Examine the five words and/or phrases and determine the relationship among the majority of words/phrases. Choose the one option that does not fit the pattern. A. abyssal clay B. calcareous ooze C. coarse lithogenic sediment D. manganese nodule E. siliceous ooze

Answer:

C- ±37.6°

Explanation:

This is the case of a single slit diffraction. In this case, the dark fringe occurs at

sin θ = ± mλ/a

where m = 1

θ = ± 30°

Hence we have sin ± 30° = ± λ/a

±0.5 = ± λ/a

Hence, we have a = 2λ

For a circular aperture, the condition for the first dark fringe is

D = diameter of circle = a = 2λ

so we have sin θ1 = 1.22λ/a

sin θ1 = 1.22λ/2λ

hence sin θ1 = 0.61

θ1 = sin⁻¹0.61 = ±37.6°

Answer:

40 ft/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = -32 ft/s² = a (downward is taken a negative)

Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -32\times -25+0^2}\\\Rightarrow v=40\ ft/s[/tex]

The speed of Isaac as he reached the ground is 40 ft/s

"The final velocity of Isaac when he hits the ground can be calculated using the kinematic equation for an object in free fall under the influence of gravity with an initial velocity of zero:

[tex]\[ v_f^2 = v_i^2 + 2a\Delta y \][/tex]

Plugging in the known values:

[tex]\[ v_f^2 = 0^2 + 2(-32 \text{ ft/sec}^2)(25 \text{ ft}) \] \[ v_f^2 = 0 + (-2)(32)(25) \] \[ v_f^2 = -1600 \][/tex]

Since velocity cannot be negative, we take the positive square root of the absolute value of [tex]\( v_f^2 \)[/tex] to find [tex]\( v_f \)[/tex]:

[tex]\[ v_f = \sqrt{1600} \] \[ v_f = 40 \text{ ft/sec} \][/tex]

Therefore, Isaac is traveling at 40 feet per second when he hits the ground.

The answer is: [tex]40 \text{ ft/sec}.[/tex]

Answer:

39.4 g

39.4 cm³

1.09137 g/cm³

Explanation:

[tex]\rho[/tex] = Density of water = 1 g/cm³

Mass of water displaced will be the difference of the

[tex]m=43-3.6\\\Rightarrow m=39.4\ g[/tex]

Mass of water displaced is 39.4 g

Density is given by

[tex]\rho=\dfrac{m}{v}\\\Rightarrow v=\dfrac{m}{\rho}\\\Rightarrow v=\dfrac{39.4}{1}\\\Rightarrow v=39.4\ cm^3[/tex]

So, volume of bone is 39.4 cm³

Average density of the bird is given by

[tex]\rho=\dfrac{43}{39.4}\\\Rightarrow \rho=1.09137\ g/cm^3[/tex]

The average density is 1.09137 g/cm³

The spin quantum number (ms) describes the orientation of the spin of the electron: TRUE

The magnetic quantum number (ml) describes the size and energy associated with an orbital. An orbital is the path that an electron follows during its movement in an atom: FALSE

The angular momentum quantum number (l) describes the orientation of the orbital: FALSE

The principal quantum number (n) describes the shape of an orbital: FALSE

Explanation:

The statement contains mixed truths and falsehoods. The spin quantum number describes the orientation of electron spin, the magnetic quantum number pertains to orbital orientation, and the angular momentum quantum number concerns orbital shape, while the principal quantum number describes orbital size and energy.

The statement is false. The spin quantum number (ms) does describe the orientation of the spin of the electron, either up or down. The magnetic quantum number (ml) describes the orientation of the orbital in space. An orbital is defined as a region in space where there's a high probability of finding an electron, not their path. The angular momentum quantum number (l) determines the shape of the orbital. The principal quantum number (n) describes the size and energy level of an orbital.

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Answer:

New potential energy will be 784 J

So option (b) will be correct answer

Explanation:

We have given height of the shelf h = 10 m

Potential energy E = 980 J

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

We know that potential energy [tex]E=mgh[/tex]

So [tex]980=m\times 9.8\times 10[/tex]

m = 10 kg

Now new height h = 8 m

So new potential energy [tex]E=mgh=10\times 9.8\times 8=784J[/tex]

So option (B) will be the correct answer

Answer:

option (b) is correct.

Explanation:

h1 = 10 m

U 1 = 980 J

Let mg be weight of mass.

U1 = m x g x h1

980 = mg x 10

mg = 98

Now h = 8 m

U2 = mg x h = 98 x 8

U2 = 784 J

thus, the potential energy is 784 J.

Answer : The original activity will be, 303 millicuries.

Explanation :

Half-life = 15 hr

First we have to calculate the rate constant, we use the formula :

[tex]k=\frac{0.693}{t_{1/2}}[/tex]

[tex]k=\frac{0.693}{15hr}[/tex]

[tex]k=4.62\times 10^{-2}\text{ hr}^{-1}[/tex]

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant  = [tex]4.62\times 10^{-2}\text{ hr}^{-1}[/tex]

t = time passed by the sample  = 24 hr

a = initial amount of the reactant  = ?

a - x = amount left after decay process = 100 millicuries

Now put all the given values in above equation, we get

[tex]24=\frac{2.303}{4.62\times 10^{-2}}\log\frac{a}{100}[/tex]

[tex]a=302.97\text{ millicuries}\approx 303\text{ millicuries}[/tex]

Therefore, the original activity will be, 303 millicuries.

Explanation:

The shorter wavelength electromagnetic waves from sun are absorbed by earth material in form of short wavelength and the radiated wavelength are longer ones. Also higher energy waves are of shorter wavelength and lower energy waves have longer wavelength. So, they are absorbed as short wavelength and radiated back as long wavelength.

Final answer:

About 8,960 atoms of carbon-14 were present in the object when it was made 18,000 years ago, based on its current carbon-14 content and the known half-life of carbon-14.

Explanation:

To calculate how many atoms of carbon-14 (14C) were present in an object that was created 18,000 years ago, we use the half-life of 14C, which is 5,730 years. Given that there are 1,000 atoms of 14C present now, we need to find out how many half-lives have passed to work backwards and determine the initial quantity of 14C atoms.

First, we divide the age of the object by the half-life of 14C:

18,000 years / 5,730 years per half-life ≈ 3.14 half-lives

Now, we apply the formula for exponential decay, taking into account the number of half-lives:

Initial Quantity = Present Quantity × (2⁰ of half-lives)

Initial Quantity = 1000 atoms × (2≈ 3.14)

Initial Quantity ≈ 1000 atoms × 8.96

Initial Quantity ≈ 8960 atoms of 14C were present when the object was made.

Answer:

2. False

Explanation:

Entropy can be interpreted as a measure of the random distribution of a system. According to the second law of thermodynamics a system in an unlikely condition will have a natural tendency to reorganize itself to a more probable condition, this reorganization will result in an increase in entropy (disorder).

Answer:

The final pressure will be 0.67 atm.

Explanation:

Using Boyle's law  

[tex] {P_1}\times {V_1}={P_2}\times {V_2}[/tex]

Given ,  

V₁ = 10 L

V₂ = 15 L

P₁ = 1.0 atm

P₂ = ?

Using above equation as:

[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]

[tex]{1.0}\times {10}={P_2}\times {15} atm[/tex]

[tex]{P_2}=\frac{{1.0}\times {10}}{15} atm[/tex]

[tex]{P_2}=0.67\ atm[/tex]

The final pressure will be 0.67 atm.

Answer:

A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is 0.67 atm.

Explanation:

Boyle's law states that in constant temperature the  variation volume of gas is inversely proportional to the applied pressure.

The formula is,

[tex]\rm \bold{ P_1\times V_1= P_2\times V_2}[/tex]

Where,

[tex]\rm \bold P_1[/tex] is initial pressure = 1 atm

[tex]\rm \bold P_2[/tex] is final pressure =  ? (Not Known)

[tex]\rm \bold V_1[/tex] is initial volume  = 10 L

[tex]\rm \bold V_2[/tex] is final volume = 15 L

Now put the values in the formula,

[tex]\rm 1\times 10 = P_2\times 15\\\\\rm P_2 = \frac{10}{15\\} \\\\\rm P_2 = 0.67[/tex]

Therefore, the answer is 0.67 atm.

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Answer:

186 lbs per man.

Explanation:

If we assume that we have 35 men in the elevator, and the elevator will be overloaded if the total weight is in excess of 6510-lbs, so we just need to take average:

6510 lbs / 35 = 186 lbs per man.

Have a nice day!

Answer:

Elastically

Explanation:

A rock that has deformed Elastically under stress keeps its new shape when the stress is released.

In elastic deformation the original shape of the object is regained when the stress is removed. Whereas in plastic deformation the original shape is parmanently deformed with the application of stress.

A rock that deforms plastically under stress will retain its new shape when the stress is released, indicating it has surpassed its yield point.

A rock that has deformed plastically under stress will keep its new shape when stress is released. When rocks are subjected to stress, they can undergo elastic, brittle, or plastic deformation. In the elastic state, rocks, like a rubber band, will return to their original shape once the imposed stress is removed, provided the stress does not exceed the elastic limit of the rocks. Beyond this limit, the material will experience plastic deformation, leading to a permanent change in shape as mineral bonds break, shift, and reform.

This behavior depends on several factors, including stress type, rock type, depth, and environmental conditions (pressure and temperature). At greater depths, where the conditions are high pressure and temperature, rocks are more likely to deform plastically. Furthermore, when rocks exhibit plastic deformation, it indicates they have gone past their yield point and will not revert to their original form even if the stress is no longer applied.

Answer:

D. 6 m/s²

Explanation:

Acceleration is change in velocity over time.

a = Δv / Δt

a = (60 m/s − 0 m/s) / 10 s

a = 6 m/s²

The acceleration of the car is 6m/s².

To find the correct option among all the options, we need to know about the acceleration.

Acceleration = ∆V/∆t = 60/10= 6 m/s².

Thus, we can conclude that the option (D) is correct.

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Answer:

240 cm

Explanation:

Gradpoint

Answer:

Visibility conditions.

Explanation:

Safe speed is a speed at which the operator of the boat can take effective action to avoid stops within the distance. To calculate the safe speed visibility conditions (fog, rain, mist, and darkness) should be included. We should also include some other factors given below:

i) Traffic density

ii) Type of vessels in the area

iii) Wind

iv) Water conditions  

The Navigation Rules dictate that a safe speed is determined by factors such as visibility, traffic density, vessel maneuverability, weather conditions, and proximity to navigational hazards.

According to the Navigation Rules, several factors should be taken into account in determining a safe speed. These include the visibility, the traffic density, the maneuverability of the vessel in immediate circumstances, the state of wind, sea, and current, and the proximity of navigational hazards. For example, on a clear day with sparse traffic and calm seas, higher speeds might be safe. However, in heavy traffic or poor visibility, the navigation rules would call for lower speeds to ensure safety.

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Answer:

True

Explanation:

when oncoming vehicle driver is at a distance of 500 feet from you, then it is advisable to dim your high beam light so that it cannot blind oncoming vehicle driver. However, of the oncoming driver is at 200 to 300 ft then you must use low beam light so to pass the oncoming driver safely from your side. Hence the statement is true.

Answer:

1)

75 kmh⁻¹

2)

75 kmh⁻¹

Explanation:

1)

[tex]v_{ab}[/tex] = Speed of train from station A to station B = 80 kmh⁻¹

[tex]d_{ab}[/tex] = distance traveled from station A to station B

[tex]t_{ab}[/tex] = time of travel between station A to station B

we know that

[tex]Time = \frac{distance}{speed}[/tex]

[tex]t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}[/tex]

[tex]d_{bc}[/tex] = distance traveled from station B to station C

[tex]v_{bc}[/tex] = Speed of train from station B to station C = 60 kmh⁻¹

[tex]t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}[/tex]

Total distance traveled is given as

[tex]d = d_{ab} + d_{bc}[/tex]

Total time of travel is given as

[tex]t = t_{ab} + t_{bc}[/tex]

Average speed is given as

[tex]v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }[/tex]

Given that :

[tex]d_{ab} = 4 d_{bc}[/tex]

So

[tex]v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}[/tex]

2)

[tex]v_{ab}[/tex] = Speed of train from station A to station B = 80 kmh⁻¹

[tex]t_{ab}[/tex] = time of travel between station A to station B

[tex]d_{ab}[/tex] = distance traveled from station A to station B

we know that

[tex]distance = (speed) (time)[/tex]

[tex]d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}[/tex]

[tex]d_{bc}[/tex] = distance traveled from station B to station C

[tex]v_{bc}[/tex] = Speed of train from station B to station C = 60 kmh⁻¹

[tex]t_{bc}[/tex] = time of travel for train from station B to station C

we know that

[tex]distance = (speed) (time)[/tex]

[tex]d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}[/tex]

Total distance traveled is given as

[tex]d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}[/tex]

Total time of travel is given as

[tex]t = t_{ab} + t_{bc}[/tex]

Average speed is given as

[tex]v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}[/tex]

Given that :

[tex]t_{ab} = 3 t_{bc}[/tex]

So

[tex]v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}[/tex]

Answer:

a) [tex]t=2.6\ s[/tex]

b) [tex]s=43.7747\ m[/tex]

Explanation:

Given:

length of inclined roof, [tex]l=54\ m[/tex]

inclination of roof below horizontal, [tex]\theta=17^{\circ}C[/tex]

acceleration of hammer on the roof, [tex]a_r=2.87\ m.s^{-2}[/tex]

height from the lower edge of the roof, [tex]h=46.5\ m[/tex]

Now, we find the final velocity when leaving the edge of the roof:

Using the equation of motion:

[tex]v^2=u^2+2.a_r.l[/tex]

[tex]v^2=0^2+2\times 2.87\times 54[/tex]

[tex]v=17.6057\ m.s^{-1}[/tex]

The direction of this velocity is 17° below the horizontal.

∴Vertical component of velocity:

[tex]v_y=v.sin\ \theta[/tex]

[tex]v_y=17.6057\times sin\ 17^{\circ}[/tex]

[tex]v_y=5.1474\ m.s^{-1}[/tex]

a.

So, the time taken to fall on the ground:

[tex]h=ut+\frac{1}{2} g.t^2[/tex]

here:

initial velocity, [tex]u=v_y=5.1474\ m.s^{-1}[/tex]

putting respective values

[tex]46.5=5.1474\times t+0.5\times 9.8\times t^2[/tex]

[tex]t=2.6\ s[/tex]

b.

Horizontal component of velocity, [tex]v_x=v.cos\ \theta=17.6057\ cos\ 17^{\circ}=16.8364\ m.s^{-1}[/tex]

Since there is no air resistance so the horizontal velocity component remains constant.

∴Horizontal distance from the edge of the roof where the hammer falls is given by:

[tex]s=v_x.t[/tex]

[tex]s=16.8364\times 2.6[/tex]

[tex]s=43.7747\ m[/tex]

Using equations of motion, it's calculated that it takes about 3.07 seconds for the hammer to fall to the ground from the roof. The hammer also travels approximately 89.2 meters horizontally from the roof to the ground.

The question is related to physics concepts of motion under gravity and kinematics. For simplicity, let's ignore air resistance.

The time it takes for the hammer to fall to the ground from the edge of the roof can be calculated using the equation of motion: h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s²), and t is the time. Solving the equation for time (t): t = sqrt(2h/g). Substituting the given values, we get t = sqrt((2*46.5)/9.8) ≈ 3.07 s.

The horizontal distance travelled by the hammer can be calculated using the formula: distance = speed × time. The horizontal speed of the hammer when it falls off the roof will be the same speed it had just as it left the roof due to the roof slope acting on it with constant acceleration. This can be gotten from the equation v = u + at, where u is the initial velocity (0 m/s), a is the acceleration (2.87 m/s²) and t is the time. The time here is the time it takes for the hammer to slide down the roof, gotten by time = distance/speed = 54.0/v. Solving all these gives a distance of approximately 89.2 m.

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Answer:

a. Clouds at night prevent the heat radiation of the earth from escaping back into the space by continuously reflecting them back.

b. If during the day the earth has been heated well by the sun and then if the night is cloudy then that night is not so cold.

c. Day time clouds prevent the heating of the earth by reflecting back the radiations of the sun back into the space.

Explanation:

During the day if there is sunshine and then if the night is cloudy then that night will not be so cold because during the day the earth has absorbed the heat of sun's radiation which it radiates at night but when there is cloud at night it acts as a reflecting screen and reflects the heat radiations back to the earth.

But  if there is cloud during the day time it prevents the radiations from the sun to fall on the earth by reflecting them back into the space preventing the rise in temperature of the earth.

Answer:

Explanation:

Given

mass [tex]m=130 kg[/tex]

Force [tex]F=800 N[/tex]

coefficient of static friction [tex]\mu _s=0.8[/tex]

coefficient of static friction [tex]\mu _k=0.5[/tex]

maximum Static Friction will be given by

[tex]F_s=\mu _smg[/tex]

[tex]F_s=0.8\times 130\times 10=1040 N[/tex]

initially  we need to provide a force of 1040 N to move the Pig . As soon Pig starts moving kinetic friction will come into play.

But maximum force is less than Maximum static friction force so it is impossible to move pig.

Answer:

C

Explanation:

The distance that the pig moves up will be less than the distance that the strongman pulls down.

Explanation: The strongman is further from the fulcrum than the pig is, so the distance that he pulls down will be greater than the distance that the pig moves up.

Answer:

(A) 9.5 m/s

(B) 5.225 m

Explanation:

vertical height (h) = 4.7 m

horizontal distance (d) = 9.3 m

acceleration due to gravity (g) = 9.8 m/s^{2}

initial speed of the fish (u) = 0 m/s

(A) what is the pelicans initial speed ?

s = ut + [tex](\frac{1}{2}) at^{2}[/tex]

since u = 0

s =  [tex](\frac{1}{2}) at^{2}[/tex]

t = [tex]\sqrt{\frac{2s}{a} }[/tex]

where a = acceleration due to gravity and s = vertical height

t = [tex]\sqrt{\frac{2 x 4.7 }{9.8} }[/tex] = 0.98 s

speed of the fish = distance / time = 9.3 / 0.98 = 9.5 m/s

initial speed of the pelican = 9.5 m/s

(B) If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?

vertical height = 1.5 m

pelican's speed = 9.5 m/s

s = ut + [tex](\frac{1}{2}) at^{2}[/tex]

since u = 0

s =  [tex](\frac{1}{2}) at^{2}[/tex]

t = [tex]\sqrt{\frac{2s}{a} }[/tex]

where a = acceleration due to gravity and s = vertical height

t = [tex]\sqrt{\frac{2 x 1.5 }{9.8} }[/tex] = 0.55 s

distance traveled by the fish = speed x time = 9.5 x 0.55 = 5.225 m

The initial speed of the pelican is 9.6 m/s. If the pelican is flying at the same speed but only 1.5 m above the water, the fish will travel approximately 5.3m horizontally before it hits the water.

We can begin by solving for the time the fish takes to hit the water. Using the equation of motion, h = 0.5gt², where h represents height and g represents the acceleration due to gravity, we find it takes approximately 0.97 seconds for the fish to hit the water. During this time, the fish has traveled 9.3m horizontally. Therefore, the initial speed of the pelican can be determined by v = d/t, which gives us a value of 9.6 m/s.

Next, if the initial speed remained the same, but the height was reduced to 1.5 meters, we can use the same process. The time it would take for the fish to hit the water would be approximately 0.55 seconds (obtained from the equation h = 0.5gt²), and using this time in the equation v = d/t, we find the fish would travel approximately 5.3m horizontally before it hits the water.

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A1) The reason why making h₁ very small would cause d to be small is; Because the horizontal component of the launch velocity would be very small.

A2) The reason why making h₂ very small would cause d to be small is;

Because the time of flight it will take the object to get to the floor would be very small and as a result, the object would not possess enough time to move horizontally before the vertical motion.

B) The equation for d in terms of h₁ and h₂ is;

d = 2√(h₁ × h₂)

A) 1) The reason why making h₁ very small would cause d to be small is because the horizontal component of the launch velocity would be very small.

A) 2) The reason why making h₂ very small would cause d to be small is because the time of flight it will take the object to get to the floor would be very small and as a result, the object would not possess enough time to move horizontally before the vertical motion.

B) Formula for Launch Velocity is;

V = √(2gh₁)

h₁ was used because the top of the slide from where the student released the block has a height of h₁.

Also, the time it takes to fall which is time of flight is given by the formula;

t = √(2h₂/g)

h₂ was used because the height of the table the object is on before falling is h₂.

Now, we know that d is distance from edge of the table and formula for distance with respect to speed and time is;

distance = speed × time

Thus;

d = √(2gh₁) × √(2h₂/g)

g will cancel out and this simplifies to give;

d = 2√(h₁ × h₂)

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(A)

(1)  The reason for making [tex]h_{1}[/tex] very small is due to smaller value of horizontal component of launch velocity.

(2) The reason for making [tex]h_{2}[/tex] very small is due to smaller value of time of flight.

(B) The distance (d) covered by the block is  [tex]2\sqrt{h_{1}h_{2}}[/tex].

Given data:

The mass of block is, m.

The height of table from the top of slide is,  [tex]h_{1}[/tex].

The height of table at the end of slide is, [tex]h_{2}[/tex].

The height of room is, H.

(A)

(1)

If the launch velocity of the block is v, then its horizontal component is very small, due to which adjusting the height  [tex]h_{1}[/tex]  to be very small will cause the d to be small.

(2)

The height  [tex]h_{2}[/tex]  is dependent on the time of flight, and since the time of flight taken by the block to get to the floor is very less, therefore the block will not get sufficient time to accomplish its horizontal motion. That is why making [tex]h_{2}[/tex] very small will cause d to be smaller.

(B)

The expression for the distance covered by the block is,

[tex]v=\dfrac{d}{t}\\d = v \times t[/tex] ..............................(1)

Here, v is the launch speed of block and t is the time of flight.

The launch speed is,

[tex]v^{2}=u^2+2gh_{1}\\v=\sqrt{u^2+2gh_{1}}\\v=\sqrt{0^2+2gh_{1}}\\v=\sqrt{2gh_{1}}[/tex]

And the time of flight is,

[tex]h_{2}=ut+\dfrac{1}{2}gt^{2}\\h_{2}=0 \times t+\dfrac{1}{2}gt^{2}\\h_{2}=0+\dfrac{1}{2}gt^{2}\\t=\sqrt\dfrac{2h_{2}}{g}[/tex]

Substituting the values in equation (1) as,

[tex]d = v \times t\\d = \sqrt{2gh_{1}}\times \sqrt\dfrac{2 h_{2}}{g}}\\d=2\sqrt{h_{1}h_{2}}[/tex]

Thus, the distance (d) covered by the block is  [tex]2\sqrt{h_{1}h_{2}}[/tex].

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The mechanical energy of the ball-Earth-floor system at the instant when the ball left the floor is of 7 J.

At any instant, the energy of an object by virtue of its motion or position is known as the mechanical energy of the object. In other words, mechanical energy is either the potential energy or the kinetic energy at any time instant.

Given data -

The initial gravitational potential energy is, U = 10 J.

The final gravitational potential energy is, U' = 7 J.

Now, considering the given case when the ball loses energy in the rebound due to ball deformation, heat loss from the bounce. The total mechanical energy is not conserved so mechanical energy as it bounces off the floor is equal to transformed potential energy at a maximum height of second bounce which is 7 J.

Thus, we can conclude that the mechanical energy of the ball-Earth-floor system at the instant when the ball left the floor is of 7 J.

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Answer:

P=133.71mmHg

Explanation:

the standard free energy of formation for liquid ethanol is -174/9 kj/mol and that for gaseous ethanol is -168.6 kj/mol. calculate the vapour pressure of ethanol at

assumption:

is that temperature is at 25C, at standard pressure of 1bar(750mmHg)

ethanol is an ideal gas

The free energy of ethanol liquid does not vary with pressure,

C2H5OH(l)⟶C2H5OH(g)

free energy of formation on the reactant side is  -174.9 kj/mol

fro the product side is  -168.6 kj

∅Gvap-∅G(l)=-168.6kj/mol-(-174.9kj/mol)

+6.3kj/mol

∅G=∅Gvap+RTlnK-∅Gliq

∅G=0

0=+6.3kj/mol+8.314Jk/mol/k(298)InK

-6.3/(RT)=Lnk

taking the exponential of both sides

[tex]e^{-6300/(8.314*298)} =K[/tex]

0.178=k

k=p/[tex]p^{0}[/tex]

P^0=refers to the pressure of ethanol vapour at its standard state

partial pressure , which is 750 mmHg

P=0.178*750

P=133.71mmHg

Final answer:

The vapor pressure of ethanol at a given temperature can be calculated using the Clausius-Clapeyron equation.

Explanation:

The vapor pressure of a substance is related to its standard free energy of formation and temperature. To calculate the vapor pressure of ethanol at a given temperature, we can use the Clausius-Clapeyron equation:

ln(P₂/P₁) = △Hvap/R × (1/T₁ - 1/T₂)

Where P₁ and P₂ are the vapor pressures at temperatures T₁ and T₂ respectively, △Hvap is the enthalpy of vaporization, R is the gas constant, and ln denotes the natural logarithm.

By substituting the given values for △Hvap = -174/9 kJ/mol, T₁ = 20.0 °C (293 K), and T₂ = desired temperature, we can solve for P₂.

Answer:

Explanation:

A) Initial speed, u = 12 m/s

   Acceleration, a = -7 m/s²

   Final velocity, v = 0 m/s

   We have equation of motion v² = u² + 2as

                0² = 12² + 2 x -7 x s

                s = 10.29 m

  He need 10.29 m to stop.

  Remaining distance = 43 - 10.29 = 32.71 m

  We have

            Remaining distance = Reaction Time x Initial velocity

            32.71 = Reaction Time x 12

            Reaction Time = 2.73 seconds.

B) Distance traveled in 1.1333 s = 1.1333 x 12 = 13.60 m

   Remaining distance = 41 - 13.6 = 27.4 m

    Initial speed, u = 12 m/s

   Acceleration, a = -7 m/s²

   Displacement, s = 27.4 m

                      v² = 12² + 2 x -7 x 27.4

                      v² =  -239.6

   Not possible.

   Motorist will not hit deer.

  He will not reach the deer.

Answer:

Q = 12.5 kJ

Explanation:

The expression to use to calculate Heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

H°: enthalpy of reaction (kJ/mol)

n: moles

Now, as it was stated in the comments, the question is incomplete, and here is the missing part:

Given:

2A + B  A2B (1)

H° = – 25.0 kJ/mol

2A2B  2AB + A2 (2)

H° = 35.0 kJ/mol

With these two reactions, we can calculate the heat.

Now, with the above two reactions, we need to get the general reaction (The one the question is giving), so, let's use (1) and (2) and do the sum of them:

2A + B -------> A2B   H°1 = -25 kJ/mol

2A2B --------> 2AB + A2   H°2 = 35 kJ/mol

Now, we sum both equations, we can see that one A2B cancels out with one A2B from equation 2, so, the equation gives:

2A + B + 2A2B -------> 2AB + A2

And the enthalpy, it's just summed:

H°3 = -25 + 35 = 10 kJ/mol

Now with this value we can calculate heat:

Q = 10 * 2.5 = 25 kJ

However, in the reaction we have 2A, so it's not 1:1 mole ratio, but instead is 1:2, so this result we have to divide it between 2 so:

Q = 25 / 2 = 12.5 kJ

The heat required for the reaction is zero because the enthalpy changes for the two parts of the reaction cancel each other out.

To calculate the heat required for the reaction, we need to know the enthalpy change (ΔH°) for the reaction 2a + b → a2b. However, the enthalpy change for this reaction is not provided in the question. Assuming that the enthalpy change for this reaction is known, we can proceed with the calculation.

 Let's denote the enthalpy change for the reaction 2a + b → a2b as ΔH°. The reaction of interest is:

[tex]\[ 2a + b + a_2b \rightarrow 2ab + a_2 \][/tex]

 This reaction can be broken down into two parts:

1. The reaction of 2 moles of a with 1 mole of b to form 1 mole of a2b:

with an enthalpy change of H°.

[tex]\[ 2a + b \rightarrow a_2b \][/tex]

2. The reaction of 1 mole of a2b to form 2 moles of ab:

[tex]\[ a_2b \rightarrow 2ab \][/tex]

Since the reaction of a2b to form 2ab does not involve any additional reactants, it can be considered as the reverse of the first part, but with twice the amount of ab produced. Therefore, the enthalpy change for this part would be -H° (since the reverse reaction has the opposite sign of the forward reaction), and for 1 mole of a2b reacting, it would be -2ΔH° (because 2 moles of ab are formed).

Now, we are given 2.50 moles of a, and we assume there is excess b and a2b. The reaction will proceed until all of a is consumed. Since the reaction stoichiometry is 2 moles of a to 1 mole of a2b, 1.25 moles of a2b will be required to react with the 2.50 moles of a (since 2.50 moles of a require 1.25 moles of a2b).

The overall enthalpy change for the reaction will be the sum of the enthalpy changes for the two parts:

 1. For the reaction of 2.50 moles of a with b to form a2b:

[tex]\[ \Delta H_1 = 2.50 \text{ moles} \times \Delta H^\circ \][/tex]

2. For the reaction of 1.25 moles of a2b to form ab:

[tex]\[ \Delta H_2 = 1.25 \text{ moles} \times (-2 \Delta H^\circ) \][/tex]

The total enthalpy change (ΔH_total) is the sum of ΔH_1 and ΔH_2:

[tex]\[ \Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 \][/tex][tex]\[ \Delta H_{\text{total}} = 2.50 \Delta H^\circ - 2 \times 1.25 \Delta H^\circ \][/tex]

[tex]\[ \Delta H_{\text{total}} = 2.50 \Delta H^\circ - 2.50 \Delta H^\circ \][/tex]

[tex]\[ \Delta H_{\text{total}} = 0 \][/tex]

 Surprisingly, the total enthalpy change for the reaction is zero. This is because the amount of heat released in the formation of a2b is exactly equal to the amount of heat absorbed in the formation of ab from a2b. Therefore, no additional heat is required for the overall reaction to proceed.

 The final answer is:  [tex]\[ \boxed{0} \][/tex]

The heat required for the reaction is zero because the enthalpy changes for the two parts of the reaction cancel each other out. This conclusion is based on the assumption that the enthalpy change for the reaction 2a + b → a2b is known and that the reaction proceeds as written. If the enthalpy change for the formation of a2b is not known, then it would be necessary to look up the standard enthalpy of formation for a2b to perform the calculation.

Answer:

7

Explanation:

The balanced reaction is as follows:

[tex]2CO+2NO \rightarrow 2CO_2 + N_2[/tex]

Coefficient of CO = 2

Coefficient of NO = 2

Coefficient of CO2 = 2

Coefficient of N2 = 1

Sum of all coefficient = 2 + 2+ 2+ 1

                                    = 7

Answer:255.255.255.224

Explanation:

/20 has a subnet mask of 255.255.240.0

We have 20 on bits for the network and 16 off bits for the host. Since the designer wants 20 to 30 hosts in each subnet we will be having 5 off bits for the host. Therefore we will be having a /27..

Which will be 255.255.255.224

Final answer:

The suitable subnet mask for creating subnets with 20-30 hosts within the 10.0.240.0/20 network is 255.255.255.224 or /27, which allows for 32 IP addresses, 30 of which can be used for hosts.

Explanation:

The correct subnet mask to meet the requirement for new subnets that support between a minimum of 20 hosts and a maximum of 30 hosts within the 10.0.240.0/20 network is 255.255.255.224 or /27. This subnet mask allows for 32 IP addresses per subnet, with 30 usable addresses for hosts when excluding the network and broadcast addresses. Subnet masks 255.255.224.0 and 255.255.240.0 provide too many hosts per subnet, and subnet mask 255.255.255.240 or /28 provides only 16 IP addresses, which would be insufficient for the required minimum of 20 hosts.

Answer:

Approximately [tex]\rm 11.2 \; N \cdot kg^{-1}[/tex] at that distance from the center of the planet.

Option A) The low value of [tex]g[/tex] near the cloud top of Saturn is possible because of the low density of the planet.

Explanation:

The value of [tex]g[/tex] on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:

[tex]\displaystyle \frac{G \cdot M \cdot m}{R^2}[/tex],

where

To find an equation for [tex]g[/tex], divide the equation for gravity by the mass of the object:

[tex]\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}[/tex].

In this case,

Calculate [tex]g[/tex] based on these values:

[tex]\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}[/tex].

Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.

Refer to the equation for [tex]g[/tex]:

[tex]\displaystyle g = \frac{G \cdot M}{R^2}[/tex].

The mass of the planet is in the numerator. If two planets are of the same size, [tex]g[/tex] would be greater at the surface of the more massive planet.

On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since [tex]R[/tex] is in the denominator of [tex]g[/tex], increasing the value of [tex]R[/tex] while keeping [tex]M[/tex] constant would reduce the value of [tex]g[/tex]. That explains why the value of [tex]g[/tex] near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately [tex]9.81\; \rm N \cdot kg^{-1}[/tex].

As a side note, [tex]5.82\times 10^7\rm \; m[/tex] likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of [tex]g[/tex] near the cloud tops of Saturn is approximately [tex]\rm 11.2 \; N \cdot kg^{-1}[/tex].

concept is gravitational force. Saturn's low value of g on its surface is possible because of its low average density, which is less than the density of water.

The value of g on the surface of Saturn is determined by its mass and radius. The low value of g on Saturn is possible because of its low average density. Saturn's mass is much larger than Earth's, but its density is much lower, resulting in a lower value of g on its surface.

The low density of Saturn, which is only 0.7 g/cm³, is less than the density of water. This means that Saturn would be light enough to float if placed in water. Therefore, despite its large mass, the low density of Saturn allows for a low value of g on its surface.

https://brainly.com/question/18961003

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Answer: 0.471 m/s

Explanation:

We are given the followin equation:

[tex]Re=\frac{D v \rho}{\eta}[/tex] (1)

Where:

[tex]Re[/tex] is the Reynolds Number, which is adimensional and indicates if the flow is laminar or turbulent

When [tex]Re<2100[/tex] we have a laminar flow

When [tex]Re>4000[/tex] we have a turbulent flow

When [tex]2100<Re<4000[/tex] the flow is in the transition region

[tex]D=2R[/tex] is the diameter of the pipe. If the pipe ha a radius [tex]R=8(10)^{-3} m[/tex] its diameter is [tex]D=2(8(10)^{-3} m)=0.016 m[/tex]

[tex]v[/tex] is the average speed of the fluid

[tex]\rho=1060 kg/m^{3}[/tex] is the density of the fluid

[tex]\eta=4(10)^{-3} Pa.s[/tex] is the viscosity of the fluid

Isolating [tex]v[/tex]:

[tex]v=\frac{Re \eta}{D \rho}[/tex] (2)

Solving for [tex]Re=2000[/tex]

[tex]v=\frac{(2000)(4(10)^{-3} Pa.s)}{(0.016 m)(1060 kg/m^{3})}[/tex] (3)

Finally:

[tex]v=0.471 m/s[/tex]