Identify each action as a wave erosion war wind erosion

Answer:

(A) Series and Parallel

Explanation:

Circuit Component: These are electrical devices that makes up the circuit. They include, resistors, capacitors, inductors, voltmeters, ammeters, cell/batteries, earth connection, bulb, switch, connecting wire etc.

These component can either be connected in series or in parallel.

(1) Series Connection: This can be refered as end to end connection of electric component.

(2) Paralel Connection: This can be refered as the side to side connection of electric component.

From the question above,

A electric component in a circuit can be combined in series and in parallel.

The right option is (A) Series and Parallel

Solution :

1. We know that : Buoyant force = weight of the liquid displace

                                                  = volume displaced x density of the fluid

Now volume of the man = [tex]$\frac{\text{mass}}{\text{density}}$[/tex]

Mass = weight / g

         [tex]$=\frac{940}{9.8}$[/tex]

         = 95.92 kg

And density = 1000 [tex]kg/m^3[/tex]

Therefore,

[tex]$\text{volume} = \frac{\text{mass}}{\text{density}}$[/tex]

           [tex]$=\frac{95.92}{1000}$[/tex]

           = 0.0959 [tex]m^3[/tex]

We know density of air = 1.225 [tex]kg/m^3[/tex]

∴ Mass of air displaced = 0.0959 x 1.225

                                       = 0.1175 kg

Weight of the air displaced = 1.1515 N

Therefore, the buoyant force = 1.1515 N

2). As the balloon is not accelerated, the net force acting on it is zero.

Thus the weight that acts downwards = buoyant force upwards

So, the weight of the air displaced = weight of the balloon

                                                          = 17000 N

Therefore, the mass of the air displaced = volume of the air displaced (volume of the balloon) x density of air

[tex]$\frac{17000}{9.8} = \text{volume of air} \times 1.225$[/tex]

[tex]$\text{Volume of air displaced} = \frac{1700}{9.8 \times 1.225}$[/tex]

                                     = 1416.0766 [tex]m^3[/tex]

Answer:

C . Magnetic dipoles is the correct

Answer:

b). movement of charged particles.

Explanation:

These charges create the nagnetic dipoles.

Answer:

a. Find the graph in the attachment

b. 720 kΩ

c. The ratio V/I gives us our resistance which is 720 kΩ

Explanation:

a) plot a graph of V against I.

To plot the graph of V against I, we plot the corresponding points against each other. With the voltage V measured in volts and the current I measured in mA, the plotted graph is in the attachment.

b) Determine the wire's resistance , R.

The resistance of the wire is determined as the gradient of the graph.

R = ΔV/ΔI = (V₂ - V₁)/(I₂ - I₁)

Taking the first two corresponding measurements. V₁ = 72 V, I₁ = 0.1 mA, V₂ = 144 V and I₂ = 0.2 mA

R = (144 V - 72 V)/(0.2 - 0.1) mA

R = 72 V/0.1 mA

R = 72 V/(0.1 × 10⁻³ A)

R = 720 × 10³ V/A

R = 720 kΩ

c) State ohm's law and try to relate it your results.​

Ohm's law states that the current flowing through a conductor is directly proportional to the voltage across it provided the temperature and all other physical conditions remain constant.

Mathematically, V ∝ I

V = kI

V/I = k = R

Since the ratio V/I = constant, from our results, the ratio of V/I for each reading gives us the resistance. Since we have a linear relationship between V and I, the gradient of the graph is constant and for each value of V and I, the ratio V/I is constant. So, the ratio V/I gives us our resistance which is 720 kΩ.

Since V/I is constant, we thus verify Ohm's law.

Answer:

4√6 rad/s

Explanation:

Since the spring is initially stretched a length of x = 4 in when the 4 lb mass is placed on it, since it is in equilibrium, the spring force, F = kx equals the weight of the mass W = mg.

So, W = F

mg = kx where m = mass = 4lb, g = acceleration due to gravity = 32 ft/s², k = spring constant and x = equilibrium displacement of spring = 4 in = 4 in × 1ft /12 in = 1/3 ft

making k the spring constant subject of the formula, we have

k = mg/x

substituting the values of the variables into the equation, we have

k = mg/x  

k = 4 lb × 32 ft/s² ÷ 1/3 ft

k = 32 × 4 × 3

k = 384 lbft²/s²

Now, assuming there is no friction and no external force, we have an undamped system.

So, the natural frequency for an undamped system, ω = √(k/m) where k = spring constant = 384 lbft²/s² and m = mass = 4 lb

So, substituting the values of the variables into the equation, we have

ω = √(k/m)

ω = √(384 lbft²/s² ÷ 4 lb)

ω = √96

ω = √(16 × 6)

ω = √16 × √6

ω = 4√6 rad/s

Answer:

1. A. Safety measures in handling machines

2. C. One lane, follow the traffic signal

3. D. Internal and external forces contribute to the formation of various surface features of the earth.

4. D. Tsunami

Explanation:

1. Safety measures are precautions that must be taken to avoid injuries when handling some potentially dangerous equipment. They are meant to protect us from harm.

2. If a part of the road is to be sealed up temporarily, road users should be alerted that there is just one lane ahead. Therefore, they should follow the traffic signal.

3. The pressure exerted on the earth through phenomena like plate tectonics, earthquakes, etc., causes the eruption and formation of different features on the earth.

4. A tsunami is a giant ocean wave caused by earthquakes and volcanic eruptions that occur under the sea. They occur with a lot of speed and cause an overflow of water onto land.

Answer:

K = 2 10⁻⁸ J

Explanation:

Let's solve this exercise in parts, we start by finding the charge on each plate of the capacitor

          C = Q / ΔV

           C = ε₀ A / d

          ε₀ A / d = Q / ΔV

          Q = ε₀ A ΔV / d        (1)

indicate the potential difference ΔV₁ = 12 V, the distance between the plates d₁ = 3 mm = 0.003 m,  

as the power supply is disconnected and the capacitor is ideal the charge remains constant

in the second part we separate the plates at d₂ = 5 mm = 0.005 m, using equation 1

          ΔV₂ = [tex]\frac{Q d_2}{ \epsilon_o A}[/tex]

we substitute the equation for Q

         ΔV₂ = [tex]\frac{d_2}{\epsilon_o A} \ \frac{\epsilon_o A \Delta V }{d_1}[/tex]

         ΔV₂ = [tex]\frac{d_2}{d_1} \ \Delta V_1[/tex]

in the third part we use the concepts of energy

starting point. Test charge near positive plate

          Em₀ = U = q ΔV₂

final point. Test charge near negative plate

          Em_f = K

energy is conserved

          Em₀ = Em_f

          q ΔV₂ = K

          K = q ΔV₁ [tex]\frac{d_2}{d_1}[/tex]

we calculate

          K = 1 10⁻⁹  12  0.005/0.003

          K = 2 10⁻⁸ J

The question is incomplete. The complete question is :

A wire 0.6 m long and with mass m = 11 g is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude B = 0.4 T. What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards? The magnetic field is directed into the page.

Solution :

Given :

Length of the wire, L = 0.6 m

Mass of the wire length, m = 11 g

                                             = [tex]11 \times 10^{-3}[/tex] kg

Magnetic field , B = 0.4 T

Know we know that :

ILB = mg

or [tex]$I=\frac{mg}{BL}$[/tex]

 [tex]$I= \frac{(11 \times 10^{-3})(9.81)}{(0.4)(0.6)}$[/tex]

 [tex]I=0.44963\ A[/tex]

 [tex]I = 449.63 \ mA[/tex]

Answer:

Both will have the same momentum.

P = M v     momentum

v = a t   for uniform acceleration

P = M a t

But a = F / M

P = M (F / M) t = F t    so both have the same momentum

We know

[tex]\boxed{\sf E=hv}[/tex]

[tex]\\ \sf\longmapsto E=6.63\times 10^{-34}J\times 3.6\times 10^{15}s^{-1}[/tex]

[tex]\\ \sf\longmapsto E=23.86\times 10^{-19}J[/tex]

[tex]\\ \sf\longmapsto E=2.38\times 10^{-18}J[/tex]

[tex]\\ \sf\longmapsto E=2.4\times 10^{-18}J[/tex]

Answer:

D!!!!!

Explanation:

This question is incomplete, the complete question is;

A magnetic field is passing through a loop of wire whose area is 0.015 m2. The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.20 T/s.

(a) Determine the magnitude of the emf induced in the loop.

(b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B = 1.5 T, if the induced emf is to be zero? (Give the magnitude of the rate of change of the area.) (m2/s)

Answer:

a) the magnitude of the emf induced in the loop is 0.003 V

b) dA/dt = 0.002 m²/s

Explanation:

Area of the loop wire A = 0.015 m²

magnitude of the field is increasing dB/dt = 0.20 T/s

a)

Determine the magnitude of the emf induced in the loop.

V = A( dB/dt )

we substitute

V = 0.015 m² × 0.20 T/s

V = 0.003 V

Therefore,  the magnitude of the emf induced in the loop is 0.003 V

b)  the induced emf is;

V = B( dA/dt ) + A( dB/dt )

given that; induced emf is 0, B = 1.5

so we substitute

0 = [ 1.5T × ( dA/dt ) ] + [ 0.015 m² × 0.20 T/s ]

-[ 1.5T × ( dA/dt )] = 0.003 m²T/s

dA/dt = -[ 0.003 m²T/s / 1.5T ]

dA/dt = -0.002 m²/s

the negative shows that the area is decreasing

hence, dA/dt = 0.002 m²/s

Answer:

down below

Explanation:

Image 1- wheels of train showing both translatory motion as well as rotatory motion.

Image 2- rotation of ball shows both rotatory motion as well as translatory motion.

Image 3- the earth rotates about its axis, same time it revolves around the sun thus showing both rotatory motion and curvilinear motion in a fixed time. (perodic motion)

Image 4- while cutting wood, the

carpenter's saw has both

translatory motion and oscillatory

motion, as it moves down while

oscillating.

Answer:

1.2 seconds

Explanation:

Answer to the following question is 1.2 seconds

Because light from the moon takes 1.2 seconds to reach Earth, the light released from the crescent immediately before it vanishes will also take 1.2 seconds to reach Earth. As a result, the earth-shine portion of the moon will vanish 1.2 seconds after the crescent has vanished.

Answer:

0

Explanation:

The speed of the ball when it reaches the floor is 0 because when an object is at rest or in uniform motion, it has no speed/velocity

The final speed of the ball when it reaches the floor is 7.10 m/s.

The conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but only converted from one form to another or transferred from one system to another. In other words, the total amount of energy in a closed system remains constant over time, even though it may be converted from one form to another.

This principle is based on the first law of thermodynamics, which states that the total energy of a closed system is always conserved, and can only be changed by the transfer of heat, work, or matter into or out of the system. The conservation of energy has important applications in various fields of physics, including mechanics, thermodynamics, and electromagnetism, and is a fundamental principle in the understanding of the natural world.

Here in the Question,

We can use the conservation of energy to solve this problem. Initially, the ball has kinetic energy due to its motion on the tabletop, but no potential energy since it is at a constant height. When the ball rolls off the edge of the table, it loses some kinetic energy due to friction but gains potential energy as it moves upward. When it reaches the floor, it has gained potential energy but lost kinetic energy due to friction. We can assume that the energy lost due to friction is converted to thermal energy, so the total energy of the system is conserved.

Let's start by calculating the potential energy gained by the ball as it moves from the edge of the table to the floor:

ΔPE = mgh

where ΔPE is the change in potential energy, m is the mass of the ball, g is the acceleration due to gravity, and h is the vertical distance traveled by the ball.

ΔPE = (0.50 kg)(9.81 m/s^2)(1.0 m) = 4.905 J

Now we can use the conservation of energy to find the final kinetic energy of the ball, which will allow us to calculate its final speed:

KEi + ΔPEi = KEf + ΔPEf

where KEi and ΔPEi are the initial kinetic and potential energies of the ball, respectively, and KEf and ΔPEf are the final kinetic and potential energies of the ball, respectively.

Since the ball is not bouncing, we can assume that its initial and final potential energies are zero. Therefore:

KEi = KEf + ΔKE

where ΔKE is the change in kinetic energy due to friction.

We can assume that the coefficient of kinetic friction between the ball and the incline is constant, and use the work-energy principle to find ΔKE:

Wfric = ΔKE

where Wfric is the work done by friction.

The work done by friction can be expressed as:

Wfric = ffricd

where ffric is the force of friction and d is the distance traveled by the ball on the incline.

The force of friction can be expressed as:

ffric = μmg

where μ is the coefficient of kinetic friction, and m and g have their usual meanings.

Putting it all together, we get:

KEi = KEf + ffricd

KEi = KEf + μmgd

(1/2)mv^2 = (1/2)mu^2 + μmgd

v^2 = u^2 + 2gd

where u is the initial speed of the ball on the tabletop, and v is the final speed of the ball on the floor.

Plugging in the given values, we get:

v^2 = (5.0 m/s)^2 + 2(9.81 m/s^2)(1.0 m)

v^2 = 50.405

v = 7.10 m/s

Therefore, the final speed of the ball when it reaches the floor is 7.10 m/s.

To learn more about  the Law of Conservation of Momentum click:

https://brainly.com/question/30487676

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Answer:

A wavefront is the locus of all the particles which are in phase. A wavelet is an oscilation that starts from zero, then the amplitude increases and later decreases to zero

Answer:

[tex]_{18}^{39} } Ar[/tex]

Explanation:

The given equation shows the disintegration of an unstable isotope of chlorine to beta particle and Argon nucleus. The nucleus undergoes the emission of a beta particle to form a more stable nucleus of Argon.

[tex]_{17} ^{39} Cl[/tex] ⇒ [tex]_{-1}^{0} e[/tex] + [tex]_{18}^{39} } Ar[/tex]

Argon is a stable gas and is found in the group 8 on the periodic table of elements.

Answer:

Answer is below

Explanation:

39 18 Ar

Explanation:

mass=kilogram,temperature=Klevin,power=watt,density=kilogram per cubic metre

Explanation:

the unit of mass is kg , temperature is kelvin ,power is watt and density is kilogram per cubic meter.

Answer:

x = 0.9 m

Explanation:

For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive

          ∑ τ = 0

          60 1.5 - 78 1.5 + 30 x = 0

where x is measured from the left side of the fulcrum

           90 - 117 + 30 x = 0

           x = 27/30

           x = 0.9 m       

In summary the center of mass is on the side of the lightest weight x = 0.9 m

Your friend would have a better experience of this event, than you .

An eclipse is produced when a planetary body moves in front of another planetary body and is visible from a third planetary body. Considering the sun, moon, and earth's locations in relation to one another during the time of the eclipse,

there are various types of eclipses in our solar system. For instance, a lunar eclipse occurs when the earth passes between the moon and the sun.

For the solar eclipse to happen the light from the sun is obstructed by the moon observing from the earth.

The buddies left Earth because they could view the whole eclipse, but you were on the moon and only saw parts of the eclipse turn black.

To learn more about the eclipse from here, refer to the link;

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Answer:

Ecologist.

Your answer is Ecologist.

(Ecologist) is a scientist who studies the whole environment as a working unit.

Answer:

1.24

Explanation:

The resistivity of copper[tex]\rho_1=2.65\times 10^{-8}\ \Omega-m[/tex]

The resistivity of Aluminum,[tex]\rho_2=1.72\times 10^{-8}\ \Omega-m[/tex]

The wires have same resistance per unit length.

The resistance of a wire is given by :

[tex]R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}[/tex]

According to given condition,

[tex]\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24[/tex]

So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.

Answer:

23s

Explanation:

s=ut+1/2at^2

the distance (s) is 400, initial velocity (u) is 0, acceleration (a) is 1.5 therefore

400=0t+1/2(1.5)t^2

400/0.75=0.75t^2/0.75

t^2=√533.33

t=23s

I hope this helps and sorry if it's wrong

Answer:poop

Explanation:

poop

Answer:

1. The image is real

2. 5.85

3. h' = 3.05 mm

4. The image is upright

Explanation:

1. Start with the first lens and apply 1/f = 1/p + 1/q

1/5.01 = 1/13.7 + 1/q

q = 7.90 cm

Since that distance is behind the first lens, and the second lens is 62.5 cm behind the first lens, that distance is 62.5 - 7.90 = 54.6 cm in front of the second lens, and becomes the object for that lens, thus,

1/25.9 = 1/54.6 + 1/q

q = 49.3 cm behind the second lens

Using that information, since q is positive, the image is real

2. Also, using that information, you have the second answer, which is 49.3 cm

The height can be found from the two magnifications.

m = -q/p

m1 = -7.9/13.7 = -.577

m2 = -49.3/54.6 = -.903

Net m = (-.577)(-.903) = .521

Then, m = h'/h

.521 = h'/5.85

3. h' = 3.05 mm

4. For the fourth answer, since the overall magnification is positive, the final image is upright

Answer:

A green object will absorb all light except for green light and reflect blue and yellow light.

Answer:

R = 0.0015Ω

Explanation:

The formula for calculating the resistivity of a material is expressed as;

ρ = RA/l

R is the resistance

ρ is the resistivity

A is the area of the wire

l is the length of the wire

Given

l = 85cm = 0.85m

A = πr²

A = 3.14*0.0018²

A = 0.0000101736m²

ρ = 1.75 × 10-8Ωm.

Substitute into the formula

1.75 × 10-8 = 0.0000101736R/0.85

1.4875× 10-8 = 0.0000101736R

R = 1.4875× 10-8/0.0000101736

R = 0.0015Ω

Answer:

the initial velocity of the car is 12.04 m/s

Explanation:

Given;

force applied by the break, f = 1,398 N

distance moved by the car before stopping, d = 25 m

weight of the car, W = 4,729 N

The mass of the car is calculated as;

W = mg

m = W/g

m = (4,729) / (9.81)

m = 482.06 kg

The deceleration of the car when the force was applied;

-F = ma

a = -F/m

a = -1,398 / 482.06

a = -2.9 m/s²

The initial velocity of the car is calculated as;

v² = u² + 2ad

where;

v is the final velocity of the car at the point it stops = 0

u is the initial velocity of the car before the break was applied

0 = u² + 2(-a)d

0 = u² - 2ad

u² = 2ad

u = √2ad

u = √(2 x 2.9 x 25)

u =√(145)

u = 12.04 m/s

Therefore, the initial velocity of the car is 12.04 m/s

Answer:

The total distance, side to side, that the top of the building moves during such an oscillation is approximately 0.291 meters.

Explanation:

Let suppose that the building is experimenting a Simple Harmonic Motion due to the action of wind. First, we determine the angular frequency of the system ([tex]\omega[/tex]), in radians per second:

[tex]\omega = 2\pi\cdot f[/tex] (1)

Where [tex]f[/tex] is the frequency, in hertz.

If we know that [tex]f = 0.18\,hz[/tex], then the angular frequency of the system is:

[tex]\omega = 2\pi\cdot (0.18\,hz)[/tex]

[tex]\omega \approx 1.131\,\frac{rad}{s}[/tex]

The maximum acceleration experimented by the system is represented by the following formula, of which we estimate amplitude of the oscillation:

[tex]r\cdot g = \omega^{2}\cdot A[/tex] (2)

Where:

[tex]r[/tex] - Ratio of real acceleration to free-fall acceleration, no unit.

[tex]g[/tex] - Free-fall acceleration, in meters per square second.

[tex]A[/tex] - Amplitude, in meters.

If we know that [tex]\omega \approx 1.131\,\frac{rad}{s}[/tex], [tex]r = 0.019[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the amplitude of the oscillation is:

[tex]A = \frac{r\cdot g}{\omega^{2}}[/tex]

[tex]A = \frac{(0.019)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{\left(1.131\,\frac{rad}{s} \right)^{2}}[/tex]

[tex]A \approx 0.146\,m[/tex]

The total distance, side to side, is twice the amplitude, that is to say, a value of approximately 0.291 meters.