 2 all energy is stored in the inductor

Explanation:

The circuit described is an oscillating circuit where the charge of the condensation stops the inductor and vice versa, in this system the angular velocity of the oscillation is

          w = √1/LC

          2π / T =√1 / LC

          T = 2π  √LC

The energy is constant and for the initial instant it is completely stored in the capacitor

         Uc = Q₀² / 2C

In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor

        U = L I² / 2

in the intermediate instant the energy is stored in the two elements.

Since the period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor

After t = 0 the maximum energy stored in the magnetic field of the inductor is equal to \(U'=\dfrac{L I^{2}}{2}\) for the time period, half of period of oscillation  (t = T/2).

The given problem is based on the charging and discharging concepts of capacitor. An oscillating circuit is a circuit where the charge of the capacitor stops the inductor and vice versa, in this system the angular frequency of the oscillation is given as,

\(\omega =\dfrac{1}{\sqrt{LC}}\\\\\\\dfrac{2 \pi}{T} =\dfrac{1}{\sqrt{LC}}\\\\\\T = 2\pi \times \sqrt{LC}\)

here, T is the period of oscillation.

Also, the energy stored in the capacitor is constant and for the initial instant it is completely stored in the capacitor. So, the energy stored is given as,

\(U =\dfrac{Q^{2}}{2C}\)

here, C is the capacitance.

In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor. So, the expression for the energy stored in the inductor is,

\(U'=\dfrac{L I^{2}}{2}\)

here, L is the inductance and I is the current.

Note :- The period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor.

Thus, we conclude that after t = 0 the maximum energy stored in the magnetic field of the inductor is equal to \(U'=\dfrac{L I^{2}}{2}\) for the time period, half of period of oscillation  (t = T/2).

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The magnitude of the force acting at point O is determined as 240 N.

The magnitude of the force at point O is calculated by applying the principles of moment as shown below.

sum of the clockwise moment = sum of the anticlockwise moment

F₀(2 m + 2m + 2m) = 260 N (2m + 2 m ) +  200 N ( 2 m )

where;

6F₀ = 260 (4) + 200(2)

6F₀ = 1,440

F₀ = 1440 / 6

F₀ = 240 N

Thus, the magnitude of the force acting at point O is determined as 240 N.

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Answer:

a. 192 m/s

b. -17,760 kPa

Explanation:

First let's write the flow rate of the liquid, using the following equation:

Q = A*v

Where Q is the flow rate, A is the cross section area of the pipe (A = pi * radius^2) and v is the speed of the liquid. The flow rate in both parts of the pipe (larger radius and smaller radius) needs to be the same, so we have:

a.

A1*v1 = A2*v2

pi * 0.02^2 * 12 = pi * 0.005^2 * v2

v2 = 0.02^2 * 12 / 0.005^2

v2 = 192 m/s

b.

To find the pressure of the other side, we need to use the Bernoulli equation: (600 kPa = 600000 N/m2)

P1 + d1*v1^2/2 = P2 + d1*v2^2/2

Where d1 is the density of the liquid (for water, we have d1 = 1000 kg/m3)

600000 + 1000*12^2/2 = P2 + 1000*192^2/2

P2 = 600000 + 72000 - 1000*192^2/2

P2 = -17760000 N/m2 = -17,760 kPa

The speed in the smaller part of the pipe is too high, the negative pressure in the second part means that the inicial pressure is not enough to maintain this output speed.

The resistance of the electric device using a power of 650 watts is approximately 27.7 ohms.

Ohm’s law states that the potential difference between two points is directly proportional to the current flowing through the resistance.

Hence

V = IR

R = V/I

Where V is the voltage or potential difference, potential difference, I is the current and R is the resistance.

We need to first find the electric current I.

Note:

Power = Voltage × Current.

Hence:

Current I = Power/Votage

Plug in the values

I = 650 / 120

I = 13/3 A

Substituting the values of voltage and current intio the above formula, we can calculate the resistance:

R = V/I

R = 120 / (13/3)

R = 27.7 ohms

Therefore, the resistance is approximately 27.7 ohms.

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Answer:

Uhh 2 one

Explanation

The Schwarzschild radius of different materials supermassive black hole and M87black hole  are\(4.80528 \times 10^{-19} and 9.49 \times 10^{-18}\) respectively.

The distance connecting a Schwarzschild black hole's centre and event horizon is known as the Schwarzschild radius. The event horizon is positioned above the Schwarzschild radius, that starts at the black hole's centre. Black holes have a rather important characteristic called the Schwarzschild radius.

Schwarzschild radius (Rs)= 2GM/c²

G=> gravitational constant = \(6.674 \times 10^{-11} m^3/kgs^2\)

c=> speed of light =>\(3\times10^{8}m/s\)

M => solar mass

A) Rs of supermassive black hole =>

                                \(2 \times 6.674 \times 10^{-11} \times 4.0\times10^{1}0/(3\times10^{8})^{2} = 4.80528 \times 10^{-19}\)

B) Rs of M87black hole

                       \(= 2 \times 6.674 \times 10^{-11} \times 6.4 \times 10^{9} /(3 \times 10^{8})^{2} = 9.49 x 10^{-18}\)

C) Rs of intermediate mass black hole =

                  \(2 \times 6.674 \times 10^{-11} \times 1.0\times10^{4} /(3\times10^{8})^{2} = 1.48 x 10^{-23}\)

D) Rs of  the Sun =

               \(2 \times 6.674 \times 10^{-11} \times\frac { 1.99 \times 10^{30}}{(3x10^{8})^{2}} = 2.95 \times 10^{3}\)

E)  Rs of  micro black hole =>

                       \(2 \times 6.674 \times 10^-11 x\frac{ 2.0\times 10^{−8}}{(3 \times 10^8)^{2}} = 2.96 \times 10^{-35}\)

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Complete question -

Black hole radius, large to small. What is the Schwarzschild radius of

(a) the supermassive black hole with 4.0×10^10 solar masses in the Abell 85 galaxy cluster

(b) the imaged M87black hole with 6.4×10^9 solar masses

(c) an intermediate mass black hole with 1.0×10^4 solar masses

(d) the Sun, with mass 1.99×10^30 kg

(e) a micro black hole with a mass of 2.0×10^−8 kg

Answer:

I do believe the answer is D. prism

Answer:

Explanation:

either all of the above or D <3

The approximate heat transfer through 1.65 cm of air with a temperature difference of 23.0 °C is approximately 24.4 J/s.

To approximate the heat transfer through the air layer in the double-paned window, we can assume that the glass layers have a negligible impact on the heat flow. The heat transfer can be calculated using Fourier's Law of Heat Conduction, which states that the heat flow (Q) is proportional to the temperature difference (ΔT) and inversely proportional to the thickness (L) and thermal conductivity (k) of the material.

First, we need to calculate the effective thermal conductivity of the air layer due to its thickness and the thermal conductivity ratio between air and glass. Let's denote the thermal conductivity of air as k_air and the thermal conductivity of glass as k_glass. Since glass has a thermal conductivity roughly 36 times greater than air, we have k_glass = 36 * k_air.

Next, we calculate the effective thermal conductivity of the air layer as:

k_eff = (k_air * L_air) / (L_air + k_glass)

Substituting the given values, we have:

k_eff = (k_air * 0.0165 m) / (0.0165 m + 0.005 m) = 0.01309 * k_air

Now, we can calculate the heat flow per second through the air layer using the formula:

Q = (k_eff * A * ΔT) / L_air

Substituting the given values, we get:

Q = (0.01309 * k_air * 0.760 m^2 * 23.0 K) / 0.0165 m = 24.4 J/s

Therefore, the approximate heat transfer through 1.65 cm of air with a temperature difference of 23.0 °C is approximately 24.4 J/s.

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Answer:

As you travel up through the Earth's atmosphere, the composition of gases changes. Near the Earth's surface, the atmosphere consists mainly of nitrogen (about 78%) and oxygen (about 21%) with traces of other gases like carbon dioxide and argon. As you go higher, the amount of oxygen decreases, and the concentration of other gases becomes more prominent, such as helium, hydrogen, and ozone.

For a spacecraft is traveling with a velocity of Vox-5690 m/s:

To solve this problem, use the equations of motion to calculate the final velocity and the vertical component of the velocity. Assume that the initial velocity in the y-direction is zero.

Given:

Initial velocity in the x-direction (V₀ₓ) = -5690 m/s

Time of engine firing (t) = 865 s

Acceleration in the x-direction (ax) = 1.45 m/s²

Acceleration in the y-direction (ay) = -8.66 m/s²

(a) To calculate the final velocity (v), use the equation:

v = V₀ₓ + ax × t

Substituting the values:

v = -5690 m/s + 1.45 m/s² × 865 s

v = -5690 m/s + 1254.25 m/s

v ≈ 685.25 m/s

Therefore, the final velocity (v) is approximately 685.25 m/s.

(b) To calculate the vertical component of the velocity (vy), use the equation:

vy = ay × t

Substituting the values:

vy = -8.66 m/s² * 865 s

vy = -7484.9 m/s

Therefore, the vertical component of the velocity (vy) is -7484.9 m/s.

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The de- Broglie wavelength of 2.80× 10⁻¹⁰ m is associated with a momentum of 2.36 × 10⁻²⁴ Kg m/s. The kinetic energy will be 3.07 × 10⁻¹⁸ J which is equal to 19.015 ev.

De-Broglie proposed  the dual nature of matter which stated that matter can behave as both wave and particle. For a matter wave, the wavelength of the wave will be the ratio of the Planck's' constant h and its momentum p.

thus, λ = h/p

Given the wavelength = 2.80× 10⁻¹⁰ m.

momentum p = h/λ

= 6.62 × 10 ⁻³⁴ / (2.80× 10⁻¹⁰ m)

=  2.36 × 10⁻²⁴ Kg m/s

The kinetic energy ke = 1/2 mv² = p² /2m.

mass of an electron = 9.1 × 10 ⁻³⁴ Kg.

hence, Ke =  (2.36 × 10⁻²⁴ Kg m/s )² /  9.1 × 10 ⁻³⁴ Kg

                 = 3.07 × 10⁻¹⁸  J.

1 J = 6.24 × 10¹⁸ ev.

thus, KE in ev = 3.07 × 10⁻¹⁸  J ×  6.24 × 10¹⁸ ev = 19 ev.

Therefore, the kinetic energy of the electron is 19 ev.

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Newts hatch in water, but they spend the most of their lives on land, returning to water only for mating.

They have rough skin, are 8 inches long, and have stubby legs and fangs. In the Pacific Northwest, some of them are toxic.

Rough-skinned newts have brown skin on their backs and bright orange skin on their bellies. When frightened by predators, these newts coil their bodies and show off their orange necks and tails. Predators are warned to stay clear from this orange tint.

It is to be noted that none of the newts are the same traits that is color or size.

A phenotypic trait (simply trait or character state) is a distinct variety of an organism's feature that is visible and quantifiable and manifests itself in an observable manner.

For example, eye color (green, blue, brown, and hazel) is a phenotypic feature that is passed down by polygenetic inheritance.

Height, wing length, and hair color are all phenotypes. Phenotypes also include observable traits that may be evaluated in the lab, such as hormone levels or blood cell counts.

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Total weight in tons

\(\\ \sf\longmapsto 2.3\times 10^6\times 2.4\)

\(\\ \sf\longmapsto 5.52\times 10^6\)

\(\\ \sf\longmapsto 5520000ton\)

\(\\ \sf\longmapsto 5520000\times 1000\)

\(\\ \sf\longmapsto 5520000000kg\)

\(\\ \sf\longmapsto 5.52\times 10^9Kg\)

Answer:

1/2mv²=0

1/2(4kg)(v²)=0

2=-v²

square root -2=v

v=1.414

Block A of mass 1 kg and block B has a mass of 3 kg, then the blocks collide and stick together so the kinetic energy of block A is one-third of the kinetic energy of the block.

Kinetic energy is a type of power that an item or particle possesses as a result of motion. When an item undergoes work—the transfer of energy—by having subject to a net force, it accelerates and consequently obtains kinetic energy. An object in motion or particle's kinetic energy, which depends on both mass and speed, is one of its features. Any combination of motions, including translation, rotation about an axis, and vibration, may be used as the type of motion.

A body's translational kinetic energy is equal to 1/2mv², or one-half of the sum of its mass, m, and the square of its velocity, v.

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GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

GFHFG

The average force the beam exerts on the pile driver while the pile driver is brought to rest is 878 KN.

The word "force" has a specific meaning in science. At this level, calling a force a push or a pull is entirely appropriate. A force is not something an object "has in it" or that it "contains." One thing experiences a force from another. There is no distinction between living and non-living things in the concept of a force.

The newton, abbreviated N, is the SI unit of force. The force-relevant base units are The symbol for the length unit known as the meter is m. the kilogram (kg), is a unit of mass. The sign for the unit of time is s or the second.

Given:-

Mass (m) = 2100Kg

The distance it moves the pilling (h) = 0.12 m

The distance over which the driver falls freely (d) = 5.00 m

\(F=\frac{(mg)(h+d)}{d}\)

Substitute values,

Where, g = 9.80 m/s^2

\(F=\frac{(2100kg)(9.80m/s^2)(0.12m+5.00m)}{0.12m}\)

\(F=\frac{(2100kg)(9.80m/s^2)(5.12m)}{0.12m}\)

\(F= 8.78*10^5N\)

\(F=878KN\)

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Answer:

1950 kpa

Explanation:

9 ÷ .6 = 15 times the speed. so 15 times the pressure 130 x15 = 1950 kpa

Answer:

  none

Explanation:

Newton's first law says an object in motion will stay in motion at the same speed and direction unless acted upon by some force.

No force is necessary for the object to keep its speed and direction on a frictionless surface.

Answer:

3. Ice.

4. Stay the same size but with different internal pressure.

5. Stars in the galaxy

Explanation:

Hope it helped!!

Answer:

unchanged

Explanation:

the box itself has same mass and volume with or without the shoes

(a) The mass of the puck is 0.16 kg

(b)  The amplitude of the oscillation is 4.3 cm.

The mass of the puck is calculated by applying the formula for the period of the oscillation.

T = 2π √ ( m /k )

T / 2π = √ ( m /k )

T² / 4π² = m/k

m = k ( T² / 4π² )

where;

The mass of the puck is calculated as;

m = 4.5 ( 1.2² / 4π²)

m = 0.16 kg

The amplitude of the oscillation is calculated as follows;

a = ω²A

where;

ω = √ ( k /m )

ω² = k / m

from this equation "a = ω²A" make amplitude "A" the subject of the formula.

A = a / ω²

A = (a) / (k / m)

A = ( 1.2 ) / ( 4.5 / 0.16)

A = (1.2) / ( 0.03556)

A = 0.043 m = 4.3 cm

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Answer:

Explanation:

Writing out the Newton's Law pf Cooling:

dT/dt = -k * (T - A),

where T is the temperature of the coffee, A is the room temperature, and k is a positive constant.

If the coffee cools from 100°C to 90°C in 1 minute at a room temperature of 25°C,

T = 100

A = 25

dT = 100 - 90 = 10

dt = 1

Putting the figures into the equation:

10/1 = -k * (100 - 25)

k = -10/75°C

After 4 minutes, dT/4 = 10/75 (100 - 25) = 10

dT = 40

Temperature after 4 minutes = 100 - 40 = 60°C

The temperature of a cup of coffee varies according to Newton's Law of Cooling, the temperature of the coffee after 4 minutes is approximately 67°C.

To tackle this problem, we can apply Newton's Law of Cooling's differential equation and solve it using variable separation.

dT/dt = -k(T - A)

At t = 0 (initial condition): T = 100°C

At t = 1 minute: T = 90°C

dT/dt = -k(T - A)

At t = 0: dT/dt = -k(100 - 25)

So,

-10 = -k(75)

k = 10/75

Separating variables and integrating, we have:

1/(T - A) dT = -k dt

∫(1/(T - A)) dT = ∫(-k) dt

ln|T - A| = -kt + C

ln|100 - 25| = 0 + C

ln|75| = C

So, the equation will be:

ln|T - A| = -kt + ln|75|

ln|(T - 25)/(75)| = -kt

Now,

ln|((T - 25)/(75))| = -(10/75)(4)

|((T - 25)/(75))| = \(e^{(-40/75)\)

T - 25 = ± 75 *  \(e^{(-40/75)\)

T = 25 ± 75 *  \(e^{(-40/75)\)

T ≈ 25 ± 42.42

Therefore, the temperature of the coffee after 4 minutes is approximately:

T ≈ 25 + 42.42 = 67.42°C

Thus, the temperature of the coffee after 4 minutes is approximately 67°C.

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Answer:

Scintillators are materials that are able to convert high energy radiation such as X or gamma-rays to a near visible or visible light. They are widely used as detectors in medical diagnostics, high energy physics and geophysical exploration (ref. Knoll).

https://web.stanford.edu › scintillators

What are scintillator materials? - Stanford: Advanced Optical Ceramics Laboratory

Answer:I don’t know

Explanation:

The work done in the spring is calculated to be 2.8 J

Hooke's law states that, the extension of a given material is directly proportional to the applied force as long as the elastic limit is not exceeded . First, we must bear in mind that the material must remain within the elastic limit for us to apply the Hooke's law in solving the problem.

Now;

From Hooke's law;

F = Ke

F = force applied

K = force constant

e = extension

F = W = mg =  2.50 - kg * 9.8 m/s^2 = 24.5 N

K = 24.5 N/ 2.76 * 10^-2

K = 888 N/m

e = F/K

F = W =  1.25 - kg * 9.8 m/s^2 = 12.25 N

e = 12.25 N/ 888 N/m = 0.014 m or 1.4 cm

Work done by an external agent = 1/2 Kx^2

= 0.5 * 888 * (8 * 10^-2)^2

= 2.8 J

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