1. Draw four illustrations of a globe and paper that are positioned to yield equatorial, transverse, oblique, and polar aspect projections. Label the equator in each. Use your textbook or lecture material if you need a reference.2. On any map, why is there distortion at areas that do not fall on lines of tangency or secancy?

Answer:

The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.

Explanation:

The angular velocity of the object ([tex]\omega[/tex]), in radians per second, is given by the following expression:

[tex]\omega(t) = 5\cdot t^{2}[/tex] (1)

Where [tex]t[/tex] is the time, measured in seconds.

The change in the angular displacement ([tex]\Delta \theta[/tex]), in radians, is found by means of the following definite integral:

[tex]\Delta \theta = \int\limits^{4}_{2} {5\cdot t^{2}} \, dt[/tex] (2)

Then we proceed to integrate on the function in time:

[tex]\Delta \theta = \frac{5}{3}\cdot (4^{2}-2^{2})[/tex]

[tex]\Delta \theta = 20\,rad[/tex]

The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.

Complete question is;

A microwave oven operates at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. Assume that microwave energy is generated uniformly on the upper surface of the cavity and propagates directly

downward toward the base. The base is lined with a material that completely absorbs microwave energy. The total microwave energy content of the cavity is 0.50 mJ.

Answer:

Power ≈ 600,000 W

Explanation:

We are given;

Frequency; f = 2400 Hz

height of the oven cavity; h = 25 cm = 0.25 m

base area; A = 30 cm by 30 cm = 0.3m × 0.3m = 0.09 m²

total microwave energy content of the cavity; E = 0.50 mJ = 0.5 × 10^(-3) J

We want to find the power output and we know that formula for power is;

P = workdone/time taken

Formula for time here is;

t = h/c

Where c is speed of light = 3 × 10^(8) m/s

Thus;

t = 0.25/(3 × 10^(8))

t = 8.333 × 10^(-10) s

Thus;

Power = (0.5 × 10^(-3))/(8.333 × 10^(-10))

Power ≈ 600,000 W

Explanation:

15.556 metres per second

Answer:

0.5

Explanation:

because the block is attached to the pulley of the string

Explanation:

[tex]\underline{\boxed{\large{\bf{Option \; A!! }}}} [/tex]

Here,

We have to find the equivalent resistance of the circuit.

Here, [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] are connected in series, so their combined resistance will be given by,

[tex]\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] is connected in parallel combination with [tex]\rm { R_3} [/tex], so their combined resistance will be given by,

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ [/tex]

Reciprocating both sides,

[tex]\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex], [tex]\rm { R_2} [/tex] and [tex]\rm { R_3} [/tex] is connected in series combination with [tex]\rm { R_4} [/tex]. So, equivalent resistance will be given by,

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ [/tex]

Henceforth, Option A is correct.

[tex]\underline{\boxed{\large{\bf{Option \; B!! }}}} [/tex]

Here, we have to find the amount of flow of current in the circuit. By using ohm's law,

[tex] \longrightarrow [/tex] V = IR

[tex] \longrightarrow [/tex] 3 = I × 3.33

[tex] \longrightarrow [/tex] 3 ÷ 3.33 = I

[tex] \longrightarrow [/tex] 0.90 Ampere = I

Henceforth, Option B is correct.

[tex] \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ [/tex]

Answer:

Time

Explanation:

The answer to the question is actually time. Time is not needed when you calculate the mass or volume of an object, a square, sphere, rectangle, or any other 3D shape. You must also calculate the length to know what numbers you will be multiplying by. The answer to the question is time.

Answer:

Answer to the following question is as follows;

Explanation:

The puck's real altitude is lower than ones projection. That's because the mechanism may not be completely frictionless. Electricity is nevertheless wasted owing to particle interactions such as friction, which might explain why the present the results is lower than predicted.

Answer:

The number of revolutions is 44.6.

Explanation:

We can find the revolutions of the wheel with the following equation:

[tex]\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}[/tex]

Where:

[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s              

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]

Where:

[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s

[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]

Hence, the number of revolutions is:

[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]

Therefore, the number of revolutions is 44.6.

I hope it helps you!

Explanation:

Given that,

Amplitude, A = 2.5 nm

Wavelength,[tex]\lambda=1.8\ m[/tex]

The speed of the wave, v = 36 m/s

At time t = 0 the left end of the string has its maximum upward displacement.

(a) Let f is the frequency. So,

[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{36}{1.8}\\\\f=20\ Hz[/tex]

(b) Angular frequency of the wave,

[tex]\omega=2\pi f\\\\=2\pi \times 20\\\\=125.7\ rad/s[/tex]

(c) The wave number of the wave[tex]=\dfrac{1}{\lambda}[/tex]

[tex]=\dfrac{1}{1.8}\\\\=0.56\ m^{-1}[/tex]

Answer:

0.3 N

Explanation:

Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.

The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N

F = Kq₁q₂ / r²

Where

From Coulomb's law,

F = Kq₁q₂ / r²

Cross multiply

Fr² = Kq₁q₂

Kq₁q₂ = constant

F₁r₁² = F₂r₂²

With the above formula, we can obtain the final force as follow:

F₁r₁² = F₂r₂²

1.2 × r² = F₂ × (2r)²

1.2r² = F₂ × 4r²

Divide both side by 4r²

F₂ = 1.2r² / 4r²

F₂ = 0.3 N

Learn more about Coulomb's law:

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Answer:

The number of turns of wire needed is 573.8 turns

Explanation:

Given;

maximum emf of the generator, = 190 V

angular speed of the generator, ω = 3800 rev/min =

area of the coil, A = 0.016 m²

magnetic field, B = 0.052 T

The number of turns of the generator is calculated as;

emf = NABω

where;

N is the number of turns

[tex]\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s[/tex]

[tex]N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns[/tex]

Therefore, the number of turns of wire needed is 573.8 turns

Relative density is the ratio of the density of a substance to the density of a given material.

Answer:

(a) Iron > plastic > ice

(b) Same on all

(c) Iron downwards, plastic net force zero, ice upwards.

(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.

Explanation:

Three spheres have same volume , plastic, ice and iron.

(a) The weight is given by

Weight = mass x gravity = volume x density x gravity

As the density of iron is maximum and the density of ice is least so the order of the weight is

Weight of iron > weight of plastic > weight of ice

(b) Buoyant force is given by

Buoyant force = Volume immersed x density of fluid x g

As they have same volume, density of fluid is same so the buoyant force is same on all the spheres.

(c) Net force is

F = weight - buoyant force  

So, the net force on the iron sphere is downwards

On plastic sphere is zero as the density of plastic sphere is same as water. On ice sphere it is upwards.

(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.  

Answer:

560.54 Ω

Explanation:

Applying,

V = IR'............... Equation 1

Where V = Voltage of the battery, I = currrent, R' = Total resistance of the resistors

make R' the subject of the equation

R' = V/I............ Equation 2

From the question,

Given: V = 5 V, I = 2.23 mA = 2.23×10⁻³ A

Substitute these values into equation 2

R' = 5/(2.23×10⁻³ )

R' = 2242.15 Ω

Since the fours resistor are connected in series and they are equal,

Therefore the values of each resistor is

R = R'/4

R = 2242.15/4

R = 560.54 Ω

Answer:

Static friction depends on the mass of the object.

Explanation:

Friction is the force between two surfaces in contact. The force of friction between two surfaces in contact depends on;

1) nature of the object and the surface(how rough or smooth the surfaces are)

2)surface area of the object and the surface

3) mass of the object

Since;

F=μmg

Where;

μ= coefficient of static friction

m= mass of the object

g= acceleration due to gravity

Hence, as the mass of the object increases, the magnitude of static friction force between an object and a surface increases and vice versa.

Answer:

The correct answer is "[tex]4.49\times 10^{10} \ joules[/tex]".

Explanation:

According to the question,

The work will be:

⇒ [tex]Work=-\frac{kQq}{R}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (18-30)\times 3\times 0.2}[/tex]

              [tex]=-\frac{1}{4 \pi \varepsilon \times (-12)\times 3\times 0.2}[/tex]

              [tex]=\frac{0.3978}{\varepsilon }[/tex]

              [tex]=4.49\times 10^{10} \ joules[/tex]

Thus the above is the correct answer.    

We have that the workdone  is mathematically given as

W=4.49*10e10 J

From the question we are told

Generally the equation for the workdone   is mathematically given as

W=-kQq/R

Therefore

0.3978/ε0 =-1/(4πε0*(18-30)*3*0.2

Hence

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Answer:

Decreases

Explanation:

because its moving against gravitational attraction and at maximum height its velocity will be and it will decrease until it reaches maximum height and the start to increase again

Answer:

the current that flows through the component is 2.42 A

Explanation:

Given;

resistance of the electrical component, r = 53 Ω

the voltage of the source, V = 128 V

The current that flows through the component is calculated using Ohm's Law as demonstrated below;

[tex]V = IR\\\\I = \frac{V}{R} = \frac{128 \ V}{53 \ ohms} = 2.42 \ A[/tex]

Therefore, the current that flows through the component is 2.42 A

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

[tex]F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz[/tex]

Therefore, the frequency of this mode of vibration is 138.87 Hz

Answer:

Atmospheric pressure at Badwater is 1.01022 atm

Explanation:

Data given:

1 atmospheric pressure (Pi) = 1.01 * 10[tex]^{5}[/tex] Pa

Elevation (h) = 86m

gravity (g) = 9.8 m/s2

Density of air P = 1.225 kg/m3

Therefore pressure at bad water Pb = Pi + Pgh

Pb = (1.01 * 10[tex]^{5}[/tex]) + (1.225 * 9.8 * 86)

Pb = (1.01 * 10[tex]^{5}[/tex]) + 1032.43 = 102032 Pa

hence:

Pb = 102032 /1.01 * 10[tex]^{5}[/tex] = 1.01022 atm

Answer:

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Explanation:

If you accelerate at 2 m/s^2 for 10 seconds, at the end of the 10 seconds you are moving at a rate of 20 m/s.

V(f) = V(i) + a*t

Final velocity = initial velocity + acceleration x time

Your average velocity will be half of your final, because you accelerated at a constant rate. So your average velocity is 10 m/s.

Distance = velocity x time, so 10 m/s X 10 s = 100 m

Answer:

100 m

Explanation:

Given,

Initial velocity ( u ) = 0 m/s

Acceleration ( a ) = 2 m/s^2

Time ( t ) = 10 sec s

To find : Displacement ( s ) = ?

By 2nd equation of motion,

s = ut + at^2 / 2

= ( 0 ) ( 10 ) + ( 2 ) ( 10 )^2 / 2

= 0 + ( 2 ) ( 100 ) / 2

= 200 / 2

s = 100 m

Answer:

Acosθ

Explanation:

The x-component of a vector is defined as :

Magnitude * cosine of the angle

Maginitude * cosθ

The magnitude is represented as A

Hence, horizontal, x - component of the vector is :

Acosθ

Furthermore,

The y-component is taken as the sin of the of the angle multiplied by the magnitude

Vertical, y component : Asinθ

Answer: hello your question lacks some vital information below is the complete question

Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces 1 × 106 J of electricity with the hot reservoir at 500 K during Day One and then produces 1 × 106 J of electricity with the hot reservoir at 600 K during Day Two. The thermal pollution was

answer:

Total thermal pollution = 2.5 * 10^6 J

Explanation:

Low temperature reservoir = 300 K

hot reservoir temperature = 500 K

Electrical energy produced by plant ( W ) = 1 * 10^6 J

lets assume ; Q1 = energy absorbed , Q2 = energy emitted

W = Q1 - Q2  or  Q2 = Q1 - W  ( we will apply this as the formula for determining thermal pollution )

For day 1

T1 = 500k , T2 = 300k

applying Carnot engine formula

W / Q1 = 1 - T2/T1

∴ Q1 = 10^6 / ( 1 - (300/500)) = 2.5 * 10^6 J

thermal pollution ; Q2 = Q1 - W = ( 2.5 * 10^6 - 1 * 10^6 ) = 1.5 * 10^6 J

for Day 2

T1 = 600k,  T2 = 300k

Q1 = 10^6 / ( 1 - (300/600)) = 2 * 10^6 J

Thermal pollution; Q2 = Q1 - W  = 1 * 10^6 J

Therefore the Total thermal pollution =  1 * 10^6  + 1.5 * 10^6  = 2.5 * 10^6 J

Answer:

Elements have a characteristic number of electrons and protons.Both Hydrogen(H) and oxygen(O) are two different elements.

••••••••••••••••

Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of .Hydrogen(H) and Oxygen(O) both qr4 naturally gases,but they react to form water(H2O),which is liquid compound.

•••••••••••••••

A mixture is made of atleast two parts》 solid,liquid or gas.The difference is that it's not a chemical substance that's bonded by other elements. 

------------------------------

Hope it helps...

Have a great day!!!

Answer: Elements have a characteristic number of electrons and protons. Both Hydrogen(H) and oxygen(O) are two different elements. Compounds are chemical substances where the atoms atoms of two different elements are combined together. It is made of.Hydrogen(H) and Oxygen(O) both qr4 naturally gases, but they react to form water(H2O), which is liquid compound. A mixture is made of at least two parts solid, liquid, or gas. The difference is that it's not a chemical substance that's bonded by other elements.

Answer:

a)[tex]|\Delta E|=4.58\: J[/tex]  

b)[tex]F=61.90\: N[/tex]

Explanation:

a)

We can use conservation of energy between these heights.

[tex]\Delta E=mgh_{2}-mgh_{1}=mg(h_{2}-h_{1})[/tex]  

[tex]\Delta E=0.608*9.81(0.6026-1.37)[/tex]

Therefore, the lost energy is:

[tex]|\Delta E|=4.58\: J[/tex]  

b)

The force acting along the distance create a work, these work is equal to the potential energy.

[tex]W=\Delta E[/tex]

[tex]F*d=mgh[/tex]

Let's solve it for F.

[tex]F=\frac{mgh}{d}[/tex]

[tex]F=\frac{0.608*9.81*1.37}{0.132}[/tex]

Therefore, the force is:

[tex]F=61.90\: N[/tex]

I hope is helps you!

Answer:

a

Explanation:

Answer:

the answer is c

'

Explanation:

Hi there!

a.

Use the formula d = st to solve:

d = 20 × 0.5 = 10m

150 - 10 = 140m away when brakes are applied

b.

Use the following kinematic equation to solve:

vf² = vi² + 2ad

Plug in known values:

0 = 20² + 2(150)(a)

Solve:

0 = 400 + 300a

-300a = 400

a = -4/3 (≈ -1.33) m/s² required

c.

Use the following kinematic equation to solve:

vf = vi + at

0 = 20 - 4/3t

Solve:

4/3t = 20

Multiply both sides by 3/4 for ease of solving:

t = 15 sec

Answer:

I = 0.0857 A

Explanation:

Given that,

Power consumed by the cellphone, P = 300 mW

The voltage of the battery, V = 3.5 V

Let I is the current flowing through the cell-phone. We know that,

P = VI

Where

I is the current

So,

[tex]I=\dfrac{P}{V}\\\\I=\dfrac{300\times 10^{-3}}{3.5}\\\\I=0.0857\ A[/tex]

So, the current flowing the cell-phone is 0.0857 A.

Answer:

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Explanation:

The bulk modulus of water ([tex]B[/tex]), in newtons per square meters, can be estimated by means of the following model:

[tex]B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}}[/tex] (1)

Where:

[tex]\rho_{o}[/tex] - Water density at 10.9 °C, in kilograms per cubic meter.

[tex]\rho_{f}[/tex] - Water density at 40 °C, in kilograms per cubic meter.

[tex]\Delta P[/tex] - Pressure change, in pascals.

If we know that [tex]\rho_{o} = 999.623\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{f} = 992.219\,\frac{kg}{m^{3}}[/tex] and [tex]B = 2.2\times 10^{9}\,\frac{N}{m^{2}}[/tex], then the bulk modulus of water is:

[tex]\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)[/tex]

[tex]\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)[/tex]

[tex]\Delta P = -16294943.19\,Pa \,(-160.819\,atm)[/tex]

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Answer:

you will get huge electricity bills ............