Which statement is true?
A)
All waves travel at the same speed through any medium.
B)
All waves travel at the same speed if no medium is present.
Electromagnetic waves require a medium, but mechanical waves do not.
D)
Mechanical waves require a medium, but electromagnetic waves do not

The specific heat capacity of the metal, given that 27.777 g of water at 22.1 °C was mixed with the metal is 9.7×10⁻² Cal/gºC

Step 1: Obtain the heat absorbed by the water. This is shown below:

Q = MCΔT

= 27.777 × 1 × 7.2

= 199.9944 Cal

Step 2: Determine the specific heat capacity of the metal using the heat absorbed by the water. Details below:

Q = MCΔT

-199.9944 = 29.292 × C × -70.6

-199.9944 = -2068.0152 × C

Divide both sides by -2068.0152

C = -199.9944 / -2068.0152

= 9.7×10⁻² Cal/gºC

Thus, the specific heat capacity of the metal is 9.7×10⁻² Cal/gºC. None of the options are correct.

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Answer:

a=  2.97m/s²

Explanation: mg = (M+m)a, with M the mass of the g

a= 100(230+100)g .

Answer: the answer is corect of what everone else said

Explanation:

Answer:

A. 30 feet

Explanation:

Including reaction time, the stopping distance is more than 20 feet at 10 miles per hour, at 20 miles per hour it will be about 63 feet. Option B is correct.

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity.

Given conditions;

10 miles per hour ⇒ 20 feet

1 mile per hour ⇒ 2 feet

20 miles per hour ⇒ 20 × 2 feet

20 miles per hour ⇒ 20 × 2 feet

20 miles per hour ⇒40 feet

If the stopping distance is more than 20 feet at 10 miles per hour at 20 miles per hour it must be more than 40 feet.So that the correct answer is 63 feet.

Hence option B is correct.

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The speed of the roller coaster at the bottom of the hill if the track was frictionless is 34.04 m/s.

Given that the weight of the roller coaster is 2000 kg and the initial vertical drop of the ride is 59.3 m. We are to find the speed of the roller coaster at the bottom of the hill if the track was frictionless.We know that the roller coaster will lose potential energy due to the vertical drop. Assuming there is no friction, the potential energy will be converted into kinetic energy at the bottom of the hill.Considering the conservation of energy between the potential and kinetic energy, we can set the initial potential energy equal to the final kinetic energy. We can use the formula to calculate potential energy, which is PE = mgh where m = 2000 kg, g = 9.8 m/s², and h = 59.3 m. Therefore,PE = 2000 kg × 9.8 m/s² × 59.3 m = 1,157,924 JWe can use the formula to calculate kinetic energy, which is KE = 1/2mv² where m = 2000 kg and v is the final velocity. Therefore,KE = 1/2 × 2000 kg × v².The total energy remains constant as we know there is no friction. Therefore the final kinetic energy will be equal to the initial potential energy,1,157,924 J = 1/2 × 2000 kg × v²v² = (2 × 1,157,924 J) / 2000 kgv² = 1157.924v = √1157.924v = 34.04 m/s.

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the answer is B

i jus took the assement

Answer:

B

Explanation:

The physics behind the movement of electric charges between parallel plates is based on the principles of electrostatics. Electric charges are either positive or negative, and they are affected by electric fields.

Electric fields are created by a difference in electric potential, which is measured in volts. When a voltage is applied to a set of parallel plates, the charges within the plates will be affected by the electric field, and will move in response to it.

When a voltage is applied to a set of parallel plates, the positive charges in the plate connected to the positive voltage will be attracted to the negative voltage, while the negative charges in the plate connected to the negative voltage will be attracted to the positive voltage.

The movement of charges between the plates is also affected by the presence of any obstacles or resistances in the electric field, such as resistance in the wire. This can slow down the movement of charges and result in a decrease in the current flowing through the circuit.

In all, the movement of charges between electric parallel plates is the result of the electric field created by a difference in electric potential, and the movement of charges is called drift velocity. The movement is also affected by the presence of resistance.

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Energy is that which has the ability to cause change in matter.

Energy is the quantitative property that is transferred to a body or to a physical system, recognizable in the performance of work and in the form of heat and light. Energy is a conserved quantity—the law of conservation of energy states that energy can be converted in form, but not created or destroyed.

So in simple definition we can say that energy is that which has the ability to cause change in matter.

Based on the given statements we can classify them as;

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What the equation given?

Maximum velocity occurs at the equilibrium position

So

Now

Now

As we know the formula

\(\\ \rm\rightarrowtail V_max=A\omega\)

These expressions can be used

The maximum speed of the oscillating object will be given by \(V_{max}=Aw\)

An oscillation is defined as the repitative periodic motion of any object about its mean or equilibrium position.

The given equation is as follows:

x(t)=Acost

Maximum velocity occurs at the equilibrium position x=0

x(0)=Acos0

x(0)=A

Hence the maximum velocity will be  \(V_{max}=Aw\)

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Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

\(\rho (r) \ \alpha \ \delta (r -R)\)

\(\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)\)

\(\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)\)

To find the constant k, we  examine the total charge Q which is:

\(Q = \int \rho (r) \ dV = \int \sigma \times dA\)

\(Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2\)

∴

\(\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2\)

\(\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2\)

\((2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2\)

Thus;

\(k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2\)

\(k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2\)

\(k * R^2= \sigma \times R^2\)

\(k = R^2 --- (2)\)

Hence, from equation (1), if k = \(\sigma\)

\(\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}\)

\(\mathbf{\rho (r) =0 \ \ at \ \ r<R \ \ or \ \ r>R}\)

To verify the units:

\(\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}\)

↓         ↓            ↓

c/m³    c/m³  ×   1/m            

Thus, the units are verified.

The integrated charge Q

\(Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr\)

\(Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr\)

\(Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr\)

\(Q = 4 \pi \sigma *R^2\)    since  \(( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )\)

\(\mathbf{Q = 4 \pi R^2 \sigma }\)

Answer: A

Explanation:

2.1 Given that channel A's carrier frequency (100 kHz) and the IF frequency are different, the lowest local oscillator frequency that might be utilised to demodulate the signal on that channel is 50 kHz (50 kHz).

2.2 The difference between the IF frequency and the channel B carrier frequency (150 kHz) determines the lowest local oscillator frequency that could be used to demodulate the signal on channel B. This frequency is 100 kHz (50 kHz).

2.3 The difference between the carrier frequency of channel A and the local oscillator frequency is zero if the local oscillator is set to 100 kHz. Thus, the envelope of the modulated signal in channel A will be represented by a DC voltage (zero frequency) at the mixer's output. The channel B transmission won't be demodulated.

2.4 The difference between the carrier frequencies of channel A and channel B, assuming the local oscillator is set to 125 kHz, is 25 kHz, while the difference between channel B's carrier frequency and the local oscillator frequency is -25 kHz. Hence, the demodulated signal from channel A (1 kHz sinusoid) and the modulated signal will both be present in the mixer's output.

2.5 The difference between the carrier frequency of channel B and the local oscillator frequency is zero if the local oscillator is set to 150 kHz. Thus, the envelope of the modulated signal in channel B will be represented by a DC voltage (zero frequency) at the mixer's output. The channel A transmission won't be demodulated.

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The earth conducts seismic waves-- when an earthquake occurs, stations farther and farther away from the quake see the S and P waves propagated through deeper and deeper layers. By measuring the arrival time of the waves, the velocity of the waves can be found as a function of depth. There is clear evidence for several layers in the earth which both refract the waves and below which the velocities are different. This is a bit like holding a book under your desk while a friend (gently) taps the other end of the desk. If you listen closely to the sound (i.e. with your ear on the top of the desk) the sound changes noticeably if a large book is pressed up against the bottom of the desktop. You might also try to figure out how to tell if a golf ball is wound, liquid filled, or solid-- without looking at it. (You can tell if you hit one!)

Scientists can tell by observing the seismic waves that are recorded all over the surface of the earth from distant earthquakes. The seismic waves are reflected (bounced off) layers of different densities, and they are refracted (bent) when they enter layers of different densities. Some of them don't go through liquid at all (the S waves). Scientists have been monitoring earthquakes and studying the phases of seismic waves that arrive at different stations for - hmm, well I don't exactly know - but certainly at least the last 75 years, with more and more sophisticated equipment. Seismologists look at the little wiggles that are made by pens on paper, connected to seismometers, every time a wave from an earthquake anywhere in the world passes under their station. It is by studying many of these seismic records, for many years, and pooling all our knowledge, that we have been able to come up with a working model of what the inside of the earth is made of, where the boundaries between layers of different density and composition lie, and why we have earthquakes where we do.

to Something you can do to model how seismologists "listen" to earthquakes is densitieshave a friend tap on a big table while you put your ear on the table at the other end. We put sensitive instruments in the ground that act like ears so we can detect seismic waves from distant earthquakes.

(a) The maximum height to which Superman can lift the water is 10.32874 m

(b) On the Moon there is no atmosphere so no atmospheric pressure which means when the straw is placed in water water will not rise in the tube.

Pₐ = Atmospheric pressure = 101325 Pa

g = Acceleration due to gravity = 9.81 m/s²

h = Height of water

ρ = Density of water = 1000 kg/m³

If the walls of the tube do not collapse that means that maximum pressure inside will be the atmospheric pressure

Atmospheric pressure is given by

Pₐ = ρgh

height = Pₐ /ρg

h = 101325 / 1000 * 9.81

h = 10.328 m

The maximum height is 10.32874 m.

(b) On the Moon there is no atmosphere so no atmospheric pressure which means when the straw is placed in water water will not rise in the tube.

Therefore, the maximum height to which Superman can lift the water is 10.32874 m.

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[NOTE: THIS IS AN INCOMPLETE QUESTION. THE COMPLETE QUESTION IS: Superman attempts to drink water through a very long vertical straw. With his great strength, he achieves maximum possible suction. The walls of the tubular straw don't collapse.

(a) Find the maximum height through which he can lift the water.

(b) Still thirsty, the Man of Steel repeats his attempt on the Moon, which has no atmosphere. Find the difference between the water levels inside and outside the straw.]

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

C should be the correct one

Answer: Distance traveled in time t is s

Explanation: Self Explanatory

Answer:

25

Explanation:

magnitude and (b) direction (as an angle relative to the x axis) of the velocity

Answer:

The work done in bringing the plates together is 5.9 x 10⁻¹⁰ J.

Explanation:

Given;

potential difference between the plates, V = 10 V

length of each square side of the plates, L = 20 cm = 0.2 m

area of the plates, A = 0.2 x 0.2 = 0.04 m²

separation of the plates, d = 3 cm = 0.03 m

The work done in bringing the plates together is calculated as;

W = ¹/₂qV

\(W = \frac{1}{2} (\frac{\epsilon_0 A }{d}V) \times V\\\\W = \frac{\epsilon_0 A V^2}{2d}\\\\W = \frac{8.85 \times 10^{-12} \ \times \ 0.04\ \times \ 10^2}{2(0.03)} \\\\W = 5.9 \times 10^{-10} \ J\)

Therefore, the work done in bringing the plates together is 5.9 x 10⁻¹⁰ J.

The wavelength of light that should be used to obtain fringes that are 0.0045 m wide after reducing the distance between the screen and the slit by half is 2.25 * 10^7 Å.

Huygens's Principle states that every point on a wavefront can be considered as a source of secondary spherical wavelets that spread out in all directions with the same speed as the original wave. The new wavefront is formed by the envelope of these secondary wavelets at a later time.

Now, let's consider a Young's double-slit experiment. In this experiment, when light passes through two narrow slits, it creates an interference pattern on a screen behind the slits. The fringe width is the distance between two consecutive bright or dark fringes in the pattern.

Given that the fringe width obtained is 0.6 cm and the wavelength of light used is 4500 Å (Angstroms), we can calculate the wavelength of light required to obtain fringes that are 0.0045 m wide.

We can use the formula for fringe width in Young's double-slit experiment:

w = (λ * D) / d

Where:

w is the fringe width,

λ is the wavelength of light,

D is the distance between the screen and the double slits, and

d is the distance between the two slits.

Let's calculate the value of D/d using the given information:

D/d = w / λ

= 0.006 m / 4500 Å (1 m = 10^10 Å)

= 0.006 * 10^10 / 4500 m^-1

Now, if the distance between the screen and the slit is reduced by half, the new value of D/d would be:

(D'/d) = (0.006/2) * 10^10 / 4500 m^-1

Now, we can rearrange the equation to solve for the new wavelength (λ'):

(λ' * D') / d = (D/d)

λ' = (D/d) * d / D

= [(0.006/2) * 10^10 / 4500] * (4500 / 0.006) Å

= 0.0045 m * 10^10 / 2 Å

= \(0.00225 * 10^{10\) Å

=\(2.25 * 10^7\)Å

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Answer:

120 kg•m/s.

Explanation:

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

Case 1

Mass of object = M

Velocity of object = V

Momentum = 15 kg•m/s

Case 2

Mass of object = 2M

Velocity of object = 4V

Momentum = ?

Momentum is defined as follow:

Momentum = mass × velocity

The momentum of object in case 2 can be obtained as follow:

From case 1

Momentum = mass × velocity

15 = M × V

15 = MV ....... (1)

From case 2:

Momentum = mass × velocity

Momentum = 2M × 4V

Momentum = 8MV ....... (2)

Finally , substitute the value of MV in equation 1 into equation 2.

Momentum = 8MV

MV = 15

Momentum = 8 × 15

Momentum = 120 kg•m/s

Therefore, an object with a mass of 2M and 4V would have a momentum of 120 kg•m/s

Answer:

1470J

Explanation:

Given parameters:

Mass of the person  = 50kg

height  = 3m

Unknown:

Work done  = ?

Solution:

To solve this problem, we use the expression below:

     Work done  = mass x acceleration due gravity x height

So;

     Work done  = 50 x 9.8 x 3 = 1470J

45º angle will result in the trebuchet’s furthest thrown projectile.

It is acceptable to infer from the results shown in the demonstration on Interactive Physics that as the weight of a counterweight on a trebuchet rises, so too should the projectile's range. Except for the fact that the distance did not rise linearly but rather more quadratically, the results support the theory. The results showed that the distance rose as the weight of the trebuchet's counterweight was increased in steps of 5 kilograms, starting at 20 kilograms and ending at 200 kilograms.

Each time, the projectiles were launched from the trebuchet at a 45-degree angle, and their distances typically followed the equation -8.1551E-4x2 +.304388x + 8.12756 (where x is the mass of the counterweight). The graph was thought to be more quadratic than linear because gravity has more time to work against the projectile and pull it down to the earth the longer it is in the air. Therefore, as additional mass is applied and the projectile is in the air for a longer period of time, the projectile distances would not grow as quickly. These findings back up Newton's Third Law of Motion as well as earlier, historical investigations.

Thus, Thomas should launch the trebuchet at a 45º angle to get the farthest thrown projectile.

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The mass of a pure platinum disc can be gotten by multiplying the density with the volume.

Therefore the mass is 2418.2 grams or 2.4182 kilograms.

A body's mass is an inherent quality. Prior to the discovery of the atom and particle physics, it was widely considered to be tied to the amount of matter in a physical body.

The kilogram is the primary mass unit in the SI.

The resistance of the body to acceleration in the presence of a net force can be measured as mass.

Due to the lower gravity on the Moon, an object would weigh less than it does on Earth while maintaining the same mass. This is due to the fact that mass, coupled with gravity, determines the strength of weight, which is a force.

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What is the mass of a pure platinum disk with a volume of 113 cm3? The density of platinum is 21.4 g/cm3.

Give your answer in grams and kilograms.

Answer:

(1.8 × 10^9) meters in one minute

Explanation:

To determine how far light can travel in one minute, we need to multiply its speed by the number of seconds in a minute.

The speed of light is 3 × 10^8 meters per second.

There are 60 seconds in a minute.

Therefore, the distance light can travel in one minute is:

Distance = Speed × Time

Distance = (3 × 10^8 meters per second) × (60 seconds)

Calculating this, we get:

Distance = 3 × 10^8 meters/second × 60 seconds

Distance = 18 × 10^8 meters

Distance = 1.8 × 10^9 meters

So, light can travel approximately 1.8 billion (1.8 × 10^9) meters in one minute.

Answer:

200 km/hr

Explanation:

Since he goes 80km per hour, multiply this by 2.5 or two and a half hours.

80 x 2.5 = 200 km/hr.

All of that energy is used to compress the bumper by x, which is equal to 6 inches, or 0.5 feet, and it is stored as spring potential energy in the bumper. (1/2)*k*x2 is the formula for that energy.So, (1/2)*m*u2 = (1/2)*k*x2 Simplifying and rearranging results in k = (m*u2) / x2 Based on the data provided, k = 20 000 lb/ft.

Energy is the capacity to work in physics. It could be potential, kinetic, thermal, electrical, chemical, nuclear, or some other form. Heat and work are also present—for instance when energy is transferred from one body to another.

The body's mass (m) will change by an amount equal to E/c2 whenever the body's energy changes in any form, as shown by the Albert Einstein equation, commonly stated as E = mc2.

Energy is defined as the "ability to do work, which is the ability to exert a force causing the displacement of an object." The meaning is quite clear, despite this perplexing definition: Energy is the only thing that moves things. Energy can be divided into two categories: potential and kinetics

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Answer:

suna bro i am nepali and nepali are pro

Answer: The statement that is a likely impact of stronger than normal trade winds in the Pacific Northwest to the United States is "Jet stream would be displaced southwards causing heavy rain and flooding."

Explanation: Hope this helps!