Billy left some rocks outside in the Sun. After many hours, they got hot. When he went inside for the night, he left them outside. What will happen to the rocks as they stay out all night?

Answer:

Predicting the size, location, and timing of natural hazards is virtually impossible, but now, earth scientists are able to forecast hurricanes, floods, earthquakes, volcanic eruptions, wildfires, and landslides using fractals.

Explanation:

Answer:

A

Explanation:

First of all the answer is less than the smallest resistor, so it is less than 18 ohms.

1 / r1 + 1/ r2 = 1/r

1/18 + 1/23.5 = 1/r

1/18 = 0.05555555

1/23.5 = 0.0422532

1/18 + 1/23.5 = 0.0555555 + 0.0422532

1/r = 0.0981087  

r = 1/0.0981087

r = 10.193

A

Final answer:

The equivalent resistance for two resistors in parallel, one with 18.0 ohms and the other with 23.5 ohms, is approximately 10.2 ohms, which is answer choice A.

Explanation:

When calculating the equivalent resistance for resistors in parallel, you use the formula 1/Req = 1/R1 + 1/R2 where Req is the equivalent resistance, and R1 and R2 are the resistances of the individual resistors. For the two resistors with resistances of 18.0 ohms and 23.5 ohms, the calculation would be:

1/Req = 1/18.0 + 1/23.5

1/Req = 0.0556 + 0.0426

1/Req = 0.0982

Req = 1 / 0.0982

Req ≈ 10.18 ohms

Therefore, the correct answer is A. 10.2 ohms.

Answer: acceleration will increase

According to newtons second law the relationship between the force on an object and the acceleration and an object

In convex mirrors the focus is virtual and the focal distance is negative. This is how the reflected rays diverge and only their extensions are cut at a point on the main axis, resulting in a virtual image of the real object .

The Mirror equation is:  

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]    (1)  

Where:  

[tex]f=-6cm[/tex] is the focal distance  

[tex]u=12cm[/tex] is the distance between the object and the mirror  

[tex]v[/tex] is the distance between the image and the mirror

We already know the values of [tex]f[/tex] and [tex]u[/tex], let's find [tex]v[/tex] from (1):  

[tex]v=\frac{u.f}{u-f}[/tex]    (2)  

[tex]v=\frac{(12cm)(-6cm)}{12cm-(-6cm)}[/tex]

[tex]v=-4cm[/tex]   (3)

On the other hand, the magnification [tex]m[/tex] of the image is given by the following equations:  

[tex]m=-\frac{v}{u}[/tex]   (4)

[tex]m=\frac{h_{i}}{h_{o}}[/tex]   (5)

Where:

[tex]h_{i}[/tex] is the image height  

[tex]h_{o}=15cm[/tex] is the object height

Now, if we want to find the image height, we firstlu have to find [tex]m[/tex] from (4), substitute it on (5) and find [tex]h_{i}[/tex]:

Substituting  (3) in (4):

[tex]m=-\frac{-4cm}{12cm}[/tex]  

[tex]m=\frac{1}{3}[/tex]    (6)

Substituting  (6) in (5):

[tex]\frac{1}{3}=\frac{h_{i}}{15cm}[/tex]

[tex]h_{i}=\frac{15cm}{3}[/tex]

Finally we obtain the value of the height of the image produced by the mirror:

[tex]h_{i}=5cm[/tex]

Answer:

The answer is D. on edgen

Explanation:

D. 5.0

Answer:

False

Explanation:

A bloody footprint or a fingerprint would be considered a transfer stain.A transfer stain is one which occurs from an object coming into contact with already existing bloodstains and will leave swipes transfers behind.A passive stain is one that results from a blood drop,blood flow or blood pool.

Answer:

C. g/cm³

Explanation:

The slope is measured by calculating the variation of the Y values over the X values between two points on a line.  

So, the formula is: Slope = Δy/Δx

That means that we also take the units.

In this case, the Y-axis unit is in g, while the X-axis unit is in cm³.

Dividing a Y-variation over an X-variation will give you g/cm³.

In this case, let's assume the line passes through (10,100) (not exactly, but close enough for the example), and it passes through (0,0)

So the slope would be: (100-0) g / (10-0) cm³ = 10 g/cm³

Answer:

12 J

Explanation:

The amount of energy stored in a spring stretched by a distance x is:

E = 1/2 k x², where k is the stiffness of the spring.

So the spring has 4J of energy when it is stretched 10 cm:

4 J = 1/2 k (0.10 m)²

k = 800 N/m

When the stretched to 20 cm, the amount of energy is:

E = 1/2 (800 N/m) (0.20 m)²

E = 16 J

So it takes an additional 12 J to stretch the spring an additional 10 cm.

(A) 3.9

When a dielectric is inserted between the plates of a capacitor, the capacitance of the capacitor increases according to the equation:

[tex]C' = k C[/tex] (1)

where

C' is the final capacitance

k is the dielectric constant

C is the original capacitance

The capacitance is inversely proportional to the to voltage across the plates:

[tex]C=\frac{Q}{V}[/tex] (2a)

where Q is the charge stored and V the potential difference across the plates. We can rewrite C' (the capacitance of the capacitor filled with dielectric) as

[tex]C'=\frac{Q}{V'}[/tex] (2b)

Substituting (2a) and (2b) into (1), we find

[tex]V'=\frac{V}{k}[/tex] (3)

where

V = 45.0 V is the original voltage across the capacitor

V' = 11.5 V is the voltage across the capacitor filled with dielectric

Solving for k,

[tex]k=\frac{V}{V'}=\frac{45.0 V}{11.5 V}=3.9[/tex]

(B) 22.8 V

When the dielectric is partially pulled away, the system can be assimilated to a system of 2 capacitors in parallel, of which one of them is filled with dielectric and the other one is not.

Keeping in mind that the capacitance of a parallel-plate capacitor is proportional to the area of the plates:

[tex]C \propto A[/tex]

and in this case, the area of the capacitor filled with dielectric is just 1/3 of the total, we can write:

[tex]C_1 = \frac{2}{3}C\\C_2 = \frac{1}{3}kC[/tex]

where C1 is the capacitance of the part non-filled with dielectric, and C2 is the capacitance of the part filled with dielectric. The total capacitance of the system in parallel is

[tex]C'=C_1 + C_2 = \frac{2}{3}C+\frac{1}{3}kC=(\frac{2}{3}+\frac{1}{3}k)C[/tex]

Substituting,

[tex]C'=(\frac{2}{3}+\frac{1}{3}(3.9))C=1.97 C[/tex]

This is equivalent to a capacitor completely filled with a dielectric with dielectric constant k=1.97. Therefore, using again eq.(3), we find the new voltage:

[tex]V'=\frac{V}{k}=\frac{45.0 V}{1.97}=22.8 V[/tex]

When half of one of the charges of equal magnitude is transferred to the other, the electric force between them reduces to 75% of the original force while maintaining the same distance.

In this scenario, you're dealing with the concept of electric forces between charged particles. The force between two charges is given by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This can be represented by the equation F = k * Q1 * Q2 / r^2

Initially, we have two charges both of magnitude Q, the force is F_initial = k * Q * Q / r^2. Half of one of the charges is transferred to the other, the charges become 1.5Q and 0.5Q. The new force is F_final = k * 1.5Q * 0.5Q / r^2 = 0.75 k * Q * Q / r^2.

So, the force between the charges reduces to 75% of the original force when half of one of the charges is transferred to the other while the distance remains the same.

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Answer:

4.71 eV

Explanation:

For an electromagnetic wave with wavelength

[tex]\lambda=264 nm = 2.64\cdot 10^{-7} m[/tex]

the energy of the photons in the wave is given by

[tex]E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{2.64\cdot 10^{-7}m}=7.53\cdot 10^{-19} J[/tex]

where h is the Planck constant and c the speed of light. Therefore, this is the minimum energy that a photon should have in order to extract a photoelectron from the copper surface.

The work function of a metal is the minimum energy required by the incident light in order to extract photoelectrons from the metal's surface. Therefore, the work function corresponds to the energy we found previously. By converting it into electronvolts, we find:

[tex]E=\frac{7.53\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=4.71 eV[/tex]

Answer:

800 kg/m³

Explanation:

I assume you want to find the density of the sphere?

Start with a free body diagram.  There are three forces acting on the sphere: gravity pulling the sphere down, buoyancy pushing the sphere up, and tension pulling the sphere down.

Applying Newton's second law:

∑F = ma

B - W - T = ma

Since the sphere isn't accelerating, a = 0.

B - W - T = 0

B = W + T

We know that the tension is one-fourth the weight:

B = W + W/4

B = 5/4 W

B = 5/4 mg

Buoyant force is defined as:

B = ρVg,

where ρ is the density of the fluid, V is the displaced volume, and g is acceleration of gravity.

ρVg = 5/4 mg

ρV = 5/4 m

The mass of the sphere is equal to its density times its volume.  Since the sphere is fully submerged, it's volume is the same as the volume of the displaced water.

ρV = 5/4 ρₓV

ρ = 5/4 ρₓ

ρₓ = 4/5 ρ

So the density of the sphere is 4/5 the density of the water.  Water's density is 1000 kg/m³, so:

ρₓ = 4/5 (1000 kg/m³)

ρₓ = 800 kg/m³

The submerged sphere experiences a gravitational force (its weight) and a counteracting buoyant force. The tension in the string tethering it is one-fourth of the sphere's weight. This situation could occur if the buoyant force on the sphere is three-fourths of the sphere's weight.

The subject of this question is about the physics of forces in a fluid medium - specifically, the interaction between buoyancy, weight, and tension. In this case, a sphere is fully submerged in water and secured with a string. Despite being underwater, the object still experiences the force of gravity, which pulls it downward. This weight is represented by the mass of the sphere times the acceleration due to gravity (we use 9.8 m/s² for Earth).

However, while submerged, the sphere also experiences a buoyant force due to the displacement of the water. This force is equal to the weight of the water displaced by the sphere, which acts in the opposite direction of the weight, or upward. The string provides a tension force that prevents the sphere from moving.

In this scenario, the tension force in the string is one-fourth the weight of the sphere. This could happen if the buoyant force acting on the sphere is equal to three-fourths of the sphere’s weight, leaving one-fourth of the sphere's weight to be balanced by the string.

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Req = 30.0Ω.

When two or more resistors are in series, the intensity of current that passes through each of them is the same. Therefore, if you notice, you can observe that the three previous series resistors are equivalent to a single resistance whose value is the sum of each one.

Req = R1 + R2 + R3 = 10.0Ω + 10.0Ω + 10.0Ω = 30.0Ω

The equivalent resistance if you connect three 10.0 Ω resistors in series will be 30 ohms.

In the series circuit, the amount of current flowing through any component in a series circuit is the same and the sum of the individual resistances equals the overall resistance of any series circuit.

The voltage in a series circuit, the supply voltage, is equal to the total of the individual voltage drops.

[tex]\rm R_{eq}= R_1+R_2+R_3 \\\\ R_{eq}=10+10+10 \\\\ R_{eq}=30 \ ohms[/tex]

Hence, the equivalent resistance will be 30 ohms.

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Answer:

B) 2 times what it now is

Explanation:

The acceleration due to gravity at the surface of the Earth is given by

[tex]g=\frac{GM}{R^2}[/tex]

where

G is the gravitational constant

M is the mass of the Earth

R is the Earth's radius

In this problem, the mass of the Earth is doubled:

M' = 2M

while the radius remains the same:

R' = R

so the new acceleration due to gravity would be

[tex]g'=\frac{GM'}{R'^2}=\frac{G(2M)}{R^2}=2\frac{GM}{R^2}=2g[/tex]

so, the acceleration due to gravity would become twice the current value.

Note also that the value of g does not depend on the mass of the objects involved.

Answer:

[tex]160 kg/m^3[/tex]

Explanation:

The density of the liquid is given by:

[tex]d=\frac{m}{V}[/tex]

where

m is the mass of the liquid

V is its volume

The mass of the liquid is

m = 1.0 kg

while the volume is the volume of a cylinder with height h=0.20 m and radius r=0.10 m:

[tex]V=\pi r^2 h = \pi (0.10 m)^2 (0.20 m)=6.28\cdot 10^{-3} m^3[/tex]

So, the density is

[tex]d=\frac{1.0 kg}{6.28\cdot 10^{-3} m^3}=159.2 kg/m^3 \sim 160 kg/m^3[/tex]

Answer:

C. It is converted into another form, mainly kinetic energy

Explanation:

Potential energy is energy available to be used in an object that isn't currently moving. When an object begins moving, potential energy becomes kinetic energy.

Answer:

The answer is: C. It becomes another form, mainly kinetic energy.

Explanation:

Potential energy is that which has a body due to the height at which it is.

For example, a pot placed in a window at a certain height has potential energy.

When a body falls from a certain height, and if we despise friction with the air, what happens is that its potential energy is transformed into kinetic energy.

The answer is: C. It becomes another form, mainly kinetic energy.

The principal quantum number, identified by n, gives information about the orbital energy level of the electron.

To understand it better:

According to the current model of the atom, it has a central nucleus with electrons orbiting around. These orbits are located at different energy levels that are related to the distance from the electron to the nucleus.

So, the first energy level, is considered the lowest, because it is the smallest and the one that is in average closer to the nucleus, and as n increases, the farther away from the nucleus is the orbital and therefore more energy the electron has.

It should be noted that the values ​​of n will always be positive integer numbers, for example: 1, 2, 3, 4, 5, 6,7. Although theoretically its value oscillates between 1 and infinity, until now only atoms whose maximum energetic level is 8 are known.

Answer:

The size

Explanation:

4)Both B and D.This happens becaus3 at the ceryain time, A experiences a flood tide causing the waters to come near it.And since B and D are the closest coastlines they are affected the most.

The answer is false , because they move without an electromagnetic

Answer: False

Explanation:

Answer:

216.31 (the work done by gravity is -216.31) positive for going up.

Explanation:

We look at this question first by getting the right equation for work.

Which should be... W = F x D.

From this, we can do everything, we need the Force (F) first - the question tells us that Joe is lying on his back and moves his arms upward to raise the barbell. This means that he is countering the force of graving on the object.

What is the formula for the force of gravity on an object near the earth?

Right here --- [tex]F_{grav}[/tex] = mg

m = the mass and...

g  =  the acceleration due to gravity which is 9.81 m/s2

Before we plug things in though, we need to convert everything to SI units,

the weight is in kg - so we're good to go there, but the length of Joe's arms are in "cm" we need m or meters. Converting 70 cm to m = .7 m.

Now, we just put it all together - (31.5kg)(9.81m/s2)(.7m) =  216.31 J or 216.31 N m.

Answer:

[tex]2.68\cdot 10^8 MJ[/tex]

Explanation:

The power is related to the energy by

[tex]P=\frac{E}{t}[/tex]

where

P is the power

E is the energy

t is the time elapsed

The power of this nuclear power planet is

[tex]P=3100 MW = 3.1\cdot 10^9 W[/tex]

The time we are considering is 1 day, which is

[tex]t = 1 d = 86400 s[/tex]

So we can re-arrange the previous equation to find the energy produced by the power plant in one day:

[tex]E=Pt =(3.1\cdot 10^9 W)(86400 s)=2.68\cdot 10^{14} J=2.68\cdot 10^8 MJ[/tex]

Explanation:

This described situation is known as Refraction, a phenomenon in which the light bends or changes it direction when passing through a medium with a index of refraction different from the other medium.  

In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.  

According to Snell’s Law:  

[tex]n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})[/tex] (1)  

Where:  

[tex]n_{1}=1[/tex] is the first medium index of refraction  (air)

[tex]n_{2}[/tex] is the second medium index of refraction (the value we want to know)

[tex]\theta_{1}=27\°[/tex] is the angle of the incident ray  

[tex]\theta_{2}=11\°[/tex] is the angle of the refracted ray

Now, let's find [tex]n_{2}[/tex] from (1):

[tex]n_{2}=n_{1}\frac{sin(\theta_{1}}{sin(\theta_{2}}[/tex] (2)  

Substituting the known values:

[tex]n_{2}=(1)\frac{sin(27\°)}{sin(11\°)}}[/tex]

Finally:

[tex]n_{2}=2.379\approx 2.4[/tex]

If we compare this result with the given table, the index of refraction value that is close to this number is diamond's index of refraction.

Therefore, the correct option is A: the material is diamond.

Answer:

[tex]3.4\cdot 10^{-4} kg/m[/tex]

Explanation:

The order of the harmonics for standing waves in a string is equal to the number of nodes minus 1, so

n = 5 - 1 = 4

In this case, the frequency of the 4th-harmonic is

[tex]f_4=1.5 kHz = 1500 Hz[/tex]

We also know the relationship between the frequency of the nth-harmonic and the fundamental frequency:

[tex]f_4 = 4 f_1[/tex]

so we find the fundamental frequency:

[tex]f_1 = \frac{f_4}{4}=\frac{1500 Hz}{4}=375 Hz[/tex]

The fundamental frequency is given by

[tex]f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}[/tex]

where

L = 1.2 m is the length of the string

T = 276 N is the tension in the string

[tex]\mu[/tex] is the linear density

Solving the equation for [tex]\mu[/tex], we find

[tex]\mu = \frac{T}{4L^2 f_1^2}=\frac{276 N}{4(1.2 m)^2(375 Hz)^2}=3.4\cdot 10^{-4} kg/m[/tex]

Answer:

A. The wires will be arranged in order of increasing resistance.

Explanation:

The resistance of a wire is given by

[tex]R=\frac{\rho L}{A}[/tex]

where

[tex]\rho[/tex] is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

In this problem we have several wires made of the same material (so, same [tex]\rho[/tex]) and same length (same L): so, the only quantity that changes is their thickness, so their value of A.

We see from the formula that the resistance R is inversely proportional to the cross-sectional area A: therefore, the smaller the value of A, the larger the value of R. This means that if we arrange the wires in order of decreasing thickness, we are arranging the wires in order of increasing resistance.

Answer:

Earth

Explanation:

"Light travels at a speed of 299,792 kilometers per second; 186,287 miles per second. It takes 499.0 seconds for light to travel from the Sun to the Earth, a distance called 1 Astronomical Unit."

Source : https://image.gsfc.nasa.gov/poetry/venus/q89.html

Answer:

Explanation:

The atmosphere is the gaseous portion of the earth. It consists of different molecules of gas.

The geosphere is the solid portion of the earth which include the crusts, mantle and the core.

Earth is a dynamic planet. Our planet is dynamic in the sense that it is constantly changing and all its parts interacts with one another.

The gases in the atmosphere such a CO₂, H₂O, Nitrogen oxides originates from volcanic processes from deep within the earth. Hydrothermal vents and black smokers constantly release gases into the atmosphere.

The atmosphere plays a lot of roles in determining weather and climatic conditions. Agents of denudation like wind, water and glaciers are connected to the movement of gaseous portion of the earth. As new rocks forms on the crust, wind, water and glaciers acts on them. This process plays a central role in the rock cycle. The rock cycle would not be complete without agents of denudation which are strongly connected to the workings of atmospheric gases and materials like dusts.  

Therefore we see that the geosphere and atmosphere are linked.

If the lightbulb A in the circuit shown in the image burned out, the path for the  current to flow is disrupted because one of its terminals is connected direct to the source. So, there will be no current through the lightbulbs B, C, and D, and they will turn off. Similarly it will happen, if the lightbulb D burned out.

If the lightbulb B burned out the current will continue circulating through the lightbulbs A, C, and D, because lightbulb B is connected in parallel. Similarly it will happen, if the lightbulb C burned out.

In a series circuit, if one bulb burns out, it interrupts the flow of electricity, causing all other bulbs in the circuit to not light up.

Assuming the lightbulbs are in a series circuit, if lightbulb A burned out, the circuit would be interrupted and thus the other bulbs (B, C, and D) would also not light up as there would be no continuous pathway for current to flow. Similarly, if lightbulb B, C, or D burned out, it would act like a break in the circuit and just like the previous scenario, none of the other bulbs will light. The fundamental aspect of a series circuit is that the current is the same through all components, so if one pathway is interrupted (one bulb burns out), the entire circuit is affected.

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       Answer:

► 88 Earth Days

       Explanation:

It takes Mercury about 88 Earth days to complete an orbit around the sun. It also takes Mercury 59 Earth days to do one full spin on its axis. The length of one full day on Mercury would take 176 Earth days.

Mordancy.

Final answer:

Mercury's orbital period is 88 Earth days, while its rotation period is 59 days, leading to a 3:2 spin-orbit resonance. A day on Mercury lasts 176 Earth days, and the planet exhibits unique solar movement patterns due to its strange rotation.

Explanation:

Mercury's orbital period, or the length of time for one revolution around the sun, is 88 Earth days. This orbital period is known as a Mercury year. In contrast to its revolution, Mercury's spin or period of rotation is 59 Earth days. This difference in timescales leads to an interesting dynamic where a day on Mercury (defined as the Sun returning to the same position in the sky) lasts 176 Earth days. This is because Mercury has a unique 3:2 spin-orbit resonance, rotating three times on its axis for every two revolutions it completes around the Sun.

Visual studies initially suggested that Mercury had a synchronous rotation with the Sun, akin to how the Moon rotates with Earth, leading to the belief that one side was perpetually hot and the other perpetually cold. However, this was proved incorrect when it was found that Mercury's rotation and orbit are in a stable 2:3 ratio. Mercury's Strange Rotation results in extremely long days relative to the years and unusual solar movement patterns in the sky for hypothetical observers on the surface.

Alkali metals alkaline earth metals and halogens and noble gases. I’m pretty sure

Answer: Alkali metals alkaline earth metals and halogens and noble gases. I’m pretty sure

Answer: 5 units

Let's begin by stating clear that movement is the change of position of a body at a certain time. So, during this movement, the body will have a trajectory and a displacement, being both different:

The trajectory is the path followed by the body (is a scalar magnitude).

The displacement is the distance in a straight line between the initial and final position (is a vector magnitude).

According to this, in the description of the object (figure attached) placed at 0 on a number line and moving some units to the left and some oter units to the right, we are talking about the path followed by the object, hence its trajectory. So, 13 units is its trajectory.

But, if we talk about displacement, we have to draw a straight line between the initial position of the object (point 0) to its final position (point 5).

Now, being this an unidimensional problem, the displacement vector for this object is 5 units.

Answer:

Of longitudinal waves

Explanation:

Depending on the direction of the oscillation, there are two types of waves:

- Transverse waves: in a transverse wave, the oscillations occur perpendicularly to the direction of propagation of the wave. Examples are electromagnetic waves.

- Longitudinal waves: in a longitudinal wave, the oscillations occur parallel to the direction of propagation of the wave. In such a wave, the oscillations are produced by alternating regions of higher density of particles, called compressions, and regions of lower density of particles, called rarefactions. Examples of longitudinal waves are sound waves.

Compressions and rarefactions are characteristic of sound waves, which are longitudinal waves. Compressions are regions of high pressure, while rarefactions are regions of low pressure, both created by the vibrating motion of the sound source.

Compressions and rarefactions are characteristic features of a sound wave, which is a type of longitudinal wave. When a sound source vibrates, it causes fluctuations in the pressure of the medium through which it travels, typically air. These fluctuations manifest as alternating regions of higher pressure, known as compressions, and lower pressure, known as rarefactions. Sound waves consist of these repeating patterns of compressions and rarefactions, moving away from the source of the sound in the form of a wave.

During compression, air molecules are pushed closer together, leading to a higher pressure region. Conversely, during rarefaction, air molecules are spread out, creating a lower pressure region. For example, when a speaker cone moves forward, it compresses the air in front of it, and when it moves backward, it creates a rarefaction. These disturbances travel through the air, causing our eardrums to vibrate and enabling us to perceive sound.

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