part 2 of 2
With what speed does the stone strike the
water?
Answer in units of m/s.

A heat engine cannot have a thermal efficiency of 100% For all reversible processes, the second-law efficiency is 100%. The second-law efficiency of a heat engine cannot be greater than its thermal efficiency. The second-law efficiency of a process is 100% if no entropy is generated during that process

Answer:

5300km/h

Explanation:

That's an easy question. You can solve it yourself.

The answer can be found by dividing the Distance traveled by the Time.

Now you can solve all the problem of this rule.

Answer:  The most massive car, x, will take the longest to reach 60 mph.

Explanation:  The inertia of a car is dependent on its mass.  F=ma.  Given that each car has the same force (identical engines) the one with the higher mass will accelerate more slowly.

Newton's second law allows to find the result for the friction coefficient of the street is:

Newton's second law establishes a relationship between the net force, the mass and the acceleration of the body.

           ∑ F = m a

Where bold indicates vectors, m is to mass and acceleration.

In the attached we see a free body diagram, it is a diagram of the forces without the details of the body, the x-axis is parallel to plane also shown with the positive in the direction of movement, going down the plane and the y-axis perpendicular to the plane.

Let's use trigonometry to break down the weight.

        Sin θ = [tex]\frac{W_x}{W}[/tex]  

        cos θ = [tex]\frac{W_y}{W}[/tex] / W

        Wₓ = W sin θ

         [tex]W_y[/tex]  = W cos θ

We write Newton's second law for each axis.

y-axis

         N- [tex]W_y[/tex]  = 0

         N = mg cos θ

x-axis

        Wₓ - fr = ma

Since they indicate that the body goes down at a constant speed, the acceleration is zero.

         W sin θ = fr

The friction force is the macroscopic representation of the interactions between the two surfaces and the formula.

         fr = μ N

we substitute.

          fr = μ mg cos θ

         mg sin θ = μ cos θ  

        μ = tan θ

Let's calculate.

         μ = tan 38.0

         μ = 0.78

In conclusion using Newton's second law we can find the results for the friction coefficient of the street is:

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Answer:

80kg

Explanation:

A = Fm

0.5 = 40m

m = 40/0.5

m = 80kg

Answer:

A)250-240

=10+960

=970

We have that for the Question, it can be said that the maximum velocity is

From the question we are told

A 2.00kg block is attached to a horizontal ideal spring with a spring constant k=100Nm.

The block-spring system is set on a horizontal surface with negligible friction.

Generally the equation for the Potential energy  is mathematically given as

[tex]P.E=\frac{1}{2}Rx_m^2\\\\2=\frax{1}{2}*100x_m^2\\\\x_m^2 = 0.04m\\\\x_m = 0.2m[/tex]

The PE is converted to KE, Therefore

[tex]KE = 2\\\\\frac{1}{2}MV_m^2 = 2\\\\V_m^2 = \frac{4}{M}\\\\V_m^2 = \frac{4}{2}\\\\V_m = \sqrt{2}\\\\V_m = 1.414m/s[/tex]

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Complete Question

A 2.00 kg block is attached to a horizontal ideal spring with a spring constant k = 100 N The block-spring system is set on a horizontal surface with negligible friction. A graph of the potential energy U as a function of time t for this system is shown. The maximum displacement IMAX of the block from its equilibrium position and the maximum speed Umat of the block during the motion represented by the graph are most nearly A IMAX = 2.0 m and UMAX 1.4" B UMAX = 1.4 and UMAX=0.20" с MAX 0.20 m and UMAX = 1.4" D IMAX 0.40 m and UMAX 1.4 E IMAX 0.04 m and UMAX 2.0"

y = y 0 + v 0 y t − 1 2 g t 2 . If we take the initial position y 0 to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity: v 0 y = v 0 sin θ 0 = ( 30.0 m / s ) sin 45 ° = 21.2 m / s .

Both choices are correct.

Some seismic waves travel along the Earth's surface.

Others don't.

Answer:

False

Explanation:

Edge2022

Answer:

This demonstration is often done following a discussion of the ideal gas equation of state, PV=nRT.

We begin by weighing a balloon, then blowing it up and weighing it again. In the photo shown on right, the mass indication increased from 3.4 to 3.5 grams. At this point, it is important to note that the scale measures force, even though it reports a conclusion about mass based on the force measurement.

One assumption made in reaching the conclusion is that the buoyant force on the object being weighed is negligible. In the case of the balloon, this is incorrect. The buoyant force on this balloon is equal to the weight of the air displaced.

Since the volume of air inside the balloon is essentially the same as the volume of air displaced, we should expect that the buoyant force would support the weight of the air inside the balloon: The reported mass should not go up at all, because the force required of the scale should not change.

The increase in reported mass of .1 gram is attributed to the higher density of the air inside the balloon: The tension in the balloon compresses the air inside, as attested by the pressure required to blow the balloon up. Evidently, for this experiment, the pressure inside is greater than atmospheric by about 2%.

In the picture at right, the balloon is being pressed into a pan of liquid nitrogen. (The pan is the styrofoam lid of a small lunch box.) The balloon floats lightly on the liquid nitrogen unless pressed down. Pressing down places more surface area in contact with the cold nitrogen and speeds the demonstration. It is interesting to note the buoyant force by this liquified constituent of air.

The balloon shrinks dramatically, as indicated below. When left in contact with the liquid nitrogen long enough (perhaps 5 minutes) the oxygen inside the balloon liquifies, and then the nitrogen liquefies also. Close observation of the photo at the upper left corner of the pan shows some liquid nitrogen bubbles may forming above the dark spot in the center of the pan. One can also make out a faint line at the upper left corner of the pan which is the liquid nitrogen surface. The balloon still floats, riding rather high on that surface. Evidently, some of the balloon contents remain in the gas phase, making the mass of the balloon less than the mass of the displaced liquid nitrogen.

Next, we take the shrunken balloon and place it back on the scale, as above. In this instance, the reported mass is 8.7 grams, an increase of 5.2 grams.

A look at the figure on the right shows a faint line near the bottom of the cold balloon. Above that line, the balloon contains gas; below the liquid. That line represents the top surface of the liquid air inside the balloon. With this evidence, the easy thing to say would be, "Of course, liquids are heavier than gases," but that would be incorrect. We assert that the amount of air inside the balloon has not changed and that the mass of that air is not dependent on temperature.

If these assertions are true, then the force of gravity on the balloon has not changed. The scale reading is determined by the force which it must exert on the balloon in order to keep it stationary. Evidently, the required force is larger when the balloon is shrunken. The reason is that the buoyant force (upward) has decreased to practically zero, leaving the scale alone to balance the downward force by gravity.

From the data, we can say that the change in the buoyant force is equal to the weight associated with the apparent change in mass. The weight of 5.2 grams is about .052 newtons. The buoyant force is less now because the balloon displaces less air. If we could measure the change in volume of the balloon as DV, then the buoyant force would be (r g DV) upwards, where r is the density of air that was displaced by the balloon, and g is the gravitational field strength, 9.8 Newton/kg.

Note that the .052 newton force is not the weight of the air inside the balloon. Rather, it is the weight of the air that was displaced by the balloon. If we ignore the compression of air inside the balloon, the two numbers are the same. However, the two samples are completely different.

We can estimate the volume of the balloon by assuming that the hand in the photograph is about .1meters across. For purposes of estimation, we say that the volume shrank to almost zero when the balloon was cold so that the change in volume was nearly equal to the original volume. Plugging in numbers gives fair agreement with the book value of 1kg/cubic meter for the density of air.

The value for the density of air is secondary to two main features of this demonstration:

Large changes in temperature produce the large changes in volume that are indicated by the ideal gas equation.

The mass of air in a volume equal to the volume of a balloon can be determined provided that the buoyant force is understood.

Explanation:

time =4.5sec

gravity =9.8m/s square

so to determime the height we can use the formula

1/2 of gravity times time square

which results in 99.25

Please find attached photograph for your answer.

Hope it helps.

Do comment if you have any query.

Answer:

A

Explanation:

Answer:

PE=0

Explanation:

If the tank is on the ground, there is 0 height which means the equation of PE=mh will be PE=12543(0)=0, hope this helps!

Answer:

Potential

Explanation:

The internal energy is the total amount of kinetic energy and potential energy of all the particles in the system. When energy is given to raise the temperature, particles speed up and gain kinetic energy.

Hope this helps :)

Answer:

i think its going to be 150 because its half of 300

Explanation:

Answer:

Explanation:

A decrease in the density of something is rarefaction. ... Most of the time, rarefaction refers to air or other gases becoming less dense. When rarefaction occurs, the particles in a gas become more spread out. You may come across this word in the context of sound waves.

Answer:

Refer to the attachment

There are many reasons actually...

◇ Clayey soil is easy to mould than the other soils.

The other soils (like sandy soil & loamy soil) are grainy or dry in nature & cannot be moulded. In the ancient times, when technology & the know-hows were still not introduced, people used clayey soul to make earthern pots to store stuff. This practice still continues even today.

◇ It keeps water cool.

Ever wondered why water stored in the earthenware remains cool even on a sunny day? It's because the heat required for evaporation is taken from water inside the pot & thus it cools the water stored inside.

◇ Clayey soil is also moist, durable, more plastic like & easy to retain making it ideal for potters to make earthenware.

__________

Clay soil is used to make earthenwares because, clay can be easily molded to desired sizes and shapes. Clay absorbs water moiety and make the pot very cool inside.

Clay is a type of soil made by the sedimentation process near water resources.  They are rich in minerals. Clay is rich in silicates and intermolecular force in the soil is comparatively less and can be easily molded any shape and size.

When clay is fired to a high enough temperature, the glass-forming materials in the ceramic form a liquid glass. This glass bridges the gaps between the clay particles. As the pottery cools at the end of the firing process, the glass between the clay particles solidifies again.

Clay pots make the water cooler inside. Hence, clay is used to make earthen wares.

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#SPJ2

Answer:

weight

Explanation:

weight depends on gravity

Answer:

Step 1: Calculate the amount of sodium hydroxide in moles

Volume of sodium hydroxide solution = 50.0 ÷ 1,000 = 0.05 dm3

Rearrange:

Concentration in mol/dm3 = amount of solute in molvolume in dm3

Amount of solutein mol = concentration in mol/dm3 × volume in dm3

Amount of sodium hydroxide = 0.300 × 0.05

= 0.01 5mol

Step 2: Find the amount of hydrochloric acid in moles

The balanced equation is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

So the mole ratio NaOH:HCl is 1:1

Therefore 0.015 mol of NaOH reacts with 0.015 mol of HCl

Step 3: Calculate the concentration of hydrochloric acid in mol/dm3

Volume of hydrochloric acid = 25.00 ÷ 1000 = 0.025 dm3

Concentration in mol/dm3 = amount of solute in molvolume in dm3

Concentration in mol/dm3 = 0.015/0.025

= 0.6 mol/dm3

Step 4: Calculate the concentration of hydrochloric acid in g/dm3

Relative formula mass of HCl = 1 + 35.5 = 36.5

Mass = relative formula mass × amount

Mass of HCl = 36.5 × 0.6

= 21.9 g

So concentration = 21.9 g/dm3

Hi there!

This is an example of a recoil collision.

Let m1, v1 be Mary

Let m2, v2 be Jim

Utilize the conservation of momentum:

m1v1 + m2v2 = m1v1' + m2v2'

The initial velocities are zero (at rest), so:

0 = m1v1' + m2v2'

Let east be positive, so Jim recoils with a negative velocity:

0 = m1v1' - m2v2'

Set the two equal and solve:

m1v1' = m2v2'

m1v1'/m2 = v2'

(50)(2.5)/65 = v2' = 1.92 m/s to the WEST

Hi there!

Since we are dealing with such large distances, we can approximate.

Recall that the equation for speed is:

displacement / time = velocity

OR:

d/t = v

Begin by converting one hour to seconds:

1 hr  = 3600 sec

Now, we can solve for velocity:

(-2.0 × 10⁷) / 3600 = -5555.6 m/s ⇒ D: -5.55 × 10³ m/s

Answer:

180 meters

Explanation:

The average distance between the two stars is 792 light years

Let the mass of the first star be [tex]m_1[/tex]

Let the mass of the second star be [tex]m_2[/tex]

The combined mass of the two stars, [tex]m_1+m_2=5.5[/tex] solar masses

The orbital period of the stars, P = 12 years

Average distance between the two stars, D = ?

The average distance between the two stars can be calculated using Kepler's equation

[tex]D=(m_1+m_2)P^2[/tex]

Substitute [tex]m_1+m_2=5.5[/tex] and P = 12 into the formula [tex]D=(m_1+m_2)P^2[/tex]

[tex]D=5.5(12^2)[/tex]

D = 5.5(144)

D = 792 light years

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Answer:

we can say when UV light is incident. Essentially, when it strikes glass or incident upon gloss, it will be absorbed. The energy has to go somewhere. Um, and in this case, this radiation is then converted to thermal energy and the glass raises and temperature.

Explanation:

Got the information from the internet

Explanation:

Exoplanets are very hard to see directly with telescopes. They are hidden by the bright glare of the stars they orbit. So, astronomers use other ways to detect and study these distant planets.

Exoplanets are difficult to detect because they are a star 'wobbles' as it orbits the center of mass, changing the wavelength of light it emits.