A cubical box with edges of length k centimetres is to be enlarged so that the dimensions of the larger box are k + 2 centimetres, k + 3 centimetres, and k centimetres. The volume of the larger box is how many cubic centimetres greater than the volume of the original box?

Answer:

Explanation:

Applied force, F = 18 N

Coefficient of static friction, μs = 0.4

Coefficient of kinetic friction, μs = 0.3

θ = 27°

Let N be the normal reaction of the wall acting on the block and m be the mass of block.

Resolve the components of force F.

As the block is in the horizontal equilibrium, so

F Cos 27° = N

N = 18 Cos 27° = 16.04 N

As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .

The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N   .... (1)

The vertically downward force acting on the block is mg - F Sin 27°

                                                      = mg - 18 Sin 27° = mg - 8.172    ... (2)

Now by equating the forces from equation (1) and (2), we get

mg - 8.172 = 6.42

mg = 14.592

m x 9.8 = 14.592

m = 1.49 kg

Thus, the mass of block is 1.5 kg.  

Answer:

[tex]E_{k}=15204Joules[/tex]

[tex]v=22.2m/s[/tex]

Explanation:

We can apply energy conservation law:

Initial Energy = Final Energy

In the beginning there is only potential energy, and in the end there is only kinetic energy.

[tex]E_{k}=E_{p}=mgh=61.5*9.81*25.2=15204Joules[/tex]

And:

[tex]E_{k} = \frac{1}{2}mv^2[/tex]

[tex]v=\sqrt{2*E_{k}/m}=\sqrt{2*15204/61.5}=22.2m/s[/tex]

Answer:

The answer to your question is: vo = 25 m/s

Explanation:

data

a = -7.5 m/s²

d = 42 m

vf = 0 m/s

vo = ?

Formula

vf² = vo² - 2ad

Substitution

0² = vo² - 2(7.5)(42)

We clear vo from the equation

vo² = 2(7.5)(42)  

vo² = 630               simplifying

vo = 25 m/s            result

Answer:

The correct equation for measuring the average microscopic weight  for 3 isotopes is multiply the rate of abundance by each weight and add them.

Explanation:

To calculate the average microscopic mass of element using weights and relative abundance we have to follow the following steps.

This gives the required  average microscopic weight of the three isotopes.

Answer:

y = k/x

Explanation:

y = k/x is a graph of a hyperbola that has been rotated about the origin.

The equation y = k/x  represents a generic equation suggested by a graph showing a hyperbola

A right circular cone and a plane are intersected at an angle in analytical geometry to form a hyperbola, which is a conic section that intersects both cone halves. It results in two distinct unbounded curves that are mirror images of one another.

While a set of points in a plane that are equally spaced apart from a directrix or focus is known as parabolas. The difference of distances between a group of points that are present in a plane to two fixed points and is a positive constant is how the hyperbola is defined.

The general equation representing a hyperbola

y= k/x

Thus,a graph with a hyperbola suggests the equation y = k/x as a general equation.

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Answer:

34.5 m at 55.5 degrees above the positive x-axis

Explanation:

Resolving a vector means finding its components along two perpendicular axis: most commonly, they are chosen as the x and y axis.

In this problem, we have a vector whose components are:

x -component: 19.5 m

y- component: 28.4 m

The two components are perpendicular to each other: this means that we can find the magnitude of the vector by using the Pythagorean theorem

[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{19.5^2+28.4^2}=34.5 m[/tex]

The direction, instead, can be found by using the following formula:

[tex]\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1} (\frac{28.4}{19.5})=55.5^{\circ}[/tex]

Answer:

Explanation:

We use the harmonic motion position equation:

[tex]x(t) = A\cos(\omega t+\phi)[/tex]

where A = 0.350 and for t = 0

[tex]x(0) A = A\ cos(\phi)[/tex]

so: [tex]\phi = 0[/tex]

and also:

[tex]\omega = \frac{2\pi}{T} = \frac{2\pi}{4.10} = 1.532 rad/s[/tex]

so we have:

x(t)=0.350cos(1.532 t)

For t = 3.403 s

x(3.403)=0.350cos(1.532 (3.403)) = 0.348 m

The block attached to a spring is undergoing simple harmonic motion. By fitting the provided variables into the displacement equation for such a motion, it is found that the block's position 3.403 seconds after release is -0.279 meters.

The problem describes a scenario involving simple harmonic motion (SHM). For such a motion, the displacement, x(t) from equilibrium position could be expressed as x(t) = A * cos(ωt + φ), where 'A' is the amplitude, 'ω' = 2π/T is the angular frequency, 'T' is the period of the motion, t is the given time, and 'φ' is the phase constant. Since the motion begins at the amplitude, φ = 0.

Amplitude given is 0.35 m and period is 4.10 s. Thus, 'ω' = 2π / 4.10 = 1.5 rad/s. Putting these values into the displacement equation, x(t) = 0.350 m * cos(1.5 rad/s * 3.403 s) = -0.279 m. Therefore, the position of the mass 3.403 s after it's released is -0.279 m.

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Answer:

[tex]d = 0.018 m[/tex]

Explanation:

Charge on the plates of the capacitor due to transfer of electrons is given as

[tex]Q = Ne[/tex]

here we know that

[tex]N = 2.1 \times 10^9[/tex]

so we have

[tex]Q = (2.1 \times 10^9)(1.6 \times 10^{-19})[/tex]

[tex]Q = 3.36 \times 10^{-10} C[/tex]

now we have electric field between the plates is given as

[tex]E = \frac{Q}{A\epsilon_0}[/tex]

here we have

[tex]1.5 \times 10^5 = \frac{3.36 \times 10^{-10}}{A(8.85 \times 10^{-12})}[/tex]

[tex]A = 2.53 \times 10^{-4} m^2[/tex]

now we have

[tex]A = \frac{\pi d^2}{4}[/tex]

[tex]2.53 \times 10^{-4} = \frac{\pi d^2}{4}[/tex]

[tex]d = 0.018 m[/tex]

The  the diameters of the disks of the parallel-plate capacitor are  0.018 m.

Given parameters:

Number of electron transferring from one disk to another: n = 2.1×10⁹.

Electric field strength: E =  1.5×10⁵ N/C .

We have to find: the diameters of the disks: d = ?

Charge of an electron is: e = 1.6 × 10⁻¹⁹ C.

So, charge transferring from one disk to another:

Q = ne = 2.1×10⁹×1.6 × 10⁻¹⁹ C = 3.36×10⁻¹⁰ C

For parallel-plate capacitor: electric field strength is given by:

E = Q/ε₀A

Where:

A = area of each disks = πd²/4

ε₀ = permittivity of free space = 8.85 × 10⁻¹² Si units.

Hence,

 A = Q/ε₀E

⇒ πd²/4 =  Q/ε₀E

⇒ d = √( 4Q/πε₀E)

Putting numerical values we get:

d = √( 4 × 3.36×10⁻¹⁰/π×8.85 × 10⁻¹² × 1.5×10⁵ ) m.

= 0.018 m.

Hence, the diameters of the disks are 0.018 m.

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Answer:

[tex]0.283\ m^3/s[/tex]

Explanation:

Given that,

1.00 in = 2.54 cm

We know that,

1 feet = 12 inches

[tex]1\ feet =12\times0.0254[/tex]

[tex]1\ feet =0.3048\ m[/tex]

We calculate the number

We need to converted [tex]ft^3/min[/tex] to [tex]m^3/sec[/tex]

The number required is

[tex]600ft^3/min = \dfrac{600\times(0.3048)^3}{60}[/tex]

[tex]600ft^3/min=0.283\ m^3/s[/tex]

Hence, This is the required solution.

The average speed of air in the duct is 11.23 m/s.

To find the average speed of air in the duct, we need to calculate the volume flow rate. The volume of the house's interior is given as 13.0 m × 20.0 m × 2.75 m = 715.0 m³. The air is carried through the duct every 15 minutes, so we need to convert the time to seconds: 15 minutes × 60 seconds/minute = 900 seconds. Therefore, the volume flow rate is 715.0 m³ / 900 s = 0.794 m³/s.

The main uptake air duct has a diameter of 0.300 m, which we can use to find the cross-sectional area of the duct: A = (π/4) × (0.300 m)² = 0.0707 m².

Now, we can calculate the average speed of air in the duct: average speed = volume flow rate / cross-sectional area = 0.794 m³/s / 0.0707 m² = 11.23 m/s.

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Answer:

[tex]\frac{dQ}{dt}= 4312 W[/tex]

Explanation:

As we know that base of the slab is given as

[tex]A = 11 \times 8[/tex]

[tex]A = 88 m^2[/tex]

now we know that rate of heat transfer is given as

[tex]\frac{dQ}{dt} = \frac{kA}{x} (T_2 - T_1)[/tex]

here we know that

[tex]k = 1.4 W/m k[/tex]

Also we have

[tex]x =0.20[/tex]

[tex]\frac{dQ}{dt} = \frac{1.4(88)}{0.20}(17 - 10)[/tex]

[tex]\frac{dQ}{dt}= 4312 W[/tex]

The rate of heat loss through the concrete slab of the basement is 4312 watts, calculated using Fourier's Law of thermal conduction with the given dimensions and temperatures.

To calculate the rate of heat loss through the concrete slab of a basement, we can use the formula for thermal conduction, which is given by Fourier's Law of heat conduction:
Q = \frac{k \cdot A \cdot (T_{hot} - T_{cold})}{d}
Where:
Q is the rate of heat transfer (W),
k is the thermal conductivity of the material (W/m·K),
A is the cross-sectional area through which heat is being transferred (m²),
T_{hot} is the temperature on the hot side of the material (°C),
T_{cold} is the temperature on the cold side (°C),
and d is the thickness of the material (m).
Using the given values:
k = 1.4 W/m·K,
A = 11 m * 8 m = 88 m²,
T_{hot} = 17°C,
T_{cold} = 10°C,
d = 0.20 m,
the rate of heat loss Q can be calculated as follows:
Q = \frac{1.4 \cdot 88 \cdot (17 - 10)}{0.20} = \frac{1.4 \cdot 88 \cdot 7}{0.20}
Q = 4312 W
The basement's heat loss through the concrete slab is 4312 watts. If the gas furnace operates at an efficiency of 0.9 (90%), the actual energy output required from the furnace would be higher.

Answer:

a) [tex]t=6.37s[/tex]

b) [tex]t=3.3333s[/tex]

Explanation:

The knowable variables are the initial hight and initial velocity

[tex]s_{o}=80ft[/tex]

[tex]v_{os}=64ft/s[/tex]

The equation that describes the motion of the ball is:

[tex]s=80+64t-16t^{2}[/tex]

If we want to know the time that takes the ball to hit the ground, we need to calculate it by doing s=0 that is the final hight.

[tex]0=80+64t-12t^{2}[/tex]

a) Solving for t, we are going to have two answers

[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]

a=-16

b=64

c=80

[tex]t=-1.045 s[/tex] or [tex]t=6.378s[/tex]

Since time can not be negative the answer is t=6.378s

b) To find the time that takes the ball to pass the top of the building on its way down, we must find how much does it move too

First of all, we need to find the maximum hight and how much time does it take to reach it:

[tex]v_{y}=v_{o}+gt[/tex]

at maximum point the velocity is 0

[tex]0=64-32.2t[/tex]

Solving for t

[tex]t=1.9875 s[/tex]

Now, we must know how much distance does it take to reach maximum point

[tex]s=0+64t-16t^{2} =64(1.9875)-12(1.9875)^{2} =80ft[/tex]

So, the ball pass the top of the building on its way down at 160 ft

[tex]160=80+64t-16t^{2}[/tex]

Solving for t

[tex]t=2s[/tex] or [tex]t=3.333s[/tex]

Since the time that the ball reaches maximum point is almost t=2s that answer can not be possible, so the answer is t=3.333s for the ball to go up and down, passing the top of the building

Answer:

(a) 5 s

(b) 4s

Explanation:

height of building, s = 80 feet

The equation of motion is given by

[tex]s=80+64t-16t^{2}[/tex]

(a) Let it takes t time to reach the ground. A it reach the ground, s = 0

So, put s = 0 in the given equation, we get

[tex]0=80+64t-16t^{2}[/tex]

[tex]t^{2}-4t-5=0[/tex]

(t + 1)(t - 5) = 0

t = -1 s, t = 5 s

As time cannot be negative, so t = 5 s

Thus, the time taken by the ball to reach the ground is 5 s.

(b) Let it crosses the building in t second, s put s = 80 feet in the above equation, we get

[tex]80+64t-16t^{2}=80[/tex]

t(4 - t)= 0

t = 0s or t  4 s

So, it takes 4 second to cross the building.

Answer:

7.5 cm

Explanation:

In the figure we can see a sketch of the problem. We know that at the bottom of the U-shaped tube the pressure is equal in both branches. Defining [tex] \rho_A: [/tex] Ethyl alcohol density and [tex] \rho_G: [/tex] Glycerin density , we can write:

[tex] \rho_A\times g \times h_1 + \rho_G \times g \times h_2 = \rho_G \times g \times h_3 [/tex]

Simplifying:

[tex] \rho_A\times h_1 = \rho_G \times (h_3 - h_2) (1) [/tex]

On the other hand:

[tex] h_1 + h_2 = \Delta h + h_3 [/tex]

Rearranging:

[tex] h_1 - \Delta h = h_3 - h_2 (2) [/tex]

Replacing  (2) in (1):

[tex] \rho_A\times h_1 = \rho_G \times (h_1 - \Delta h) [/tex]

Rearranging:

[tex] \frac{h_1 \times (\rho_A - \rho_G)}{- \rho_G} = \Delta h [/tex]

Data:

[tex] h_1 = 20 cm; \rho_A = 0.789 \frac{g}{cm^3}; \rho_G = 1.26 \frac{g}{cm^3} [/tex]

[tex] \frac{20 cm \times (0.789 - 1.26) \frac{g}{cm^3}}{- 1.26\frac{g}{cm^3}}  = \Delta h [/tex]

[tex] 7.5 cm =  \Delta h [/tex]

To find the difference in height between glycerin and alcohol in a U-tube, apply the principle of communicating vessels using their respective densities. Due to glycerin's higher density, its height in equilibrium will be lower than the alcohol's for equal pressures at the base of each arm of the tube.

The difference in height between the top surface of the glycerin and the top surface of the alcohol can be determined by using the principle of communicating vessels and the densities of the two liquids. Since the two liquids do not mix, they will each exert a pressure at the bottom of their respective sides of the U-tube based on their densities and heights.

The density of glycerin is 1.26 g/cm3, and the density of ethyl alcohol is 0.789 g/cm3. Since the pressures at the bottom of the tubes must be equal and the pressure exerted by a column of liquid is given by the product of its density, gravitational acceleration, and height, we can set up the following equation:

density of glycerin × height of glycerin = density of alcohol × height of alcohol.

Using this relationship, we can calculate the heights. However, since the question states that the ethyl alcohol column's height is already 20 cm, we only need to verify if the glycerin column's height will remain at 20 cm or change. As glycerin is denser than ethyl alcohol, for the pressures to be equal, the height of the glycerin column should be less than the height of the alcohol column. However, additional information like the specific heights after the alcohol is added would be necessary to provide a numerical answer.

Answer:

1.5mi/hr, 4.5mi/hr

Explanation:

Given:

45 min = 0.75h, 15 min = 0.25h

(1) 0.75v₁ = 0.25v₂

(2) v₁ = v₂ - 3

Solve for v₁ and v₂:

0.75(v₂ - 3) = 0.25v₂ = 0.75v₂ - 2.25 = 0.25v₂

0.5v₂ = 2.25

v₂ = 4.5

v₁ = 1.5

Answer:

uphill speed = 1.5 mph, downhill speed = 4.5 mph

Explanation:

We are asked to find Julian’s uphill speed and downhill speed. Let’s let r represent Julian’s uphill speed in miles per hour. In 45min=34hr, he rides uphill a distance rt=34r miles. Since his downhill speed is 3 miles per hour faster, we represent that as r+3. In 15min=14hr, he rides downhill a distance (r+3)t=r+34 miles. We can set these distances equal to find

34r=r+34

Multiplying both sides by 4 and then subtracting r from both sides yields

3r2r=r+3=3

Solving for r=32, Julian’s uphill speed is 1.5 miles per hour and his downhill speed is r+3=4.5 miles per hour.

[tex]\boxed{n=2.92\times 10^{10} \ stars}[/tex]

Order-of-magnitude estimates are useful when we want to get a very crude estimate of a quantity. Sometimes, to get the exact calculation of a problem is very difficult, or impossible, so we use order of magnitud in order to get a rough approximation. In this exercise, we need to find the order of magnitude of the number of stars in the Milky Way. We know:

The distance from the Sun to the nearest star is:

[tex]4 \times 10^{16}m[/tex]

The Milky Way galaxy is roughly a disk of diameter:

[tex]10^{21}[/tex]

The Milky Way galaxy is roughly a disk of thickness:

[tex]10^{19}[/tex]

So we can approximate volume of the Milky Way:

[tex]V=\pi r^2 h \\ \\ r=\frac{10^{21}}{2}m=5 \times 10^{20}m \\ \\ h=10^{19}m \\ \\ \\ V=\pi(5 \times 10^{20})^2(10^{19}) \\ \\ V=7.85\times 10^{60}m^3[/tex]

Now, let's estimate a rough density for the Milky Way. It is well known that in a sphere with a radius of  [tex]4\times 10^{16}m[/tex] there is a star, which is the sun. So the density of the Milky way is:

[tex]\rho =\frac{n}{V} \\ \\ \rho:Density \ of \ Milky \ way \\ \\ n: Number \ of \ stars \\ \\ V:Volume[/tex]

For one star [tex]n=1[/tex] so we know the data at the neighborhood around the Sun, so the volume is a sphere:

[tex]V=V_{sphere}=\frac{4}{3}\pi r^3 \\ \\ V_{sphere}=\frac{4}{3}\pi (4\times 10^{16})^3 \\ \\ V_{sphere}=2.68\times 10^{50}m^3 \\ \\ So: \\ \\ \rho=\frac{1}{2.68\times 10^{50}}=3.73\times 10^{-51}stars/m^3[/tex]

Finally, the numbers of stars can be found as:

[tex]n=\rho V \\ \\ V:Volume \ of \ the \ Milky \ Way \\ \\ p:Density \ of \ the \ Milky \ Way \\ \\ n:Number \ of \ stars \\ \\ n=(3.73\times 10^{-51})(7.85\times 10^{60}) \\ \\ \boxed{n=2.92\times 10^{10} \ stars}[/tex]

Answer:

The raisins provide 8 grams of protein.

Explanation:

First wrote the information as a equation:

[tex]C=4h+9f+4p[/tex]

Where: C is calories, h carbohidrates, f fats and p protein

The information given for the raisin, 148 calories, 0 fat and 29 carbohidrates. That into a equation.

[tex]148=4(29)+9(0)+4p[/tex]

We clear p from the equation

14[tex]\frac{148-4(29)}{4} =p=8 gr[/tex]

A motorcycle has a constant acceleration of 2.5 m/s², the motorcycle takes 4 seconds to change its speed by 10 m/s.

The kinematic equation can be used to determine how long it will take the motorcycle to alter its speed. Initial velocity ([tex]v_0[/tex]), final velocity (v), acceleration (a), and time (t) are related by the following equation:

[tex]\[v = v_0 + at\][/tex]

Given that the acceleration (a) is 2.5 m/s² and both velocity and acceleration are in the same direction, we can use this equation to solve for time (t).

(a) Changing speed from 21 m/s to 31 m/s:

Initial velocity ([tex]\(v_0\)[/tex]) = 21 m/s

Final velocity (v) = 31 m/s

Acceleration (a) = 2.5 m/s²

31 = 21 + 2.5t

2.5t = 31 - 21

2.5t = 10

[tex]\[t = \dfrac{10}{2.5} = 4 \, \text{seconds}\][/tex]

(b) Changing speed from 51 m/s to 61 m/s:

Initial velocity ([tex]\(v_0\)[/tex]) = 51 m/s

Final velocity (v) = 61 m/s

Acceleration (a) = 2.5 m/s²

61 = 51 + 2.5

2.5t = 61 - 51

2.5t = 10

[tex]\[t = \dfrac{10}{2.5} = 4 \, \text{seconds}\][/tex]

Thus, in both cases, the motorcycle takes 4 seconds to change its speed by 10 m/s.

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The time required for the motorcycle to change its speed by 10 m/s, under a constant acceleration of 2.5 m/s², is 4 seconds in both scenarios: from 21 m/s to 31 m/s, and from 51 m/s to 61 m/s.

In order to answer your question about the time required for the motorcycle to change its speed, we use the formula for acceleration, which is change in velocity divided by time (a = Δv/Δt). Thus, we can solve for time with Δt = Δv / a.

(a) For a speed change from 21 m/s to 31 m/s, Δv = 31 m/s - 21 m/s = 10 m/s. Substituting into the formula, we find Δt = 10 m/s / 2.5 m/s² = 4 seconds.

(b) For a speed change from 51 m/s to 61 m/s, Δv = 61 m/s - 51 m/s = 10 m/s. Again substituting into the formula, we find Δt = 10 m/s / 2.5 m/s² = 4 seconds.

Therefore, the time required for the motorcycle to change its speed by 10 m/s, under a constant acceleration of 2.5 m/s², is 4 seconds in both cases.

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The equation which has the fastest runner is Distance = 2 * Time.

The correct answer is Option D.

Given data:

The equation that describes the fastest runner among the given options is D. distance = 2⋅time.

This equation suggests that the distance traveled by the runner is directly proportional to twice the time taken. In other words, if two runners start at the same time and one of them has a faster speed, they will cover a greater distance in the same amount of time. This equation aligns with the principles of speed and distance in physics.

Option D reflects that the distance covered by a runner is determined by the product of time and speed. When the speed is greater (as indicated by the coefficient of 2), the distance covered in a given time will be larger, making this equation representative of the fastest runner. The other options, A, B, and C, do not consider the concept of speed and do not accurately describe the behavior of a fastest runner in relation to distance and time.

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Final answer:

The fastest runner is described by Equation D, which represents a speed of 2 meters per second (2 m/s), as it has the highest coefficient of time in the distance = speed × time equation.

Explanation:

The equation that describes the fastest runner is the one where the coefficient of time (t) in the distance = speed × time equation is the greatest, as this coefficient represents the speed of the runner. The faster the runner, the greater the distance they would cover in a given time. Comparing the given equations:

A. distance = 0.5×time (speed of 0.5 m/s)

B. distance = 0.33×time (speed of 0.33 m/s)

C. distance = time (speed of 1 m/s)

D. distance = 2×time (speed of 2 m/s)

The fastest runner is described by Equation D, where the distance is twice the time, indicating a speed of 2 meters per second (2 m/s).

Answer:

a) v = 2.733 m/s

b) t = 6.58s

c) a = -0.415 m/s^2

d) v = 0.789 m/s

Explanation:

We can see it in the pics

Answer:

Negative acceleration

Explanation:

We know that

v= u+at

Where v is the final velocity

u is the initial velocity velocity

a is the acceleration

t is time

When final velocity of train will be zero .

v=0

0= u+at

a=- u/t

So from above we can say that acceleration will be negative .When acceleration is positive then train will never come in rest position.

Final answer:

The slope of the catenary curve where it meets the right pole is -1. The angle between the line and the pole is -45 degrees.

Explanation:

The shape of the telephone line between two poles is known as a catenary. To find the slope where it meets the right pole, we can use the formula:

slope = -c/a

where 'c' is the distance from the vertex to the right pole and 'a' is half the distance between the poles. In this case, c = 7m and a = 7m. Substituting these values into the formula, we get:

slope = -7/7 = -1

Therefore, the slope of the curve where the line meets the right pole is -1.

To find the angle between the line and the pole, we can use the formula:

angle = arctan(slope)

Substituting the slope (-1) into the formula, we get:

angle = arctan(-1)

Using a calculator, we find that the angle is approximately -45 degrees.

Answer:

0.961

Explanation:

m1 = 300 g = 0.3 kg

u1 = 1 m/s

m2 = 600 g = 0.6 kg

u2 = - 0.75 m/s

Let after collision they move together with velocity v.

By using the conservation of linear momentum

Total momentum before collision = Total momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) v

0.3 x 1 - 0.6 x 0.75 = (0.3 + 0.6) v

0.3 - 0.45 = 0.9 v

v = - 0.166 m/s

Total initial Kinetic energy

[tex]K_{i}=0.5m_{1}u_{1}^{2}+0.5m_{1}u_{1}^{2}[/tex]

[tex]K_{i}=0.5\times 0.3\times 1\times 1+0.5 \times 0.6 \times 0.75 \times 0.75[/tex]

[tex]K_{i}=0.31875 J[/tex]

Total final Kinetic energy

[tex]K_{f}=0.5\left ( m_{1}+m_{2} \right )v^{2}[/tex]

[tex]K_{f}=0.5\times 0.9 \times 0.166 \times 0.166[/tex]

[tex]K_{f}=0.0124 J[/tex]

fraction of kinetic energy lost

[tex]\frac{K_{i}-K_{f}}{K_{f}}=\frac{0.31875-0.0124}{0.31875}=0.961[/tex]

Two pieces of clay are moves directly, and collide the momentum before and after the collision remain same. The fraction part of the total initial kinetic energy lost during the collision is 0.961.

Kinetic energy is the energy of of the body, which it posses due to force of motion. The kinetic energy of a body is half of the product of mass times square of its velocity.

Given information-

The mass of piece one is 300 grams.

The mass of the second piece is 600 grams.

The speed of the first piece is 1 m/s.

The speed of the second piece is 0.75 m/s.

By the conservation of momentum, we know that the momentum of two body before the collision is equal to the momentum after the collision. As the momentum is product of mass times velocity. Thus,

[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]

Here, [tex]m_1,m_2[/tex] is the mass of body one, body 2 respectively and [tex]v_1,v_2[/tex] are the velocities of body one, body two before the collision respectively.

Put the values to find the value of velocity after the collision as,

[tex]0.3\times1+0.6\times0.75=(0.3+0.6)v\\v=-0.166 m/s[/tex]

Negative sine indicates the direction after the collision is changed.

Now the kinetic energy lost is the ratio of difference of initial kinetic energy and final kinetic energy to the initial kinetic energy.

Thus the kinetic energy lost is,

[tex]\Delta KE=\dfrac{k_i-k_f}{k_f}[/tex]

The kinetic energy is the half of the mass time square of its velocity. Thus,

[tex]\Delta KE=\dfrac{(\dfrac{1}{2}0.3\times0.1^2)-(\dfrac{1}{2}0.6\times0.75^2)}{\dfrac{1}{2}0.3\times0.1^2}\\\Delta KE=0.961[/tex]

Thus the fraction part of the total initial kinetic energy lost during the collision is 0.961.

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Answer: The student’s understanding of all four terms (speed, velocity, scalar, and vector) is correct.

Explanation:

Let's start by explaining that a vector is one that has a numerical value along with its units (called a module) and a direction, while a scalar is only determined with a number and its corresponding units, without direction.

Then, speed is the distance an object travels in a given time. That is, it only takes into account the distance traveled, dividing it by time to know how fast it moves, therefore it is a scalar.

Instead, velocity refers to the time it takes for an object to move in a certain direction. So, by involving the direction of movement, velocity is a vector.

In short, the speed does not take into account the direction of the object, while the velocity does.

Therefore, as the student understands this four concepts, the correct option is:

The student’s understanding of all four terms (speed, velocity, scalar, and vector) is correct.

Answer:

(D) 0.250 mole

Explanation:

From molecular weight we know that 1 mole of P4O10 wights 284 grams. So

[tex]14.2 g P4O10 \times \frac{1 mole P4O10}{284 g P4O10} = 0.05 moles P4O10 [/tex]

In order to form 1 mole of P4O10 it is necessary 5 moles of O2,  so that the number of oxygen atoms are the same.

Knowing that 0.05 moles of P4O10 is formed we need 5 x 0.05 = 0.25 moles of O2

Answer:

weight of sealed tube that is filled with iodine gas is 27 gm

Explanation:

As tube is closed therefore mole of gas is enclosed in a tube, thus it will only converted into gas ( Iodine Gas). In the gas form it will going to exert pressure on the wall of a tube. From conservation of mass principle weight of tube remain same, there will be no change in the weight of gas. therefore weight of sealed tube that is filled with iodine gas is 27 gm

The weight of the sealed tube filled with iodine gas will remain 27.0 grams, as the principle of conservation of mass dictates that the total mass in a closed system does not change.

The principle of conservation of mass states that matter cannot be created or destroyed. Therefore, the total mass before and after the iodine sublimates remains the same in a closed system. The total mass of the tube and iodine before heating is 27.0 grams, including the 1.0-gram iodine sample and the tube. After the iodine sublimates, it remains within the sealed tube; thus, there is no change in mass. So, the weight of the sealed tube filled with iodine gas will remain 27.0 grams.

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A maximum of four bonds can be used between two atoms.

Explanation:

Ionic bond is one of the strongest bond used in chemical reaction where the valence electrons are bonded and shared between the atoms. Sometimes four covalent bonds between two atoms might make them unstable as the bonds are formed in their outermost shell to make them to form or complete a total of 8 in their shell. For example carbon has 3 as maximum.

Answer:

C) The amount of water released through the dam and the dam's height account for the potential energy

Explanation:

The gravitational potential energy of an object is given by:

[tex]U=mgh[/tex]

where

m is the mass of the object

g is the acceleration of gravity

h is the height of the object relative to the ground

As we see from the equation, the potential energy depends on two factors related to the object:

- its mass

- its height above the ground

So for the water in the dam, its potential energy depends on

- the mass of the water (= the amount of water released)

- the height of the water (= the dam's height)

So the correct answer is

C) The amount of water released through the dam and the dam's height account for the potential energy

Answer:

Turbulent

Explanation:

The Reynolds number is:

Re = ρvL/μ

where ρ is the fluid density,

v is the velocity,

L is the characteristic length (for pipe, it's the diameter),

and μ is the dynamic viscosity.

Given:

ρ = 0.805 g/cm³

v = 0.48 ft/s = 14.63 cm/s

L = 2.067 in = 5.250 cm

μ = 0.43 cp = 0.0043 g/cm/s

Re = (0.805 g/cm³) (14.63 cm/s) (5.250 cm) / (0.0043 g/cm/s)

Re ≈ 14400

Flow is considered turbulent if the Reynolds number is greater than 4000.  So this is turbulent.

Answer:

There are 7.8 x 10¹³ molecules ghrelin in 1 l blood

Explanation:

Molarity (M) means the number of moles of a solute in one-liter solution. Then, 1.3 x 10⁻¹⁰ M means that there are 1.3 x 10⁻¹⁰ moles of ghrelin in 1 l of blood.

Since 1 mol of something are 6.022 x 10²³ something, then:

1.3 x 10⁻¹⁰ mol ghrelin * 6.022 x 10²³ molecules ghrelin / 1 mol ghrelin

= 7.8 x 10¹³ molecules ghrelin

Answer:

the final  pressure = 1120 torr

Explanation:

In the starting, when a 10-L drum consist 6 L of ether at 18°C

Volume of gas  [tex]= ( 10 L - 6 L ) = 4 L[/tex]

Total pressure of gas which is equal to Atmospheric Pressure = 760 Torr

vapor pressure of liquid (ether) = 400 Torr

vapor pressure of air

[tex] = ( 760 Torr - 400 Torr ) = 360 Torr[/tex]

when the drum top is sealed & dented to 8 L capacity

Volume of gas [tex] = ( 8 L - 6 L ) = 2 L[/tex]

Total pressure of gasses

[tex] = 760 Torr x ( 4 L / 2 L ) = 1560 Torr[/tex]

vapor pressure of ether  [tex]= 400 Torr x 2 = 800 Torr[/tex]

vapor pressure of air [tex]= 360 Torr x 2 = 720 Torr[/tex]

But some of the liquid (ether) vapor will condensed into liquid just to maintain the vapor pressure of ether at 400 Torr )

therefore, the final  pressure

[tex]=  [ 400\ Torr (ether ) + 720\ Torr (air) ] = 1120 Torr[/tex]

Answer:

final total pressure is 1120 torr

Explanation:

given data

volume v1 = 10 L

volume v2 = 6 L

temperature = 18°C

atmosphere pressure P = 760 torr

volume v3 = 8 L

vapor pressure = 400 torr

to find out

pressure inside the dented

solution

we know here at 6 L volume by small amount of vaporize no temperature will change

and

gas volume in drum is = 10 - 6 = 4L

and

atmosphere pressure = total pressure inside drum

and

partial pressure of air is = 760 - 400

partial pressure of air is  = 360 torr  

so

at 8 L volume sealed and sealed then gas volume inside drum is

gas inside = 760 × [tex]\frac{4}{2}[/tex]

gas inside = 1560 torr

so partial pressure either side is

partial pressure = 2× 400  = 800 torr

partial pressure = 2× 360 = 720 torr

so we keep vapor pressure here 400 torr

so

final total pressure = 400 + 720

final total pressure is 1120 torr

The question is weird. It doesn't matter whether it was a man or a woman, or whether he went on a diet, or whether he gained or lost mass or stayed the same, or what his mass was later. In fact, it doesn't even matter WHAT the number 100.6 represents. The answer would be the same in any case.

100.6 = 1.006 x 10^2

If a man finds that he has a mass of 100.6 kg. He goes on a diet, and several months later he finds that he has a mass of 96.4 kg, then his mass in the scientific notation would be 1.006 × 10² kg.

In positional notation, significant figures refer to the digits in a number that is trustworthy and required to denote the amount of something, also known as the significant digits, precision, or resolution.

Only the digits permitted by the measurement resolution are trustworthy, therefore if a number expressing the result of a measurement (such as length, pressure, volume, or mass) has more digits than the number of digits permitted by the measurement resolution, only these can be significant figures.

Thus, If a man discovers that he weighs 100.6 kg. He starts a diet, and after a few months discovers he weighs 96.4 kg, his mass would be written as 1.006 10² kg in the scientific notation.

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