The table below describes some features of methods used to generate electricity. Name method 4.

Phytochemicals are compounds that are produced by plants ("phyto" means "plant"). They are found in fruits, vegetables, grains, beans, and other plants. Some of these phytochemicals are believed to protect cells from damage that could lead to cancer.

Answer:

When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.

Explanation:

As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.

Blueshifts happens when the source of the wave and the observer are moving closer towards one another.

Assume that the wave is of frequency [tex]f\; {\rm Hz}[/tex] at the source. In other words, the source of the wave sends out a peak after every [tex](1/f)\; {\text{seconds}}[/tex].

Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.

The source of this wave sends out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. It would appear to the observer that consecutive peaks arrive every [tex](1/f)\; {\text{seconds}}\![/tex]. That would correspond to a frequency of [tex]f\; {\rm Hz}[/tex].

On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of [tex]v \; {\rm m \cdot s^{-1}}[/tex].

Again, the source of this wave would send out a peak after each period of [tex](1/f)\; {\text{seconds}}[/tex]. However, by the time the source sends out the second peak, the source would have been [tex]v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}[/tex] closer to the observer then when the source sent out the first peak.

When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than [tex](1/f)\; {\text{seconds}}[/tex]. The observed frequency of this wave would be larger than the original [tex]f\; {\rm Hz}[/tex].

Explanation:

First, we convert the energy from eV to Joules:

[tex]2.90\:\text{eV}×\left(\dfrac{1.6×10{-19}\:J}{1\:\text{eV}} \right)[/tex]

[tex]= 4.64×10^{-19}\:\text{eV}[/tex]

We know from definition that

[tex]E=h\nu = \dfrac{hc}{\lambda}[/tex]

so the wavelength of the photon is

[tex]\lambda = \dfrac{hc}{E} = 4.28×10^8\:\text{m}[/tex]

Answer: See attached pic. Hope this helps.

Explanation:

Answer:

The intensity of 4 lawn movers is 86 dB.

Explanation:

Intensity of one lawnmower = 80 dB

Let the intensity is I.

Use the formula of intensity

[tex]dB = 10 log\left ( \frac{I}{Io} \right )\\\\80=10log\left ( \frac{I}{Io} \right )\\\\10^8 = \frac{I}{10^{-12}}\\\\I = 10^{-4} W/m^2[/tex]

Now the intensity of 4 lawn movers is

[tex]dB = 10 log\left ( \frac{4I}{Io} \right )\\\\dB=10log\left ( \frac{4\times10^{-4}}{10^{-12}} \right )\\\\dB = 86 dB\\[/tex]

Answer:

0.5766422350752*10^-24 N

Explanation:

Couloumb's law states that states that there is an electrical force acting on 2 static charges. The magnitude is directly proportional to the product of the 2 charges.

Strength of electrons : q1 = q2 = 1.602 x 10-19. C

Substitute and solve:

F = (9*10^9)(1.602 x 10-19)(1.602 x 10-19) / (0.02)^2

Done.

Answer:

  v = 2.75 10⁴ m / s

Explanation:

For this exercise we must use Kepler's third law which is an application of Newton's second law to the solar system

            F = ma

where force is the force of gravity

            F = [tex]G \frac{m M}{r^2}[/tex]

acceleration is centripetal

             a = [tex]\frac{v^2}{r}[/tex]

we substitute

           G m M / r² = m v² / r

           [tex]\frac{GM}{r}[/tex] = v²

           v = [tex]\sqrt{GM/r}[/tex]

indicate that the radius of the orbit is r = 1.17 AU, let's reduce to the SI system

            r = 1.17 AU (1.496 10¹¹ m / 1 AI) = 1.76 10¹¹ m

let's calculate

         v = [tex]\sqrt{\frac{6.67 \ 10^{-11} 1.991 \ 10^{30} }{ 1.76 \ 10^{11}} }[/tex]Ra (6.67 10-11 1.991 10 30 / 1.76 10 11

         v = [tex]\sqrt{7.5454 \ 10^8 }[/tex]ra 7.5454 10 8

         v = 2.75 10⁴ m / s

Answer:

[tex]n=3.8[/tex]

Explanation:

From the question we are told that:

Magnetic Field [tex]B=0.01T[/tex]

Current [tex]I=1.00[/tex]

Wire Diameter [tex]d_w=0.5*10^3m[/tex]

Layers Diameter [tex]d_l=1*10^2m[/tex]

Length [tex]l=0.2m[/tex]

Generally the equation for number of layers is mathematically given by

[tex]n=\frac{Bd_w}{\mu_o I}[/tex]

Where

[tex]Vacuum\ permeability=\mu_0[/tex]

[tex]n= \frac{0.01*0.5*10^3m}{4 \pi *10^{-7}*1 }[/tex]

[tex]n=3.8[/tex]

Answer:

idek

Explanation:

Answer:

Test 1

1.True

2.True

3.True

4.False

5.True

6.True

7.False

8.True

9.True

10.True

yung iba nasa pic

Answer:

1.67 N

Explanation:

Applying,

F = u(dm/dt)+m(du/dt)................ Equation 1

Where F = force, m = mass of the vehicle, u = speed.

Since u is constant,

Therefore, du/dt = 0

F = u(dm/dt)............... Equation 2

From the question,

Given: u = 10 m/s, dm/dt = 10 kg/min = (10/60) kg/s

Substitute these values into equation 2

F = 10(10/60)

F = 100/60

F = 1.67 N

Answer:

h = 2755102 m = 2755.102 km

Explanation:

According to the given condition:

Potential Energy = Energy Consumed by Bulb

[tex]mgh = Pt\\\\h = \frac{Pt}{mg}[/tex]

where,

h = height = ?

P = Power of bulb = 75 W

t = time = (2 h)(3600 s/1 h) = 7200 s

m = mass of bulb = 20 g = 0.02 kg

g = acceleration due to gravity = 9.8 m/s²

Therefore,

[tex]h = \frac{(75\ W)(7200\ s)}{(0.02\ kg)(9.8\ m/s^2)}[/tex]

h = 2755102 m = 2755.102 km

Answer:

P =40.69 atm

Explanation:

We need to find the approximate pressure at a depth of 400 m.

It can be calculated as follows :

P = Patm + ρgh

Put all the values,

[tex]P=1\ atm+1025 \times 9.81\times 400\times 9.8692\times 10^{-6}\ atm/Pa\\\\P=40.69\ atm[/tex]

So, the approximate pressure is equal to 40.69 atm.

Answer:

answer here

Explanation:

the unit of work is fundamental unit because it doesn't depend on other units.

__________________

Thx

Answer:

The SI unit of work is joule (J)

Explanation:

Joule is a derived unit. ∴ unit of work is a derived unit

Answer:

d

Explanation:

Ya gon find the Kenitic Energy first

K=½mv²===> K=½×0.2×(0.2)²===> 0.1(0.04)===> 0.004

and now the replacement:

0.004=½×0.4V²====> v²=0.02===> V=0.14m/s

Answer:

219 ft

Explanation:

Here we can define the value t = 0s as the moment when the car starts decelerating.

At this point, the acceleration of the car is given by the equation:

A(t) = -10 ft/s^2

Where the negative sign is because the car is decelerating.

To get the velocity equation of the car, we integrate over time, to get:

V(t) = (-10 ft/s^2)*t + V0

Where V0 is the initial velocity of the car, we know that this is 88 ft/s

Then the velocity equation is:

V(t) = (-10 ft/s^2)*t + 88ft/s

To get the position equation we need to integrate again, this time we get:

P(t) = (1/2)*(-10 ft/s^2)*t^2 + (88ft/s)*t + P0

Where P0 is the initial position of the car, we do not know this, but it does not matter for now.

We want to find the total distance that the car traveled in a 3 seconds interval.

This will be equal to the difference in the position at t = 3s and the position at t = 0s

distance = P(3s) - P(0s)

 = ( (1/2)*(-10 ft/s^2)*(3s)^2+ (88ft/s)*3s + P0) - ( (1/2)*(-10 ft/s^2)*(0s)^2 + (88ft/s)*0s + P0)

=  ( (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s + P0) - ( P0)

=  (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s = 219ft

The car advanced a distance of 219 ft in the 3 seconds interval.

Answer:

By the principle of corona discharge.

Explanation:

The charge on each ball will decreases over time due to the electrical discharge in air.

According to the principle of corona discharge, when the curvature is small, the discharge of the charge takes placed form the pointed ends.

Answer:

the speed of the nerve impulse in miles per hour is 201.59 mi/hr

Explanation:

Given;

the speed of the nerve impulse, v = 90.1 m/s

To convert this speed in meters per second to miles per hour, we use the following method;

1,609 meter = 1 mile

3,600 s = 1 hour

[tex]v(mi/h) = 90.1 \ \frac{m}{s} \times \frac{1 \ mile}{1,609 \ m} \times \frac{3,600 \ s}{1 \ hour} = (\frac{90.1 \times 3,600}{1,609} )\frac{mi}{hr} = 201.59 \ mi/hr[/tex]

Therefore, the speed of the nerve impulse in miles per hour is 201.59 mi/hr

Answer:

pretty sure its 6.2 x 10^-13

Explanation:

I looked it up I'm not a bigbrain but want to help

Answer:

The correct answer will bs "2.41 m".

Explanation:

According to the question,

M = 50 g

or,

   = 0.050 kg

[tex]\Theta = 25^{\circ}[/tex]

k = 25.9 N/m

Δx = 0.200 m

Let the traveled distance be "x".

By using trigonometry, the height will be:

⇒ [tex]h = l Sin \Theta[/tex]

hence,

⇒ [tex]Potential \ energy \ at \ the \ top=Spring \ potential \ energy[/tex]

                                       [tex]Mgh=\frac{1}{2} k(\Delta x)^2[/tex]

By putting the values, we get

             [tex]0.050\times 9.8\times lSin 25^{\circ}=\frac{1}{2}\times 25.0\times (0.200)^2[/tex]

                                              [tex]l=2.41 \ m[/tex]      

Explanation:

Here,we've been given that,

We've to find the acceleration of the object. By using the third equation of motion,

→ v² - u² = 2as

→ (420)² - (0)² = 2 × a × 145

→ 176400 - 0 = 290a

→ 176400 = 290a

→ 176400 ÷ 290 = a

→ 608.275862 m/s² = a

If you know initial speed and final speed, you can find the average speed.  Then, knowing distance, you can find the time.

KimYurii posted the first answer to this question.  

That answer is well organized, well presented, elegant and correct, and it deserves to be awarded "Brainliest" and several merit badges.

My problem is that I can never remember all the different formulas.  I guess I had to work with so many uvum in all the Physics, Geometry, and Calculus classes that I took, I filled up all the memory slots with formulas, and over the years they all eventually merged into a big glob of goo.  Now, the only formulas I can remember are the ones I had to use as an Electrical Engineer.

When I see this kind of question, I can only remember one or two simple formulas, and I reason it out like this:

Starting speed . . . zero

Ending speed . . . 420 m/s

Formula:  Average speed . . . (1/2)·(0 + 420) = 210 m/s

Distance covered . . . 145 m

Formula: Time taken = (distance) / (average speed) = (145/210) second

(Now you have the time.)

Formula: Distance = (1/2)·(acceleration)·(time²)

145 m = (1/2)·(acceleration)·(145/210 sec)²

Acceleration = 290 m / (145/210 s)²

Acceleration = 608.28 m/s²

Answer:

v₂ = 53.23 m/s

Explanation:

Given that,

The mass of a golf club, m₁ = 158 g = 0.158 kg

The initial speed of a golf club, u₁  =  48.2 m/s

The mass of a golf ball, m₂ = 46 g = 0.046 kg

It was at rest, u₂ = 0

Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s

We use the conservation of energy to find the speed of the golf ball just after impact as follows :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s[/tex]

So, the speed of the golf ball just after the impact is equal to 53.23 m/s.

Answer:

[tex]K.E_{max}=0.8973J[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.64kg[/tex]

Equation of Mass

[tex]X=0.33cos(3.17t)[/tex]...1

Generally equation for distance X is

[tex]X=Acos(\omega t)[/tex]...2

Therefore comparing equation

Angular Velocity [tex]\omega=3.17rad/s[/tex]

Amplitude A=0.33

Generally the equation for Max speed is mathematically given by

[tex]V_{max}=A\omega[/tex]

[tex]V_{max}=0.33*3.17[/tex]

[tex]V_{max}=1.0461m/s[/tex]

Therefore

[tex]K.E_{max}=0.5mv^2[/tex]

[tex]K.E_{max}=0.5*1.64*(1.0461)^2[/tex]

[tex]K.E_{max}=0.8973J[/tex]

Answer:

The other angle is 75⁰

Explanation:

Given;

velocity of the projectile, v = 10 m/s

range of the projectile, R = 5.1 m

angle of projection, 15⁰

The range of a projectile is given as;

[tex]R = \frac{u^2sin(2\theta)}{g}[/tex]

To find another angle of projection to give the same range;

[tex]5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\ the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0[/tex]

Check:

sin(2θ) = sin(2 x 75) = sin(150) = 0.5

sin(2θ) = sin(2 x 15) = sin(30) = 0.5

Answer:

100j

Explanation:

Answer:

Following are the solution to the given question:

Explanation:

Please find the complete question in the attached file.

The cost after 30 days is 60 dollars. As energy remains constant, the cost per hour over 30 days will be decreased.

[tex]\to \frac{\$60}{\frac{30 \ days}{24\ hours}} = \$0.08 / kwh.[/tex]

Thus, [tex]\frac{\$0.08}{\$0.12} = 0.694 \ kW \times 0.694 \ kW \times 1000 = 694 \ W.[/tex]

The electricity used is continuously 694W over 30 days.

If just resistor loads (no reagents) were assumed,

[tex]\to I = \frac{P}{V}= \frac{694\ W}{120\ V} = 5.78\ A[/tex]

Energy usage reduction percentage = [tex](\frac{60\ W}{694\ W} \times 100\%)[/tex]

This bulb accounts for [tex]8.64\%[/tex] of the energy used, hence it saves when you switch it off.