List what sources of uncertainty go into calculating the wavelength of the laser (no explanation necessary here). (b) Accurately report the uncertainties for these quantities. (c) Explain which of these contributes the most to the final uncertainty on the laser wavelength

Answers

Answer 1

Answer:

thanks for da 5points hoi

Explanation: thanks dawg

Answer 2

There can be uncertainty in calculating the wavelength of a laser light  due to experimental errors

All measurements have an uncertainty, in the case of direct measurements the uncertainty is equal to the precision of the given instrument.

What are uncertainity in measuring ?

Uncertainty  means the range of possible values within which the true value of the measurement lies.

What are errors?

The deviation  in the value of the measured quantity from the actual quantity or true value is called an error

(a) For the calculation of wavelength of laser light , the sources which can lead to uncertainty are

1. least count of measuring instruments like spectrometer or interferometer

2. Parallax error in the measurement

3. Error in identifying the order of fringes

4.. unable to identify the accurate  reading of Vernier or circular scales present in the measuring instruments.

5. Propagating errors

What is least count?

The least count of a measuring instrument is the smallest and accurate value in the measured quantity that can be measured by instrument.

What is propagating error?

When you have derived variables, that is, when measurements are made with different instruments, each with a different uncertainty, the way to find the uncertainty or error is that  all the errors add up. which increases the uncertainty

b. The uncertainty in measurement due to  least count depends on the instrument used for measurement f wavelength. A  Michelson's

interferometer has the least count of .0001mm. whereas spectrometer has a least count of 0.5⁰. Hence uncertainty in the measurement by Michelson's interferometer is very less as compared to any other instrument.

C. The maximum uncertainty arises due to the least count , as all other errors can be minimized by taking an average value of many observations but the least count of an instrument do not change so uncertainty within the least count arises.

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Related Questions

a. Do the waves made by the two faucets travel faster than the waves made by just one faucet?
b. How do you know this? Describe how the two-faucet wave pattern compares with the one-faucet pattern.
c. Describe what happens to the two-faucet wave pattern as the separation of the faucets is increased.

Answers

Answer:

asdasd dsa dasdasd sadas dasd asdasd asd asd dsa asdd 223 aasd ada dasd sa dasd dsaa sd adsd asasd

Explanation:

A very long, straight solenoid with a diameter of 3.00 cm is wound with 40 turns of wire per centimeter, and the windings carry a current of 0.235 A. A second coil having N turns and a larger diameter is slipped over the solenoid so that the two are coaxial. The current in the solenoid is ramped down to zero over a period of 0.40 s.

Required:
a. What average emf is induced in the second coil if it has a diameter of 3.5 cm and N = 7?
b. What is the induced emf if the diameter is 7.0 cm and N = 10?

Answers

Answer:

a) ε = 14.7 μv

b) ε = 21 μv

Explanation:

Given the data in the question;

Diameter of solenoid; d = 3 cm

radius will be half of diameter,  so, r = 3 cm / 2 = 1.5 cm = 1.5 × 10⁻² m

Number of turns; N = 40 turns per cm = 4000 per turns per meter

Current; [tex]I[/tex] = 0.235 A

change in time Δt = 0.40 sec

Now,

We determine the magnetic field inside the solenoid;

B = μ₀ × N × [tex]I[/tex]  

we substitute

B = ( 4π × 10⁻⁷ Tm/A ) × 4000 × 0.235  

B = 1.1881 × 10⁻³ T

Now, Initial flux through the coil is;

∅₁ = NBA = NBπr²  

and the final flux  

∅₂ = 0        

so, the εmf induced ε = -Δ∅/Δt = -( ∅₂ - ∅₁ ) / Δt

= -( 0 - NBπr² ) / Δt  

= NBπr² / Δt    

a)

for N = 7

ε = [ 7 × ( 1.1881 × 10⁻³ ) × π( 1.5 × 10⁻² )² ] / 0.40

ε = 14.7 × 10⁻⁶ v

ε = 14.7 μv      

b)

for N = 10

ε = [ 10 × ( 1.1881 × 10⁻³ ) × π( 1.5 × 10⁻² )² ] / 0.40

ε = 21 × 10⁻⁶ v

ε = 21 μv  

 

An airplane which intends to fly due south at 250 km/hr experiences a wind blowing westward at 40 km/hr. What is the actual speed of the airplane relative to the ground? ​

Answers

Answer:

simple is rumple a daily ok I'll be

A gymnast weighs 450 N. She stands on a balance beam of uniform construction which weighs 250 N. The balance beam is 3.0 m long and is supported at each end. If the support force at the right end is four times the force at the left end, how far from the right end is the gymnast

Answers

Answer:

   x = 9.32 cm

Explanation:

For this exercise we have an applied torque and the bar is in equilibrium, which is why we use the endowment equilibrium equation

Suppose the counterclockwise turn is positive, let's set our reference frame at the left end of the bar

          - W l / 2 - W_{child} x + N₂ l = 0

             x = [tex]\frac{-W l/2 + n_2 l}{W_{child}}[/tex]             1)

now let's use the expression for translational equilibrium

         N₁ - W - W_(child) + N₂ = 0

indicate that N₂ = 4 N₁

we substitute

           N₁ - W - W_child + 4 N₁ = 0

           5 N₁ -W - W_{child} = 0

           N₁ = ( W + W_{child}) / 5

         

we calculate

           N₁ = (450 + 250) / 5

          N₁ = 140 N

           

we calculate with equation 1

           x = -250 1.50 + 4 140 3) / 140

           x = 9.32 cm

An auto mechanic needs to determine the emf and internal resistance of an old battery. He performs two measurements: in the first, he applies a voltmeter to the battery's terminals and reads 11.9 V;11.9 V; in the second, he applies an ammeter to the terminals and reads 16.1 A.16.1 A.
What are the battery's emf E and internal resistance r?

Answers

Answer:

Hence the battery's emf E is ε = 11.9 V.

The internal resistance is r = 0.739 ohms.

Explanation:

Now we know that

Voltage V = 11.9 V.

Current I = 16.1 A.

Hence this is an ideal voltmeter there are no current flows when the Voltmeter is applied.

ε = V + I r

∵ I = 0

ε = V

ε = 11.9 V

Then the ammeter is applied.

Let's take ( r ) to be the total resistance which is equal to internal resistance.

V = I r

r = [tex]\frac{V}{I}[/tex]

 [tex]= \frac{11.9}{16.1}[/tex]

r = 0.739 ohms

The battery's emf (E) and internal resistance (r) are 11.9 Volts and 0.739 Ampere respectively.

Given the following data:

Voltage = 11.9 Volts.Current = 16.1 Amperes.

To determine the battery's emf (E) and internal resistance (r):

How to calculate emf (E).

For an ideal voltmeter, there isn't a flow of current and as such the current is equal to 0.

Mathematically, emf (E) is given by this formula:

[tex]E = V + IR[/tex]

Substituting the given parameters into the formula, we have;

[tex]E = 11.9 + 0R\\\\E = 11.9 + 0[/tex]

E = 11.9 Volts.

For the internal resistance (r):

Note: The total resistance is equal to internal resistance.

Applying Ohm's law, we have:

[tex]R = \frac{V}{I} \\\\R = \frac{11.9}{16.1}[/tex]

R = r = 0.739 Ampere.

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Simple Pendulum: A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing

Answers

Answer:

The correct answer is "4.443 sec".

Explanation:

Given:

Mass of child,

= 34 kg

Mass of swing,

= 18 kg

Length,

= 4.9 m

The time period of pendulum will be:

T = [tex]2 \pi \sqrt{4g}[/tex]

  = [tex]2 \pi \sqrt{\frac{4.9}{9.8} }[/tex]

  = [tex]4.443 \ sec[/tex]  

Answer:

The time taken to back and forth is 4.4 s .

Explanation:

Length, L = 4.9 m

let the time period is T.

Acceleration due to gravity, g = 9.8 m/s^2

Use the formula of time period

[tex]T = 2 \pi\sqrt{L}{g}\\\\T = 2 \times 3.14\sqrt{4.9}{9.8}\\\\T = 4.4 s[/tex]

define nortons theorem​

Answers

Answer:

In direct-current circuit theory, Norton's theorem is a simplification that can be applied to networks made of linear time-invariant resistances, voltage sources, and current sources. At a pair of terminals of the network, it can be replaced by a current source and a single resistor in parallel.

Physics question plz help ASAP

Answers

Answer:

Option D.

Explanation:

From the question given above, the following data were obtained:

Force applied (F) = 5 N

Extention (e) = 0.075 m

Spring constant (K) =?

The spring constant for the spring can be obtained as follow:

F = Ke

5 = K × 0.075

Divide both side by 0.075

K = 5 / 0.075

K = 67 N/m

Thus, the spring constant for the spring is 67 N/m

(a) Calculate the force needed to bring a 800 kg car to rest from a speed of 85.0 km/h in a distance of 115 m (a fairly typical distance for a nonpanic stop).
N
(b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a), i.e. find the ratio of the force in part(b) to the force in part(a).
(force in part (b) / force in part (a))

Answers

Answer:

2Al + 2H2O + 2NaOH ⟶ 3H2 + 2NaAlO2

Chất rắn màu xám bạc của nhôm (Al) tan dần trong dung dịch, sủi bọt khí là hidro (H2).

Explanation:

What is the largest known star?

Answers

Answer:

UY Scuti is slightly larger than VY Canis Majoris

Explanation:

These stars are millions of miles away and cannot be seen by the naked eye.

Beetlejuice is another large star that can be seen by the eye.

A silicon solar cell behaves like a battery with a 0.46 V terminal voltage. Suppose that 1.0 W of light of wavelength 620 nm falls on a solar cell and that 50%% of the photons give their energy to charge carriers, creating a current. What is the solar cell's efficiency that is, what percentage of the energy incident on the cell is converted to electric energy?

Answers

We have that the percentage of the energy incident on the cell that is converted to electric energy is

[tex]n=11\%[/tex]

From the question we are told that:

Voltage [tex]V=0.46V[/tex]

Power of light [tex]P=1.0W[/tex]

Wavelength [tex]w=620nm[/tex]

50 \% of the photons give their energy to charge carriers,

Generally, the equation for number of Protons is mathematically given by

[tex]N_p=\frac{P}{E}[/tex]

[tex]N_p=\frac{P \lambda}{hc}[/tex]

[tex]N_p=\frac{1}{(6.62*10^(-34)}*\frac{3*10^8}{(570*10^{-9}))}[/tex]

[tex]N_p=2.87*10^{18}[/tex]

Generally, the equation for Number of electron is mathematically given by

[tex]N_e=50 \% *n_3[/tex]

[tex]N_e=0.5*2.87*10^{18}[/tex]

[tex]N_e=1.43*10^{18}[/tex]

Therefore

Total current

[tex]I= e*N_e[/tex]

Where

e=electron Charge

Therefore

[tex]I=1.43*10^{18}*1.6*10^-{19}[/tex]

[tex]I=0.230A[/tex]

Generally, the equation for Power is mathematically given by

[tex]P=VI[/tex]

[tex]P=0.46*0.230[/tex]

[tex]P=0.1058W[/tex]

Therefore

Efficiency

[tex]n=\frac{0.1058}{1}[/tex]

[tex]n=0.1058[/tex]

[tex]n=11\%[/tex]

In conclusion

The percentage of the energy incident on the cell that is converted to electric energy is

[tex]n=11\%[/tex]

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The instrument includes a light source, which is passed through a Choose... , which isolates a single wavelength to pass through an aperture to reach the Choose... . Then, the light travels to the Choose... , which measures the intensity of light reaching it.

Answers

Answer:

Following are the response to the given question:

Explanation:

It's being used to measure the amount of light absorbed after traveling through a test tube (the amount of solar radiation received). For several quantitative estimations, this technique is widely employed. Spectrometer and Spectrometer were two devices that are used together to light intensity and light intensity.

It creates and diffuses phosphorescent light into the selected frequency, while the Spectrometer measures the strength of attenuation by the sample solution.

Diffraction beams or prisms are being used to convert polychromatic illumination into monochrome light.

Afterward, the sunlight has a certain hue. Once it reaches the specimen cuvette, it begins absorption. It falls on a sensor that transforms its intensity into such an electronic current.

Here are some ways to fill in such gaps:

In order to reach the specimen cuvette, the light from the light source must be routed via an aperture in order to be isolated by either a diffraction pattern. Light travels to the detector, which detects its intensity.

A rock is suspended by a light string. When the rock is in air, the tension in the string is 37.8 N. When the rock is totally immersed in water, the tension is 32.0 N. When the rock is totally immersed in an unknown liquid, the tension is 20.2 N. What is the Density of the unknown liquid?

Answers

When the rock is suspended in the air, the net force on it is

F₁ = T₁ - m₁g = 0

where T₁ is the magnitude of tension in the string and m₁g is the rock's weight. So

T₁ = m₁g = 37.8 N

When immersed in water, the tension reduces to T₂ = 32.0 N. The net force on the rock is then

F₂ = T₂ + B₂ - m₁g = 0

where B₂ is the magnitude of the buoyant force. Then

B₂ = m₁g - T₂ = 37.8 N - 32.0 N = 5.8 N

B₂ is also the weight of the water that was displaced by submerging the rock. Let m₂ be the mass of the displaced water; then

5.8 N = m₂g   ==>   m₂ ≈ 0.592 kg

If one takes the density of water to be 1.00 g/cm³ = 1.00 × 10³ kg/m³, then the volume of water V that was displaced was

1.00 × 10³ kg/m³ = m₂/V   ==>   V ≈ 0.000592 m³ = 592 cm³

and this is also the volume of the rock.

When immersed in the unknown liquid, the tension reduces further to T₃ = 20.2 N, and so the net force on the rock is

F₃ = T₃ + B₃ - m₁g = 0

which means the buoyant force is

B₃ = m₁g - T₃ = 37.8 N - 20.2 N = 17.6 N

The mass m₃ of the liquid displaced is then

17.6 N = m₃g   ==>   m₃ ≈ 1.80 kg

Then the density ρ of the unknown liquid is

ρ = m₃/V ≈ (1.80 kg)/(0.000592 m³) ≈ 3040 kg/m³ = 3.04 g/cm³

The voltage in an EBW operation is 45 kV. The beam current is 50 milliamp. The electron beam is focused on a circular area that is 0.50 mm in diameter. The heat transfer factor is 0.87. Calculate the average power density in the area in watt/mm2.

Answers

Answer:

[tex]P_d=6203.223062W/mm^2[/tex]

Explanation:

From the question we are told that:

Voltage [tex]V=45kV[/tex]

Current [tex]I=50mAmp[/tex]

Diameter  [tex]d=0.50mm[/tex]

Heat transfer factor [tex]\mu= 0.87.[/tex]

Generally the equation for  Power developed is mathematically given by

[tex]P=VI\\\\P=45*10^3*50*10^{-3}[/tex]

[tex]P=2.250[/tex]

Therefore

Power in area

[tex]P_a=1400*0.87[/tex]

[tex]P_a=1218watt[/tex]

Power Density

[tex]P_d=\frac{P_a}{Area}[/tex]

[tex]P_d=\frac{1218}{\pi(0.5^2/4)}[/tex]

[tex]P_d=6203.223062W/mm^2[/tex]

A rectangular loop of wire with sides 0.129 and 0.402 m lies in a plane perpendicular to a constant magnetic field (see part a of the drawing). The magnetic field has a magnitude of 0.888 T and is directed parallel to the normal of the loop's surface. In a time of 0.172 s, one-half of the loop is then folded back onto the other half, as indicated in part b of the drawing. Determine the magnitude of the average emf induced in the loop.

Answers

Answer:

[tex]0.2677\ \text{V/m}[/tex]

Explanation:

A = Area of loop = [tex]0.129\times0.402[/tex]

B = Magnetic field = [tex]0.888\ \text{T}[/tex]

t = Time taken = [tex]0.172\ \text{s}[/tex]

Electric field is given by

[tex]E=B\dfrac{dA}{dt}\\\Rightarrow E=0.888\times\dfrac{0.129\times 0.402}{0.172}\\\Rightarrow E=0.2677\ \text{V/m}[/tex]

The emf induced is [tex]0.2677\ \text{V/m}[/tex].

Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An electron at rest at point 1 is accelerated by the electric field to point 2.

Required:
Write an equation for the change of electric potential energy ΔU of the electron in terms of the symbols given.

Answers

Answer:

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

So, substituting the values of the variables into the equation, we have

ΔU = eΔV

ΔU = eΔV

ΔU = -1.602 × 10⁻¹⁹ C × 142 V

ΔU = -227.484 × 10⁻¹⁹ J

ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in diameter and has 1500 turns. When turned on, the current in the solenoid is increases linearly to 20 A in 1 second. What is the induced emf in the ring?
a) 2.0 x 10-5 v
b) 3.8 x 10-5 v
c) 1.2 x 10-3 v
d) 1.9 x 10-4 v

Answers

Answer:

the answer should be b) 3.8 x 10-5 v

I NEEED HELP IN PHYSICS PLEASE!

Answers

Answer:

in which topic you need help

Hai quả cầu kim loại nhỏ giống nhau, mang điện tích q1 = 2.10-8 C; q2 = 6.10-8 C, đặt cách

nhau một đoạn r trong không khí thì chúng đẩy nhau bằng một lực là 18.10-5 N. Cho hai quả cầu

tiếp xúc nhau rồi đưa về khoảng cách cũ thì lực tương tác giữa hai quả cầu là:

Answers

Answer:

fgggggffgggcffghhhjjkuuu of to ok with queen size of your yyyygtyttttttyyyhgggghhhhfrghjjkkk

Explanation:

jjjgggyuuuuiiii hhjiiihuyyuuugggyujjhhhhggghhhjjhhhhjhhui

Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is 60 cm2, determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle

Answers

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

a) The mass flow rate through the nozzle can be calculated with the following equation:

[tex] \dot{m_{i}} = \rho_{i} v_{i}A_{i} [/tex]

Where:

[tex]v_{i}[/tex]: is the initial velocity = 20 m/s

[tex]A_{i}[/tex]: is the inlet area of the nozzle = 60 cm²  

[tex]\rho_{i}[/tex]: is the density of entrance = 2.21 kg/m³

[tex] \dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s [/tex]  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

[tex] \rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f} [/tex]

[tex] 0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f} [/tex]

[tex] A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2} [/tex]

Therefore, the exit area of the nozzle is 23.6 cm².

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a) Mass flow rate through the nozzle: 0.265 kilograms per second, b) Exit area of the nozzle: 23.202 square centimeters.

We determine the Mass Flow Rate through the nozzle and the Exit Area of the nozzle by means of the Principle of Mass Conservation. A nozzle is a device that works at Steady State, so that Mass Balance can be reduced into this form:

[tex]\dot m_{in} = \dot m_{out}[/tex] (1)

Where:

[tex]\dot m_{in}[/tex] - Inlet mass flow, in kilograms per second.

[tex]\dot m_{out}[/tex] - Outlet mass flow, in kilograms per second.

Given that air flows at constant rate, we expand (1) by dimensional analysis:

[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex] (2)

Where:

[tex]\rho_{in}, \rho_{out}[/tex] - Air density at inlet and outlet, in kilograms per cubic meter.

[tex]A_{in}, A_{out}[/tex] - Inlet and outlet area, in square meters.

[tex]v_{in}, v_{out}[/tex] - Inlet and outlet velocity, in meters per second.

a) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex] and [tex]v_{in} = 20\,\frac{m}{s}[/tex], then the mass flow rate through the nozzle is:

[tex]\dot m = \rho_{in}\cdot A_{in}\cdot v_{in}[/tex]

[tex]\dot m = \left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)[/tex]

[tex]\dot m = 0.265\,\frac{kg}{s}[/tex]

The mass flow rate through the nozzle is 0.265 kilograms per second.

b) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex], [tex]v_{in} = 20\,\frac{m}{s}[/tex], [tex]\rho_{out} = 0.762\,\frac{kg}{m^{3}}[/tex] and [tex]v_{out} = 150\,\frac{m}{s}[/tex], then the exit area of the nozzle is:

[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex]

[tex]A_{out} = \frac{\rho_{in}\cdot A_{in}\cdot v_{in}}{\rho_{out}\cdot v_{out}}[/tex]

[tex]A_{out} = \frac{\left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)}{\left(0.762\,\frac{kg}{m^{3}} \right)\cdot \left(150\,\frac{m}{s} \right)}[/tex]

[tex]A_{out} = 2.320\times 10^{-3}\,m^{2}[/tex]

[tex]A_{out} = 23.202\,cm^{2}[/tex]

The exit area of the nozzle is 23.202 square centimeters.

What are the examples of pulley? Plz tell the answer as fast as possible plz.​

Answers

Answer:

elevators

Theatre system

construction pulley

lifts

Answer:

elevator,cargo lift system

if 6000j of energy is supplid to a machine to lift a load of 300N through a vvertical height of 1M calculatework out put​

Answers

Answer:

300J

Explanation:

Work done = Force x the distance travelled in the direction of the force

=300 x 1

=300J

What distance do I cover if I travel at 10 m/s E for 10s?

Answers

Answer:

100m

Explanation:

i think this is the answer because the formula for distance is

d=speed×time in this case the speed is 10m/s and the time is 10s therefore the distance will be

10m/s×10s

=100m

I hope this helps

Answer:

100 m

Explanation: this is when you need to find velocity and the formula for velocity is displacement by time taken.

A force of 200 N, acting at 60° to the horizontal, accelerates a block of mass 50 kg along a horizontal plane. Calculate the component of the 200N force that accelerates the block horizontally​

Answers

Answer:

Explanation:

a)     Fx = F cos (θ)

           = (200) cos(60)

           = 100 N

b)     FR = ma

       Fx + Ff = ma

      100 + Ff = (50)(1,5)

       Ff     = 75 - 100

               =  -25 N

c)    Fy = F sin θ

           = (200) sin(60)

           = 173,2 N

A capacitor consists of two metal surfaces separated by an electrical insulator with no electrically conductive path through it. Why does a current flow in a resistor-capacitor circuit when the switch is closed?

Answers

Answer:

Displacement current flows in the dielectric material(insulated region)

Explanation:

Firstly a capacitor stores charge when a capacitor is charging (or discharging), current flows in the circuit. Also, there is no charge transfer in the dielectric material in the capacitor which is contradictory to the flow of current. Hence, displacement current is the current in the insulated region due to the changing electric flux.

A scientist who studies the tiny microorganisms of the environment .

geologist
meteorologist
microbiologist
entomologist

Answers

Question:- A scientist who studies the tiny microorganisms of the environment

Answer:- Microbiologist

Explanation:-

Microbiologist means a person who studies micro sized living organisms

Microbiologist word is combination of two words Micro and biologist

Micro stands for objects which cannot be seen with the naked eyes and are very small in sizeBiologist a person who studies living forms.

Answer:

microbiologist i think

hope it helps

please mark Brainliest if you think the answer is correct

Explain why the motor turns. Consider the Lorentz (magnetic) force we have discussed in class, and how this would apply here. Why did you have to remove only half the insulation on the ends of the wire

Answers

Solution :

Owing to the continuous attraction and repulsion force caused by the magnet or the electromagnet around the core of the motor produces a unidirectional torque whose direction is given by the Lorentz force, [tex]$F=q(\vec v \times \vec B)$[/tex], and thus the torque causes the rotation of the electric motor.

Removing half the insulation from the coil makes it to rotate just half a turn. When the half insulation is removed, the coil turns half and the rest of the time the connection terminates. The rest half turn will be provided by the angular momentum. Now after this half turn by the angular momentum, the connections will again be connected and again the torque will work on it to rotate the half turn. This continues and the motor rotates.

A T-shirt is launched at an angle of 30° with an initial velocity of 25 m/s how long does it take to reach the peak? How long is it in the air for totally?

Answers

Answer:

The launched angle θ = 30 degrees, the initial velocity Vo = 20 m/s, the initial horizontal velocity Vox= ?, the initial vertical velocity Voy = ?, the time of flight t = ? the maximum height h = ?

Vox = Vo * (cos of 30 degrees)

Voy = Vo * (sin of 30 degrees)

t = 2 * (Voy / g)

h = Voy * 0.5 t - 1/2 g * (0.5t)^2

I have given the equations for you to use, just plug – in the values and then solve in a step by step manner.

Answer:

approximately 15.68 meters.

Explanation:

Here is how;

First, let's calculate the time of flight for the t-shirt. We can use the vertical motion equation:

y = y0 + v0y * t - 0.5 * g * t^2

where:

y is the vertical displacement (27.7 m)

y0 is the initial vertical position (0 m)

v0y is the vertical component of the initial velocity (v0 * sin(theta))

g is the acceleration due to gravity (9.8 m/s^2)

t is the time of flight

Plugging in the values:

27.7 = 0 + (25.8 * sin(63.6°)) * t - 0.5 * 9.8 * t^2

Simplifying the equation, we get a quadratic equation:

4.9t^2 - (25.8 * sin(63.6°))t + 27.7 = 0

Solving this quadratic equation will give us the time of flight, t. Using the quadratic formula, we find that:

t ≈ 1.23 s

Now, let's find the horizontal displacement of the t-shirt using the horizontal motion equation:

x = x0 + v0x * t

where:

x is the horizontal displacement

x0 is the initial horizontal position (0 m)

v0x is the horizontal component of the initial velocity (v0 * cos(theta))

t is the time of flight

Plugging in the values:

x = 0 + (25.8 * cos(63.6°)) * 1.23

Calculating this:

x ≈ 14.92 m

The t-shirt falls short of reaching the person by the horizontal distance of:

Shortfall = 30.6 m - 14.92 m

Calculating this:

Shortfall ≈ 15.68 m

Therefore, the t-shirt will be approximately 15.68 meters short of reaching the person.

What is the maximum speed at which a car can round a curve of 25-m radius on a level road if the coefficient of static friction between the tires and road is 0.80? ​

Answers

I assume the curve is flat and not banked. A car making a turn on the curve has 3 forces acting on it:

• its weight, mg, pulling it downward

• the normal force from contact with the road, n, pushing upward

• static friction, f = µn, directed toward the center of the curve (where µ is the coefficient of static friction)

By Newton's second law, the net forces on the car in either the vertical or horizontal directions are

F (vertical) = n - mg = 0

F (horizontal) = f = ma

where a is the car's centripetal acceleration, given by

a = v ²/r

and where v is the maximum speed you want to find and r = 25 m.

From the first equation, we have n = mg, and so f = µmg. Then in the second equation, we have

µmg = mv ²/r   ==>   v ² = µgr   ==>   v = √(µgr )

So the maximum speed at which the car can make the turn without sliding off the road is

v = √(0.80 (9.80 m/s²) (25 m)) = 14 m/s

A refrigerator has a coefficient of performance equal to 4.00. The refrigerator takes in 110 J of energy from a cold reservoir in each cycle. (a) Find the work required in each cycle. J (b) Find the energy expelled to the hot reservoir. J

Answers

Answer:

The correct answer is:

(a) 27.5 Joules

(b) 141.5 Joules

Explanation:

Given:

Energy,

[tex]Q_c = 110 \ J[/tex]

Coefficient of performance refrigerator,

[tex]Cop(refrig)=4[/tex]

(a)

As we know,

⇒ [tex]Cop(refrig) = \frac{Q_c}{Work}[/tex]

or,

⇒ [tex]Work=\frac{Q_c}{Cop(refrig)}[/tex]

              [tex]=\frac{110}{4}[/tex]

              [tex]=27.5 \ Joules[/tex]

(b)

⇒ [tex]Heat \ expelled = Heat \ removed +Work \ done[/tex]

or,

⇒ [tex]Q_h = Q_c+Work[/tex]

         [tex]=114+27.5[/tex]

         [tex]=141.5 \ Joules[/tex]

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