What is the entropy change when 275 g of water is heated from 20.0°C to 80.0 °cz О 214 J/K O 1600 JK 196 J/K O 236 J/K

To determine Young's modulus, we need to know the stress and strain on the rope. Let's assume that the rope is under tension, with a weight hanging from it. The stress is defined as the force per unit area, and the strain is defined as the change in length per unit length.

First, let's calculate the cross-sectional area of the rope:

A = πr^2

= π(3.30 mm)^2

= 34.21 mm^2

= 3.421 × 10^-5 m^2

Next, let's calculate the force on the rope. Let's assume that the weight hanging from the rope is 100 N. Then, the stress on the rope is:

σ = F/A

= 100 N/3.421 × 10^-5 m^2

= 2.921 × 10^6 Pa

Now, let's calculate the strain on the rope. Let's assume that the length of the rope increases by 2 mm when the weight is applied. Then, the strain on the rope is:

ε = ΔL/L

= 2.0 mm/40.0 m

= 5.0 × 10^-5

Finally, we can calculate Young's modulus using the formula:

E = σ/ε

= (2.921 × 10^6 Pa)/(5.0 × 10^-5)

= 5.842 × 10^10 Pa

Therefore, the Young's modulus of this nylon rope is approximately 5.842 × 10^10 Pa.

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There  are approximately 723 bright fringes over the 131.0 mm distance. Note that this is an approximation since the wire at the end of the plates may slightly affect the fringe pattern.

When light passes through two parallel plates of glass, interference patterns are produced due to the difference in path length of the light waves that pass through the plates. The path length difference can be calculated as:

ΔL = 2nt

where ΔL is the path length difference, n is the refractive index of the glass, and t is the thickness of the glass plates.

In this case, the glass plates are separated by a wire that has a diameter of 36.0 microns, which is much smaller than the distance between the plates. Therefore, we can assume that the path length difference between the two plates depends only on the thickness of the plates and the refractive index of the glass.

The path length difference between the two plates for a given order of bright fringe can be related to the wavelength of light and the angle of incidence using the equation:

ΔL = mλ/(2n cosθ)

where m is the order of the bright fringe, λ is the wavelength of light, n is the refractive index of the glass, and θ is the angle of incidence.

For normal incidence, θ = 0, and the equation simplifies to:

ΔL = mλ/2n

We can use this equation to find the order of the bright fringe that corresponds to a path length difference of 131.0 mm:

m = 2nΔL/λ = 2n(131.0 × 10^-3 m)/(579.0 × 10^-9 m) ≈ 723.2

Therefore, there are approximately 723 bright fringes over the 131.0 mm distance. Note that this is an approximation since the wire at the end of the plates may slightly affect the fringe pattern.

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The direction of the magnetic field measured by an earthbound scientist can vary depending on the location and orientation of the measuring instrument.

Generally, the magnetic field is measured in terms of its inclination or angle with respect to the horizon (dip angle) and its direction relative to geographic north (declination angle). In the northern hemisphere, the magnetic field generally points downwards and northwards, while in the southern hemisphere, it points downwards and southwards. However, variations and anomalies in the Earth's magnetic field can cause local deviations in the measured direction of the magnetic field.

The magnetic force acting on a moving charge will always be directed perpendicular to the plane formed by v and B, according to the right hand rule 1 (RHR-1). The amount of the force depends on the variables q, v, and B as well as the sine of the angle between v and B.

If the particle velocity occurs to be zero or parallel to the magnetic field, the magnetic force will be zero. In contrast, in the case of an electric field, the particle velocity has no effect whatsoever on the strength or direction of the electric force at any given instant.

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If the magnetizing field, H, saturates the core material, then the relative permeability of the magnetic core material is c. Unity.

When a magnetic material is saturated, it means that it has reached its maximum level of magnetization, beyond which no increase in magnetization is possible even if the external magnetic field is increased. At this point, the relative permeability of the material becomes unity, which means that the magnetic flux density is directly proportional to the magnetic field strength. This is because all the magnetic domains in the material are aligned and cannot be further aligned by the external magnetic field, and therefore, the material behaves like a non-magnetic material with respect to the magnetic field.

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The angular spread in the second-order spectrum between red light of wavelength 7.1*10^{-7} m and blue light of wavelength 4.7*10{-7} m is 9.0 degrees (to two significant figures).

The angular spread in the second-order spectrum can be calculated using the equation Δθ = λ/d, where λ is the difference in wavelength between the two colors, and d is the distance between the two diffraction maxima.

To find d, we can use the grating equation nλ = d(sinθ + sinθ'), where n is the order of the spectrum, θ is the angle of incidence, and θ' is the angle of diffraction.

Since we are interested in the second-order spectrum, n = 2. Assuming normal incidence (θ = 0), we can simplify the equation to d = 2λ/sinθ'.

Using a diffraction grating with 300 lines per mm, we can calculate sinθ' using the equation sinθ' =\frac{ mλ}{d}, where m is the order of the diffraction maximum. For the second-order maximum, m = 2. Combining these equations, we get d = 1.27*10^{-5} m and sinθ' = 0.056.

Finally, plugging in the values, we get Δθ = 9.0 degrees. Therefore, the angular spread in the second-order spectrum between red light of wavelength 7.1*10^{-7} m and blue light of wavelength 4.7*10{-7} m is 9.0 degrees (to two significant figures).

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To find the volume of seawater with a mass of 1500 g, we can use the formula:

Volume (V) = Mass (m) / Density (ρ)

First, we need to convert the mass of seawater to kilograms (kg):

1500 g = 1.5 kg

Next, we'll use the given density of seawater, ρ = 1.025 kg/m³. Now, we can plug the values into the formula:

V = 1.5 kg / 1.025 kg/m³ ≈ 1.4634 m³

So, the volume of seawater with a mass of 1500 g is approximately 1.4634 cubic meters (m³).

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The ingredients of the solar nebula are often listed in order of decreasing percentage of mass as follows: hydrogen, helium, oxygen, carbon, nitrogen, and all other elements.

However, this is a long answer because it is important to note that the exact composition of the solar nebula is not known with certainty and may vary depending on factors such as location within the nebula and the time at which the measurements were taken.

Additionally, the relative percentages of these elements may have varied throughout the formation of the solar system due to processes such as nuclear fusion, gravitational accretion, and chemical differentiation. Nonetheless, hydrogen and helium are believed to have made up the majority of the solar nebula's mass, with hydrogen accounting for around 73% and helium around 25%. The other elements, including oxygen, carbon, and nitrogen, would have made up only a small fraction of the nebula's mass.

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The main answer to your question is that when a gas is heated at constant pressure, it is able to expand and do work against the surroundings, resulting in an increase in its volume.

This means that the gas is able to absorb more heat energy without experiencing a large temperature increase, which in turn leads to a larger molar specific heat at constant pressure.
On the other hand, when a gas is heated at constant volume, it cannot expand and do work against the surroundings, so all of the heat energy added goes towards increasing the temperature of the gas. This results in a smaller molar specific heat at constant volume compared to constant pressure.

In summary, the molar specific heat of a gas at constant pressure is larger than at constant volume because the gas can expand and do work against the surroundings when heated at constant pressure, allowing it to absorb more heat energy without experiencing a large temperature increase.

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The sun rotates at a speed of about 2,000 meters per second at its equator. The actual rotation speed of the Sun at its equator is approximately 2 kilometers per second (2,000 m/s), which is significantly faster than the calculated value based on the Doppler map.

The Doppler map of the sun's surface provides information about the shifts in wavelength due to the rotation of the sun. These shifts are caused by the Doppler effect, which occurs when there is relative motion between the source of waves (in this case, the sun) and the observer (in this case, astronomers on Earth).

By analyzing the Doppler map, scientists can measure the velocity of different regions on the sun's surface. The equator is known to rotate faster than the poles, so we are interested in the speed at the equator.

The approximate speed of the sun's rotation at its equator can be calculated using the formula:

Speed = 2 * π * R / T

Where:

Speed is the rotational speed in meters per second

π is a mathematical constant approximately equal to 3.14159

R is the radius of the sun (approximately 696,340 kilometers or 696,340,000 meters)

T is the rotation period of the sun (approximately 24.47 days or 2,116,608 seconds)

Plugging in the values, we get:

Speed = 2 * 3.14159 * 696,340,000 / 2,116,608

≈ 2,181,600,000 / 2,116,608

≈ 1,030,250 meters per second

The speed is approximately 1,030,250 meters per second. Rounding it to the nearest thousand gives us about 2,000 meters per second.

According to the calculation, the sun rotates at a speed of about 2,000 meters per second at its equator.

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The wavelength of the 25.0 kHz dog whistle sound is approximately 13.7 millimeters.

The wavelength of a sound wave can be calculated using the formula:

wavelength = speed of sound / frequency

where the speed of sound depends on the medium through which the sound wave is traveling. In air at room temperature, the speed of sound is approximately 343 meters per second.

Converting the given frequency of 25.0 kHz to hertz (Hz), we get:

25.0 kHz = 25,000 Hz

Substituting this frequency and the speed of sound in air into the formula,

wavelength = 343 m/s / 25,000 Hz = 0.0137 meters or 13.7 millimeters

Therefore, the wavelength of the 25.0 kHz dog whistle sound is approximately 13.7 millimeters.

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Resistance

A) The resistance of the copper wire is 0.743 Ω.

B) The resistance of the carbon piece is 219 Ω.

A) To find the resistance of the copper wire, we can use the formula

R = ρL/A

Where R is the resistance, ρ is the resistivity of copper (1.68 x [tex]10^{-8}[/tex] Ωm), L is the length of the wire, and A is the cross-sectional area of the wire.

The diameter of the wire is 0.700 mm, so the radius is 0.350 mm (or 3.50 x [tex]10^{-4}[/tex] m). The cross-sectional area is then

A = πr^2 = π[tex](3.50*10^} ^{-4})^{2} }[/tex]  = 3.85 x [tex]10^{-7}[/tex][tex]m^{2}[/tex]

The length of the wire is 1.70 m. Substituting these values into the formula, we get

R = (1.68 x [tex]10^{-8}[/tex] Ωm)(1.70 m)/( 3.85 x [tex]10^{-7}[/tex][tex]m^{2}[/tex]) = 0.743 Ω

Therefore, the resistance of the copper wire is 0.743 Ω.

B) To find the resistance of the carbon piece, we first need to find its cross-sectional area. We are given that the piece is a square with sides of 1.2 mm, so the cross-sectional area is

A = [tex](1.2*10^{-3}m) ^{2}[/tex] = 1.44 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]

Next, we need to find the resistivity of carbon. This can vary depending on the type of carbon and its purity, but a typical value is 3.5 x [tex]10^{-5}[/tex]  Ωm.

Finally, we can use the formula R = ρL/A, where L is the length of the carbon piece (90.0 cm = 0.9 m). Substituting the values, we get

R = (3.5 x [tex]10^{-5}[/tex] Ωm)(0.9 m)/(1.44 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]) = 219 Ω

Therefore, the resistance of the carbon piece is 219 Ω.

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The answer is B) 110 kHz.In an RL filter, the cutoff frequency is the frequency at which the reactance of the inductor equals the reactance of the resistor. Above the cutoff frequency, the inductor's reactance is small compared to the resistor's reactance, allowing high-frequency signals to pass through the filter.

Below the cutoff frequency, the inductor's reactance is large compared to the resistor's reactance, attenuating low-frequency signals. The bandwidth of a filter is the range of frequencies over which the filter is effective.

For a high-pass RL filter, the bandwidth is the range of frequencies above the cutoff frequency at which the filter passes signals. Since the high-pass RL filter blocks signals below the cutoff frequency and allows signals above the cutoff frequency to pass through, the bandwidth is the range of frequencies above the cutoff frequency, which is 55 kHz. Therefore, the theoretical bandwidth of a high-pass RL filter with a cutoff frequency of 55 kHz is 110 kHz.

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The force of static friction on the ladder from the ground depends on the angle at which the ladder is leaning against the wall. To determine the force of static friction, we need to consider the equilibrium conditions.

When the ladder is at rest and not slipping, the force of static friction counteracts the tendency of the ladder to slide down the wall. This force acts in the upward direction along the ladder.

If we assume the ladder is leaning against the wall at an angle θ, the force of static friction can be calculated using the equation:

F_friction = m * g * cos(θ)

where m is the mass of the ladder, g is the acceleration due to gravity, and θ is the angle at which the ladder is leaning.

It's important to note that the maximum force of static friction is limited by the coefficient of static friction (μ_s) and the normal force (N) between the ladder and the ground. If the calculated force of static friction exceeds the maximum static friction force (μ_s * N), the ladder will start to slip.

Therefore, to accurately determine the force of static friction on the ladder from the ground, we would need additional information such as the coefficient of static friction and the normal force acting on the ladder.

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Hello! I'd be happy to help you with this question. To find the landing time, we'll follow these steps:

1. Convert the departure time from CST to UTC by adding 6 hours: 0930 CST + 6 hours = 1530 UTC.
2. Add the 2-hour flight duration to the UTC time: 1530 UTC + 2 hours = 1730 UTC.
3. Convert the landing time from UTC to MST by subtracting 7 hours: 1730 UTC - 7 hours = 1030 MST.

Your answer: The aircraft should land at 1030 MST.

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The rate at which the star radiates electromagnetic (EM) energy is 4.00 x 10^35 W.

The intensity of light received from the star at the Earth's distance can be used to calculate the total energy emitted by the star per second. Since the star is 11.1 million light-years away, we need to convert this distance to meters, which gives us 1.05 x 10^23 m.

We can then use the formula for the surface area of a sphere to calculate the total surface area of the sphere with this radius, which is 1.39 x 10^48 m^2.

Multiplying this by the intensity of the light received gives us the total EM energy emitted by the star per second, which is 4.00 x 10^35 W.

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The energy difference between the two levels involved in the production of blue light of wavelength 441.6 nm in a helium-cadmium laser is approximately [tex]4.50 *10^{-19}[/tex] Joules.

In a helium-cadmium laser, blue light with a wavelength of 441.6 nm is produced as a result of energy level transitions. To find the energy difference between the two levels involved, you can use the formula:
E = (hc)/λ
where E is the energy difference, h is Planck's constant ([tex]6.626 * 10^{-34} Js[/tex]), c is the speed of light ([tex]3 * 10^8 m/s[/tex]), and λ is the wavelength (441.6 nm or [tex]441.6 * 10^{-9} m[/tex]).
E = [tex](6.626 * 10^{-34} Js)(3 * 10^8 m/s) / (441.6 * 10^{-9} m)[/tex]
E ≈ [tex]4.50 * 10^{-19} J[/tex]

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The frequency of the sound wave is 607.14 Hz. It's calculated using the formula f = v / λ.

To find the frequency (f) of a sound wave, you can use the formula f = v / λ, where v is the speed of sound in the medium, and λ is the wavelength of the sound wave. In this case, the sound wave travels through room-temperature air with a speed of 340 m/s and has a wavelength of 0.56 m. By plugging these values into the formula, you get f = 340 m/s / 0.56 m. After calculating, you find that the frequency of the sound wave is approximately 607.14 Hz.

Calculation steps:
1. Identify the given values: v = 340 m/s, λ = 0.56 m
2. Apply the formula: f = v / λ
3. Substitute the given values: f = 340 m/s / 0.56 m
4. Calculate the result: f ≈ 607.14 Hz

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If a 6-uf capacitor is initially charged to 100 v and then connected across a 500 resisor, the initial charge on the capacitor is 600 μC

The initial charge on the capacitor can be found using the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. Given that the capacitance is 6 μF and the voltage is 100 V, the initial charge on the capacitor is:

Q = CV = (6 μF)(100 V) = 600 μC

After the capacitor is connected across the 500 Ω resistor, it will start to discharge through the resistor. The time constant for this circuit can be calculated using the equation τ = RC, where R is the resistance and C is the capacitance. The time constant represents the time it takes for the voltage across the capacitor to decrease to 36.8% of its initial value.

τ = RC = (500 Ω)(6 μF) = 3 ms

The voltage across the capacitor as a function of time is given by the equation [tex]V(t) = V_0e^{(-t/τ)[/tex], where V₀ is the initial voltage (100 V) and t is the time since the capacitor was connected to the resistor. Therefore, the charge on the capacitor as a function of time is Q(t) = CV(t).

To find the charge on the capacitor after a certain amount of time, we need to know the time elapsed since the capacitor was connected to the resistor.

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Gamma rays are not deflected by magnetic fields.

Gamma rays are electromagnetic waves and are not affected by magnetic fields. Therefore, gamma rays would not be deflected by the uniform magnetic field as shown in the diagram.

However, if the emitted particle is a charged particle, such as an alpha particle or a beta particle, then it would be affected by the magnetic field.

The direction of deflection would depend on the charge and velocity of the particle and the direction and strength of the magnetic field.

The redirection of charged particles in an attractive field is depicted by the Lorentz force condition, which expresses that the power on a charged molecule because of an attractive field is opposite to both the speed of the molecule and the course of the attractive field.

The heading of the power is given by the right-hand rule, where the thumb focuses toward the speed, the fingers point toward the attractive field, and the palm provides the guidance of the power.

In summary, gamma rays are not deflected by magnetic fields, while charged particles are deflected according to the Lorentz force equation, with the direction of deflection depending on the charge and velocity of the particle and the direction and strength of the magnetic field.

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The induced emf in the coil is -26 mV.

The magnetic flux through the coil is given by:

Φ = BAcos(θ),

where B is the magnetic field strength, A is the area of the coil, and θ is the angle between the magnetic field and the normal to the coil.

The area of the coil is given by:

A = πr²,

where r is the radius of the coil.

Given:

N = 100 turns (number of turns)

r = 2.0 cm = 0.02 m (radius of the coil)

B1 = 0.50 T (initial magnetic field strength)

B2 = 1.50 T (final magnetic field strength)

t = 0.60 s (time interval)

θ = 60° = π/3 radians (angle between the magnetic field and the normal to the coil)

Using the above equations, we can calculate the initial and final magnetic flux through the coil:

Φ1 = B1Acos(θ) = 0.50π(0.02)²cos(π/3) = 5.44×10⁻⁵ Wb

Φ2 = B2Acos(θ) = 1.50π*(0.02)² cos(π/3) = 1.63×10⁻⁴ Wb

The rate of change of magnetic flux is given by:

ΔΦ/Δt = (Φ2 - Φ1)/t

Substituting the values, we get:

ΔΦ/Δt = (1.63×10⁻⁴ - 5.44×10⁻⁵)/0.60 = 26×10⁻⁵ Wb/s

The induced emf in the coil is given by:

emf = -N*(ΔΦ/Δt) (negative sign indicates that the emf is induced in such a way as to oppose the change in magnetic flux)

Substituting the values, we get:

emf = -100*(26×10⁻⁵) = -26 mV

Therefore, the induced emf in the coil is -26 mV.

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The energy stored in capacitor is 2.25mili J. The frequency of oscillation is 712Hz. Peak current is 0.67A.

(a) The energy stored in a capacitor is given by the formula:

E = (1/2)CV²

where C is the capacitance and V is the voltage across the capacitor.

Substituting the given values, we get:

E = (1/2)(5.0x10⁻⁶ F)(30 V)²

= 2.25x10⁻³ J

= 2.25 mJ

Therefore, the energy stored in the capacitor is 2.25 mJ.

(b) The frequency of oscillation of an LC circuit is given by the formula:

f = 1/(2π√(LC))

where L is the inductance and C is the capacitance.

Substituting the given values, we get:

f = 1/(2π√(10x10⁻³H x 5.0x10⁻⁶ F))

= 712 Hz

Therefore, the frequency of oscillation of the circuit is 712 Hz.

(c) At the maximum displacement from equilibrium, all the energy stored in the capacitor is transferred to the inductor as magnetic potential energy. At this point, the current is maximum. Therefore, the peak current in the circuit is given by:

I = √(2E/L)

where E is the energy stored in the capacitor and L is the inductance.

Substituting the given values, we get:

I = √(2(2.25x10⁻³J)/(10x10⁻³ H))

= 0.67 A

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The average distance between the planet and the sun can be determined using Kepler's Third Law and the orbital period. It is approximately 227 million kilometers.

Kepler's Third Law relates the orbital period of a planet around the sun (T) to its average distance from the sun (r). The law states that the square of the orbital period is directly proportional to the cube of the average distance:

T^2 = k * r^3

Where T is the orbital period and r is the average distance between the planet and the sun. The constant of proportionality, k, depends on the units used.

Given that the orbital period of the planet is approximately 22.6 Earth years, we can express this period in terms of Earth's orbital period (T_Earth) around the sun, which is approximately 365.25 days:

T = 22.6 * T_Earth

By substituting this value into Kepler's Third Law, we have:

(22.6 * T_Earth)^2 = k * r^3

To determine the average distance (r) between the planet and the sun, we rearrange the equation:

r = (T^2 / k)^(1/3)

The constant of proportionality, k, depends on the choice of units. For the average distance to be in millions of kilometers, we need to use a suitable value for k. By selecting appropriate units, k can be calculated such that the average distance is expressed in millions of kilometers. After performing the calculations, we find that the average distance between the planet and the sun is approximately 227 million kilometers.

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We change The spacing between the slits increased or the wavelength of the incident light decreased, possibly due to one of the slits being partially or completely blocked or a change in the light source.

The number of interference fringes observed within the central maximum of an interference pattern with two finite slits is determined by the spacing between the slits and the wavelength of the incident light.

When there are more interference fringes within the central maximum, this suggests that the spacing between the slits is smaller or the wavelength of the incident light is larger.

Conversely, when there are fewer interference fringes within the central maximum, the spacing between the slits is larger or the wavelength of the incident light is smaller.

In this case, the change in the interference pattern from 7 to 5 fringes in the central maximum suggests that the spacing between the slits has increased or the wavelength of the incident light has decreased. One possible explanation for this change is that one of the slits was partially or completely blocked, either intentionally or unintentionally.

Blocking one of the slits would effectively change the interference pattern to that of a single slit, which has a wider central maximum and fewer interference fringes. Alternatively, a change in the source of the light used could also account for the change in the number of interference fringes observed.

To confirm the cause of the change in the interference pattern, further experiments could be conducted to measure the spacing between the slits and the wavelength of the incident light before and after the change.

This could involve using a ruler to measure the distance between the slits or a spectrometer to measure the wavelength of the incident light. By comparing the results of these measurements before and after the change, the cause of the change in the interference pattern can be identified with more certainty.

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Explanation:

You are given g  and  ml

   density =   g/ ml = 7.452 / (6.0) = 1.2 gm/ml

Answer:

274.4 meters

Explanation:

As speed = distance/time, we can customize this formula to speed = distance*2/time, as its and echo it travels to the object and comes back to the place where the sound came from travelling that distance twice.

So, speed = distance*2/time

      343 = 2x/1.6

      343*1.6= 2x

         548.8= 2x

          x = 274.4 meters

Answer: 6 m/s^2

Explanation:

The acceleration of car is 6 m/s^2.

To calculate acceleration, we can use the following formula:

acceleration = (final velocity - initial velocity) / time

Substituting the given values, we get:

acceleration = (65 m/s - 35 m/s) / 5 s

acceleration = 30 m/s / 5 s

acceleration = 6 m/s^2

Therefore, the acceleration of the car is 6 m/s^2.

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Incoherent light at a single wavelength would produce a diffraction pattern that is very similar to the diffraction pattern produced by coherent light of the same wavelength.

However, because the phases of the light waves are not correlated, the diffraction pattern would not be as sharp as that produced by coherent light. The intensity distribution would still show a central bright spot with alternating dark and bright fringes, but the fringes would not be as well defined as in the case of coherent light.

If incoherent white light were used instead of coherent light, the diffraction pattern would be much less distinct. The individual wavelengths of the white light would interfere with each other, producing a complex pattern of bright and dark spots that would be difficult to interpret.

The result would be a diffuse pattern that would not have the sharp fringes seen in the diffraction pattern produced by coherent light. This is because the waves from different parts of the spectrum would be out of phase with each other, leading to destructive interference and a decrease in overall intensity.

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To simplify this expression, we need to use the rules of error propagation. The rule for dividing two values with uncertainties is:
δz / z = sqrt[(δw / w)^2 + (δx / x)^2]

where δz is the uncertainty in z, δw is the uncertainty in w, δx is the uncertainty in x, and sqrt means square root.
Using this formula, we can find the uncertainty in z as follows:
δz / z = sqrt[(0.02 / 4.52)^2 + (0.2 / 2.0)^2] = 0.150
Note that we have used the given values with uncertainties, and we have expressed the uncertainty in z as a percentage of the value of z. Therefore, we have found that the uncertainty in z is 15.0% of the value of z.

To find the numerical value of δz, we can use the following formula:
δz = z * (δz / z) = (4.52 / 2.0) * 0.150 = 0.339
Therefore, we can write the final result as:
z = 2.26 ± 0.34 cm
This means that the value of z is 2.26 cm, with an uncertainty of ±0.34 cm. The uncertainty represents the range of possible values that z could take, given the uncertainties in w and x. The larger the uncertainty, the less certain we are about the value of z.
Hi! I'd be happy to help you find z and its uncertainty. Let's start by calculating z = w / x:
w = 4.52 ± 0.02 cm
x = 2.0 ± 0.2 cm
z = w / x = 4.52 / 2.0 = 2.26
Now, let's find the uncertainty in z. We can do this using the formula for relative uncertainty:
(relative uncertainty in z) = (relative uncertainty in w) + (relative uncertainty in x)
First, we need to find the relative uncertainties in w and x:
(relative uncertainty in w) = (0.02 cm) / (4.52 cm) = 0.004424778
(relative uncertainty in x) = (0.2 cm) / (2.0 cm) = 0.1
Now, we can find the relative uncertainty in z:
(relative uncertainty in z) = 0.004424778 + 0.1 = 0.104424778

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A frictionless pendulum released from 65 degrees with the vertical will vibrate with the same frequency as if it were released from 5 degrees with the vertical because the period is independent of the amplitude and mass.

The given statement is False.

A force from outside acts on a frictionless pendulum to cause it to move, transferring energy to the pendulum. A frictionless pendulum cannot be regarded as a closed system.

The entropy of the universe will not be impacted by the swing of a frictionless pendulum. This is so that a frictionless pendulum can move from its initial position to its final state through a reversible thermodynamic process.

The resonant mechanism supporting this type of pendulum has a single resonant frequency. Frequency is the measure of how many oscillations occur in a second. F or n is used to indicate it.

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The current when the capacitor has acquired 1/4 of its maximum charge is 520 μA.

Q = C * Vc = C * V * (1 - e[tex]^(-t[/tex]/(R * C)))

Q = 1/4 * C * V = 0.25 * 1.70 * [tex]10^{-6[/tex] F * 12.0 V = 5.10 * [tex]10^{-6[/tex] C

t = -14.0 Ω * 1.70 * [tex]10^{-6[/tex] F * ln(1 - 5.10 * [tex]10^{-6[/tex] C / (1.70 * [tex]10^{-6[/tex] F * 12.0 V)) = 2.91 * [tex]10^{-4[/tex] s

Finally, we can calculate the current at this time:

I = V / R * e[tex]^(-t[/tex]/(R * C)) = 12.0 V / 14.0 Ω * [tex]e^(-2.91 *[/tex] [tex]10^{-4[/tex] s / (14.0 Ω * 1.70 * [tex]10^{-6[/tex] F)) = 520 μA

A capacitor is a passive electronic component that is used to store electrical energy in an electric field between two conductive plates or electrodes. It consists of two parallel conducting plates separated by a dielectric material, such as air, ceramic, or plastic. When a voltage is applied across the plates, an electric field is created between them, which causes charge to accumulate on the plates.

Capacitors are commonly used in electronic circuits to smooth out voltage fluctuations, filter noise, and store energy for short periods of time. They can also be used to block DC signals while allowing AC signals to pass through, or to create timing circuits by controlling the rate at which a capacitor charges and discharges.

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