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Answer:

Do we have to choose out of those three?

Explanation:

The time between when the bat emits the sound and when it hears the echo is 0.05 s

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

Velocity of sound (v) = 343 m/s

Distance (x) = 8.42 m

We can obtain obtained the time as illustrated below:

v = 2x / t

343 = 2 × 8.42 / t

343 = 16.84 / t

Cross multiply

343 ×  t = 16.84

Divide both side by 343

t = 16.84/343

Thus, the time between  when the bat emits the sound and when it hears the echo is 0.05 s

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Answer:

0.0491

Explanation:

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:

Asch's experiment showed that about 75% of people were "yielders" who conformed and 25% were "independent" who didn't conform. Asch concludes that people ignored reality and gave an incorrect answer in order to follow the rest of the group.

Answer:

(a) 6.283 Wb (b) 69.11 Wb (c) I = 0.628 A

Explanation:

Given that,

The diameter of the loop, d = 40 cm

Radius, r = 20 cm

Initial magnetic field, B = 5 mT

Final magnetic field, B' = 55 mT

Initial magnetic flux,

[tex]\phi_i=BA\\\\=5\times 10^{-3}\times \pi \times 20^2\\\\=6.283\ Wb[/tex]

Final magnetic flux,

[tex]\phi_f=B'A\\\\=55\times 10^{-3}\times \pi \times 20^2\\\\=69.11\ Wb[/tex]

Due to change in magnetic field an emf will be generated in the loop. It is given by :

[tex]\epsilon=-\dfrac{d\phi}{dt}\\\\=\phi_f-\phi_i\\\\=69.11-6.283\\\\=62.827\ V[/tex]

Let I be the current in the loop. We can find it using Ohm's law such that,

[tex]\epsilon=IR\\\\I=\dfrac{\epsilon}{R}\\\\I=\dfrac{62.827}{100}\\\\=0.628\ A[/tex]

Hence, this is the required solution.

Answer:

True

Explanation:

The electromagnetic spectrum can be regarded as term that give description of range of light that are in existence, and this range from radio waves up to gamma rays. It can be explained as

range of frequencies that electromagnetic radiation takes with thier respective wavelengths as well as photon energies

The Types of electromagnetic radiation that make up the electromagnetic spectrum are

✓ microwaves

✓ infrared light

✓ ultraviolet light

✓X-rays,

✓gamma rays.

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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Answer:

Drinking straw, syringe, Dropper, vacuum, etc.

Explanation:

Answer:

the image will be real and inverted

Answer:

b

Explanation:

is correct .....................

i think

Answer:

Explanation: the major difference between an electromagnet and permanent magnet is that the former can have a magnetic field when electric current flows through it and disappears when the flow of the current stops. ... It will always displays the magnetic behaviour.

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:

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:

yes, because you dont want to be mean and make her feel worse about it and hiv has something do with s3x not germs so yes go

Explanation:

Answer:

Recessive, dominant

Explanation:

umm thats the answer

Answer:A

Explanation:

Answer:

The initial speed of the bullet will be:

[tex]v_{ib}=331.36\: m/s[/tex]

Explanation:

Using the momentum conservation:

[tex]p_{i}=p_{f}[/tex]

[tex]m_{b}v_{ib}=Mv[/tex] (1)

Where:

Using the conservation of energy:

[tex]\frac{1}{2}Mv^{2}=\frac{1}{2}Mv_{2}^{2}+Mgh [/tex] (2)

Where v(2) is the speed of the system at 0.8 m of vertical height.

Using the forces acting on the system we can find v(2).

The forces can be equal to the centripetal force:

[tex]T-Mg*sin(\alpha)=M\frac{v_{2}^{2}}{L}[/tex]            

α is the angle of T with respect to the horizontal, here α = 25.4°  

So, v(2) will be:

[tex]T-Mg*sin(\alpha)=M\frac{v_{2}^{2}}{L}[/tex]  

[tex]\frac{L}{M}(T-Mg*sin(\alpha))=v_{2}^{2}[/tex]  

[tex]\frac{1.4}{0.76}(5-0.76*9.81*sin(25.4))=v_{2}^{2}[/tex]  

[tex]v_{2}=1.82 \: m/s[/tex]

Using this value on equation (2) we will find v.

[tex]v^{2}=v_{2}^{2}+2gh [/tex]

[tex]v^{2}=1.82^{2}+2(9.81)(0.8) [/tex]

[tex]v=4.36\: m/s [/tex]

And finally using equation (1) we can find the initial speed of the bullet.

[tex]m_{b}v_{ib}=Mv[/tex]

[tex]0.01*v_{ib}=0.76*4.36[/tex]

[tex]v_{ib}=331.36\: m/s[/tex]

I hope it helps you!

Answer:

3.69 m/s²

Explanation:

weight = mass × Acceleration of gravity

479.96 = 129.72 × Acceleration of gravity

Acceleration of gravity = 479.96/129.72 = 3.69 m/s²

Answer:

explanation

Explanation:

1 = C

2 = A

3 = D

4 = E

5 = B

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.

Answer:

8.08 J/g °C

Explanation:

Q=m*Cp*ΔT-->

Cp=Q/(m*ΔT) -->

Cp=1600/[18*(31-20)]-->

Cp=8.08 J/g °C

Answer:

R₁ = 0.126 m

Explanation:

Let's use the definition of intensity which is the power per unit area

          I = P / A

the generated power is constant

          P = I A

power is

         P = E / t

if we perform the calculations for a given time, the wave energy is

        E = q V

we substitute

        P = [tex]\frac{q V\ A}{t}[/tex]

we can write this equation for two points, point 1 the antenna and point 2 the receiver

         V₁A₁ = V₂A₂

          A₁ = [tex]\frac{V_2}{V_1} \ A_2[/tex]

          A₁ = 0.1 10⁻³  5 10⁻⁴ /V₁

          A₁ = 5 10⁻⁸ /V₁

In general, the electric field on the antenna is very small on the order of micro volts, suppose V₁ = 1 10⁻⁶ V

let's calculate

          A₁ = 5 10⁻⁸ / 1 10⁻⁶

         A₁ = 5 10⁻² m²

the area of ​​a circle is

          A = π r²

we substitute

         π R1₁²= 5 10⁻²

         R₁ = [tex]\sqrt{ \frac{5 \ 10^{-2} }{\pi } }}[/tex]

         R₁ = 0.126 m

Answer:

put it in a volcano

Explanation:

Answer:oxidation number

Explanation:

i just took quiz;)

The ratio of the elements in a binary ionic compound are based on the element's oxidation number.

Oxidation number can be generally described as the number which is allocated to the chemical elements in a chemical combination of different compounds. The oxidation number of an element is basically the number or count of electrons that atoms in a molecule can share, lose or gain while they are forming chemical bonds with the other atoms of a different element or atom.

The oxidation state, or oxidation number is the hypothetical charge which is present on an atom if all of the bonds in an atom to different atoms were fully ionic in nature. Oxidation state describes the degree of oxidation of an atom in a chemical compound or molecule. The oxidation state may be positive, negative or zero depending upon the loss or gain of electrons.

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Answer:

1.33

Explanation:

According to snell's law,

n = sin(a)/sin(b)

Given

a = 25.0°

b = 18.5°

Substitute

n = sin25/sin18.5

n = 0.4226/0.3173

n = 1.33

Hence the index of refraction nb for this type of glass is 1.33