5. A bicyclist is finishing her repair of a flat tire when a friend rides by at a constant velocity of
3.5 m/s. Three seconds later, the bicyclist hops on her bike and accelerates at 3.6 m/s² until she
catches her friend.
a. How much time does it take until she catches her friend?
b. How far has she traveled in this time?
c. What is her speed when she catches up?

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

Explanation:-

Microbiologist means a person who studies micro sized living organisms

Microbiologist word is combination of two words Micro and biologist

Answer:

microbiologist i think

hope it helps

please mark Brainliest if you think the answer is correct

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:

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

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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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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I hope it helps you!                                                                   

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.

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

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.

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].

Answer:

Magnification = 1

Explanation:

given data

radius of curvature r = - 0.983 m

image distance u = - 0.155

solution

we get here first focal length that is

Focal length, f = R/2     ...................1

f = -0.4915 m

we use here formula that is

[tex]\frac{1}{v} + \frac{1}{u} + \frac{1}{f}[/tex]      .................2

put here value and we get

[tex]\frac{1}{v} = \frac{1}{0.155} - \frac{1}{4915}[/tex]  

so

Magnification will be here as

m = [tex]- \frac{v}{u}[/tex]

m =  [tex]\frac{0.155}{0.155}[/tex]

Answer:

The magnification is 1.5.

Explanation:

radius of curvature, R = - 0.983 m

distance of object, u = - 0.155 m

Let the distance of image is v.

focal length, f = R/2 = - 0.492 m

Use the mirror equation

[tex]\frac{1}{f}=\frac{1}{v}+\frac {1}{u}\\\\\frac{-1}{0.492}=\frac{1}{v}-\frac{1}{0.155}\\\\\frac{1}{v}=\frac{1}{0.155}-\frac{1}{0.492}\\\\\frac{1}{v}=\frac{0.492-0.155}{0.155\times 0.492}\\\\\frac{1}{v}=\frac{0.337}{0.07626}\\ \\v = 0.226 m[/tex]

The magnification is given by

m = - v/u

m = 0.226/0.155

m = 1.5

Answer:

in which topic you need help

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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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]

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

Answer:

elevators

Theatre system

construction pulley

lifts

Answer:

elevator,cargo lift system

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.

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

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  

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.

Answer:

simple is rumple a daily ok I'll be

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

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]

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]

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.

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.

Answer:

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

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.

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

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

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:

Answer:

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

Explanation:

Answer:

300J

Explanation:

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

=300 x 1

=300J

Answer:

Explanation:

jjjgggyuuuuiiii hhjiiihuyyuuugggyujjhhhhggghhhjjhhhhjhhui