⦁ A certain resistor is required to dissipate 0.25 W, what standard rating should be used?

Answer:

I cheated on

Spanish

ELA

history

Explanation:

Answer:

thx

Explanation:

Answer:

-33750 N

Explanation:

Use coulomb's law: [tex]\frac{k(q1)(q2)}{r^{2} } = \frac{(9x10^{9})(-5x10^{-6})(3x10^{-6})}{0.002^{2} } = -33750 N[/tex]

The wavelength of this radio signal is equal to 3.18 meters.

Given the following data:

Wavelength can be defined as the distance between two (2) successive crests (troughs) of a wave.

Mathematically, the wavelength of a wave is given by this formula:

[tex]\lambda = \frac{V}{F}[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]\lambda = \frac{3 \times 10^8}{9.45 \times10^7}[/tex]

Wavelength = 3.18 meters.

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The wavelength of the radio signal travel at speed of light is 3.17m.

Given the data in the question;

Wavelength the spatial period of a periodic wave. That is to say, when the shapes of waves are Wavelength , the distance over which they are repeated is called wavelength. Wavelength  is expressed as;

[tex]\lambda = \frac{v}{f}[/tex]

Where [tex]\lambda[/tex] is wavelength, f is the frequency of the wave and c is the velocity or speed of light ( [tex]c = 3*10^8m/s[/tex] )

We substitute our values into the expression above.

[tex]\lambda = \frac{c}{ f}\\ \\\lambda = \frac{3*10^8m/s}{9.45*10^7s^{-1}} \\\\\lambda = \frac{3*10^8ms/s}{9.45*10^7}\\\\\lambda = 3.17m[/tex]

Therefore, the wavelength of the radio signal travel at speed of light is 3.17m.

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Answer:

The base of taller buildings are made broader because it reduces the pressure exerted on the ground so that they do not sink in due to the extremely high pressure of the building.

Explanation:

hope thiss helpss~!

A broad base is used to engage or recruit a diverse range of individuals to broad-based environmentalism.

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The magnitude of the angular acceleration of the pendulum at the moment of release is; α = 18.45 rad/s²

We are given;

Mass of rod; m = 1 kg

Length of rod; L = 49.8 cm = 0.498 m

Mass of Disk; M = 4 kg

Radius of disk; r = 24.9 cm = 0.249 m

Let us first calculate the torque acting from the formula;

τ = mg(L/2) + MgL

Thus;

τ = (1 × 9.8 × (0.498/2)) + (4 × 9.8 × 0.498)

τ = 21.96 N.m

Using parallel axis theorem, we can find the moment of inertia about the given axis as;

I = (mL²/3) + ½MR² + ML²

Plugging in the relevant values gives;

I = (1 * 0.498²/3) + ½(4 * 0.249²) + (4 * 0.498²)

I = 1.19 kg.m²

The angular acceleration is given by the formula;

α = I/τ

α = 21.96/1.19

α = 18.45 rad/s²

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Answer:

D.Conduction, because the metal wire was in direct contact with water in both containers

Conduction, because the metal wire was in direct contact with water in both containers.​ Option D is correct.

Given that,
Students filled two cups with water. One cup contains 100mL of cold water while the other has 100 mL of hot water. A metal heat wire is placed inside the water of both cups and thermometers are attached to each cup.
It is to be identified, the system contains which type of heat transfer between the cups

Heat transfer is the process, in a thermodynamic system, when an object at a higher temperature is placed with an object at has a lower temperature, after some time the system attains equilibrium because heat flows from higher temperature to lower temperature.

In the given question,
Cups, that are connected through a metal wire, one is hot and the other is cold, Principal of conduction state that when there is a system of connected bodies the heat flows throughout the connected system till the thermal equilibrium attain. Implies heat flow from hot cup to cold cup through the wire through conduction heat transfer.

Thus, Conduction, because the metal wire was in direct contact with water in both containers.​ Option D is correct.

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#SPJ2

Answer:

Explanation:

The answer is c. I am very sure

Answer:

i think its b

Explanation:

im not very sure

Answer:

strongest are at the points of the north pole and the south pole, specifically between the red box and the letter of each pole.

Explanation:

The lines of magnetic force are drawn so that the density of lines is proportional to the intensity of the magnetic field.

Therefore, the sections where the magnetic field is strongest are at the points of the north pole and the south pole, specifically between the red box and the letter of each pole.

Answer:

Drinking straw, syringe, Dropper, vacuum, etc.

Explanation:

Answer:

I think is is

Explanation:

B and C why because i have a gut feeling

Answer:

Do we have to choose out of those three?

Explanation:

Answer:

the image will be real and inverted

Answer:

v = 2683.33 m/s

Explanation:

The magnetic force and the electric force on the electron must be the same, in order for them to cancel each other:

[tex]Electric\ Force = Magnetic\ Force\\Eq = qvBSin\theta \\\\v = \frac{E}{BSin\theta}[/tex]

where,

v = velcoity of electron = ?

E = Electric Field = 3059 V/m

B = Magnetic Field = 1.14 T

θ = Angle between velocity and magnetic field = 90°

Therefore,

[tex]v = \frac{3059\ V/m}{(1.14\ T)Sin90^o}[/tex]

v = 2683.33 m/s

Answer:

boiling water - When water heats, the heat from the burner is transferred to the container, heating the water at the bottom. This hot water rises, creating a circular motion, as colder water descends to replace it.

Answer:

Recessive, dominant

Explanation:

umm thats the answer

Answer:  The younger elliptical and lenticular galaxies had results similar to spiral galaxies like the Milky Way. The researchers found that the older galaxies have a larger fraction of low-mass stars than their younger counterparts.

Explanation:

Answer:

Homogeneous mixtures

Explanation:

I think so because homogeneous means mixed mixtures

Answer:

put it in a volcano

Explanation:

Answer:

sum of volumes = (16.6 ± 0.03) cm³

and difference of volumes = (3.8 ± 0.03) cm³

Explanation:

Here,

V₁ = (10.2 ± 0.02) cm³ and V₂ = (6.4 ± 0.01) cm³.

Now,

∆V = ± (∆V₁ + ∆V₂)

= ± (0.02 + 0.01) cm³

= ± 0.03 cm³

V₁ + V₂ = (10.2 + 6.4) cm³ = 16.6 cm³ and

V₁ - V₂ = (10.2 - 6.4) cm³ = 3.8 cm³

Thus, sum of volumes = (16.6 ± 0.03) cm³

and difference of volumes = (3.8 ± 0.03) cm³

-TheUnknownScientist

Answer:

I hope it's helpful .............

Answer:

[tex]\mathbf{v_{ground} = \sqrt{{v^2+2ghb}}}[/tex]

Explanation:

From the information given:

The avg. velocity post the window is;

[tex]v_{avg} = \dfrac{L_w}{t}[/tex]

[tex]v_b[/tex] = velocity located at the top of the window

[tex]v_b[/tex] = velocity situated at the bottom of the window  

Using the equation of kinematics:

[tex]v_b = v_t + gt[/tex]

Hence,

[tex]v_t = v_b - gt[/tex]

To determine the average velocity as follows:

[tex]v_{avg} = \dfrac{1}{2} (v_t + v_b)\dfrac{L_w}{t}= \dfrac{1}{2}(v_b - gt +v_b) \\ \\\dfrac{L_w}{t} = v_b - \dfrac{1}{2}gt \\ \\ v_b = \dfrac{L_w}{t }+ \dfrac{1}{2} gt\\ \\ = \dfrac{1}{t} \Bigg(L_w + \dfrac{1}{2}gt^2 \Bigg) \\ \\[/tex]

where;

[tex]v_b[/tex] = velocity gained when fallen through the height h.

Similarly, using the equation of kinematics, we have;

[tex]v_b^2 = 2gh \\ \\h = \dfrac{v_b^2}{2g}[/tex]

[tex]\implies \dfrac{(L_w + \dfrac{1}{2} gt^2_^2}{2gt^2}[/tex]

Thus, the velocity at the ground is;

[tex]v^2_{grround} = v_b^2 + 2ghb[/tex]

[tex]\mathbf{v_{ground} = \sqrt{{v^2+2ghb}}}[/tex]

The period of the vertical mass-spring is 0.64 s.

The angular speed of the vertical mass-spring is calculated as follows;

[tex]V_{max} = A \omega\\\\\omega = \frac{V_{max}}{A} \\\\\omega = \frac{7.02}{0.713} \\\\\omega = 9.85 \ rad/s[/tex]

The period of the vertical mass-spring is calculated as follows;

[tex]f = \frac{\omega }{2\pi} \\\\T = \frac{1}{f} \\\\T = \frac{2 \pi}{\omega } \\\\T = \frac{2\pi }{9.85} \\\\T = 0.64 \ s[/tex]

Thus, the period of the vertical mass-spring is 0.64 s.

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Complete question

To reduce the drag coefficient and thus to improve the fuel efficiency, the frontal area of a car is to be reduced. Determine the amount of fuel and money saved per year as a result of reducing the frontal area from 18 to 14 ft2. Assume the car is driven 12,000 mi a year at an average speed of 55 mi/h. Take the density and price of gasoline to be 50 lbm/ft3 and $3.10/gal, respectively; the density of air to be 0.075 lbm/ft3, the heating value of gasoline to be 20,000 Btu/lbm; and the overall efficiency of the engine to be 30 percent. Take the drag coefficient as CD=0.3 for a passenger car.

Answer:

22.22%

$57

Explanation:

From the question we are told that

Initial area of frontal area [tex]a_1=18ft^2[/tex]

Final area of frontal area [tex]a_2=14ft^2[/tex]

Distance covered a year [tex]D=12000mile[/tex]

Average speed a year [tex]V_{avg}=55mile/h[/tex]

Density [tex]\rho=50 lbm/ft3[/tex]

Price [tex]P= $3.10/gal[/tex]

Density of air [tex]\rho_{air} 0.075 lbm/ft3[/tex]

Heating value of gasoline [tex]Q=20,000 Btu/lbm[/tex]

Efficiency [tex]\eta=30\%[/tex]

Drag coefficient [tex]CD=0.3[/tex]

[tex]\triangle A=18-14ft^2=4ft^2[/tex]

Generally the equation for drag force [tex]C_D[/tex] is mathematically given as

[tex]F_D=\frac{C_DAPV^2}{2}[/tex]

[tex]F_D=\frac{0.3*18*0.075*(80.685^2)*A}{2}[/tex]

[tex]F_D=73.24Alb[/tex]

where [tex]v=55mil/h*1.467=80.685ft/s[/tex]

Generally the equation for work done W is mathematically given as[tex]W=F_D*L[/tex]

where [tex]L=12000mile*5280[/tex]

[tex]L=63360000[/tex]

[tex]W=73.24A*63360000[/tex]

[tex]W=4.6*10^9A[/tex]

Generally the equation for overall efficiency [tex]\eta[/tex] is mathematically given as

where

[tex]W_{req}=required\ gasoline\ power\ efficiency[/tex]

[tex]\eta=\frac{W}{W_{req}}[/tex]

[tex]W_{req}=\frac{W}{\eta}[/tex]

[tex]W_{req}=\frac{4.6*10^9}{0.3}[/tex]

[tex]W_{req}=1.55*10^{10}A[/tex]

Generally the equation for reduction fee with change in frontal area [tex]\triangle M[/tex] is mathematically given as

[tex]\triangle M =\triangle Vgasoline*cost[/tex]

Where

[tex]\triangle Vgasoline= volume\ reduction\ of\ gasoline[/tex]

[tex]\triangle Vgasoline=\frac{E_{req}}{H*P}[/tex]

[tex]\triangle Vgasoline=\frac{1.55*10^{10}A}{20000*778.169*32.2*50}[/tex]

if

[tex]20000btu/ibm=20000*778.169*32.2(1bm.ft^2/s)[/tex]

[tex]\triangle Vgasoline=\frac{1.55*10^{10}A}{20000*778.169*32.2*50}[/tex]

[tex]\triangle Vgasoline=0.61859A[/tex]

Therefore

[tex]\triangle M =0.61859A*3.10[/tex]

if [tex]1ft^2=7.48gal[/tex]

[tex]\triangle M =0.61859(4)*3.10*7.48[/tex]

[tex]\triangle M = \$ 57.671[/tex]

Generally the equation for reduction of fuel [tex]F_r[/tex]is mathematically given as

[tex]F_r=\frac{\triangle A}{\triangle i}*100[/tex]

where

[tex]\triangle i=18ft^2[/tex]

[tex]F_r=\frac{4}{18}*100[/tex]

Fuel reduction price by reducing front area is 22.22%

Money saved per year is $57

Answer:

The answer is C, I just guessed and got it right lol

Explanation:

Answer:

The kinetic energy of large objects can be converted into this thermal energy.

Explanation:

For example, if you drop a water balloon onto the ground, its kinetic energy is converted mostly to thermal energy. If the balloon weighs 1 kilogram and you drop it from about 2 meters, it will heat up by less.