Consider a normal shock wave moving with a velocity of 680 m/s into still air at standard atmospheric conditions (p
1 =1 atm and T
1 =288 K). a. Using the equations of Sec. 7.2, calculate T 2 ,p 2 , and u p behind the shock wave. b. The normal shock table, Table A.2, can be used to solve moving shock wave problems simply by noting that the tables pertain to flow velocities (hence, Mach numbers) relative to the wave. Use Table A.2 to obtain T 2 ,p 2 , and u p for this problem

A. the change in volume of the cube is approximately 0.120601 cm^3. B. the average rate of change of volume of the cube with respect to an edge length is approximately 12.0601 cm^3/cm. C. the equation of the line that passes through the point (1, 5) and is parallel to the tangent line to f(x) = 1/x at x = 3 is y = (-1/9)x + 14/9.

(a) To find the change in volume of the cube as x increases from 2.00 to 2.01 centimeters, we need to calculate the difference in volume between these two values.

The volume of a cube is given by V = x^3, where x is the edge length.

For x = 2.00 cm, the volume V1 = (2.00 cm)^3 = 8.00 cm^3.

For x = 2.01 cm, the volume V2 = (2.01 cm)^3 = 8.120601 cm^3.

The change in volume is ΔV = V2 - V1 = 8.120601 cm^3 - 8.00 cm^3 ≈ 0.120601 cm^3.

Therefore, the change in volume of the cube is approximately 0.120601 cm^3.

(b) The average rate of change of volume of the cube with respect to an edge length as x increases from 2.00 to 2.01 centimeters can be calculated by dividing the change in volume by the change in edge length.

ΔV = 0.120601 cm^3 (from part a)

Δx = 2.01 cm - 2.00 cm = 0.01 cm

The average rate of change of volume is ΔV/Δx = 0.120601 cm^3 / 0.01 cm ≈ 12.0601 cm^3/cm.

Therefore, the average rate of change of volume of the cube with respect to an edge length is approximately 12.0601 cm^3/cm.

(c) The instantaneous rate of change of volume of the cube with respect to an edge length at the instant when x = 2 centimeters can be found by taking the derivative of the volume function V = x^3 with respect to x and evaluating it at x = 2.

dV/dx = 3x^2

At x = 2 cm, the instantaneous rate of change of volume is dV/dx evaluated at x = 2:

dV/dx = 3(2 cm)^2 = 12 cm^2.

Therefore, the instantaneous rate of change of volume of the cube with respect to an edge length at x = 2 centimeters is 12 cm^2.

To find the equation of the line that passes through the point (1, 5) and is parallel to the tangent line to f(x) = 1/x at x = 3, we need to determine the slope of the tangent line.

The derivative of f(x) = 1/x is given by f'(x) = -1/x^2.

At x = 3, the slope of the tangent line is f'(3) = -1/(3^2) = -1/9.

Since the line we want to find is parallel to the tangent line, it will have the same slope. So the slope of the line is -1/9.

Using the point-slope form of a linear equation, we can write the equation of the line as:

y - y1 = m(x - x1),

where (x1, y1) is the given point (1, 5) and m is the slope.

Substituting the values, we have:

y - 5 = (-1/9)(x - 1).

Expanding and rearranging the equation, we get:

y = (-1/9)x + 14/9.

Therefore, the equation of the line that passes through the point (1, 5) and is parallel to the tangent line to f(x) = 1/x at x = 3 is y = (-1/9)x + 14/9.

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Given:

Mass attached to the left-hand side = 3 kg

Mass attached to the right-hand side = 5 kg

Let's find the magnitude of the instantaneous acceleration of the system when it is released from rest.

Apply the formula:

Where:

m1 = 3 kg

m2 = 5 kg

g = 9.8 m/s^2

Thus, we have:

The magnitude of the instantaneous acceleration is 2.45 m/s².

ANSWER:

2.45 m/s²

The angular magnification (M) obtained with a thin lens used as a simple magnifier when the object is at the focal point is -1.

The angular magnification is given by the formula:

M = -f / (f - d)

Where:

M is the angular magnification,

f is the focal length of the lens,

d is the distance between the object and the lens.

In this case, since the object is at the focal point, the distance between the object and the lens (d) is equal to the focal length of the lens (f). Substituting the values, we have:

M = -f / (f - f)

M = -f / 0

M = -1

Therefore, when the object is at the focal point, the angular magnification obtained with the lens is -1.

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A promising evidence of a habitat that might support life on the planet Mars is water.

There is evidence supporting the existence of liquid water on Mars from numerous sources. The finding of repeated black streaks on Martian slopes and the presence of hydrated minerals suggest the potential of seasonal or location-specific briny water flows. A necessary component of life as known is liquid water.

Additionally, Mars possesses underground ecosystems that could provide defence against radiation and severe surface conditions. Researchers have found evidence of ancient underground hydrothermal systems as well as beneath ice. These settings might offer consistent conditions for the development of microbial life.  Methane gas has been found in the Martian atmosphere, along with variations over time, and this has led to questions regarding its origin. Both geological and biological processes can result in the production of methane.

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The rms voltage required for a resistor to dissipate 11.5 W is 21.8 V.

The power (P) dissipated by a resistor is given by P = V²/R, where V is the voltage across the resistor and R is the resistance. We are given that the resistor dissipates 1.80 W when the rms voltage is 9.50 V, so we can write:

1.80 watt(W) = (9.50 V)²/R

Solving for R, we get:

R = (9.50 V)²/1.80 W = 49.97 Ω

To find the rms voltage required for the resistor to dissipate 11.5 W, we can use the same equation:

11.5 W = V²/49.97 Ω

Solving for V, we get:

V = √(11.5 W * 49.97 Ω) = 21.8 V

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Answer:

0.65 seconds

2.07m

Explanation:

Here we need to look at the vertical and the horizontal components of the initial velocity.

In terms of his hitting the water.

Define the upwards direction as positive

We know his initial upwards velocity (u) is 4.5sin(45)

His displacement (s) when he hits the water is zero

Acceleration due to gravity is -9.8

We want to find t

So using s=ut+1/2a\(t^{2}\)

0=4.5sin(45)t+1/2(-9.8)\(t^{2}\)

Take out a common t

0=t(4.5sin45-4.9t)

t=0   or t= 0.65

The first answer is not valid so the answer is 0.65.

Now we can look at the horizontal component

It takes him 0.65 seconds to hit the water, he travels at a constant horizontal velocity of 4.5cos45

Therefore using s=vt

s=4.5cos45 x 0.65= 2.07m away

The magnitude of the net electric field at the origin (0,0) cm is approximately \(83.19 × 10^6 N/C\).

To calculate the magnitude of the net electric field at the origin, we need to calculate the electric fields generated by each charge and then sum them up.

The electric field at a point due to a point charge is given by Coulomb's Law:

E = k * (q / \(r^2\))

where E is the electric field, k is the electrostatic constant (\(9 × 10^9 Nm^2/C^2\)), q is the charge, and r is the distance from the charge to the point.

Let's calculate the electric fields generated by each charge at the origin:

For the 5nC charge:

q1 = 5nC

r1 = 7 cm = 0.07 m

E1 = k * (q1 / \(r1^2\))

For the 2nC charge:

q2 = 2nC

r2 = 3 cm = 0.03 m

E2 = k * (q2 / \(r2^2\))

Now, we can calculate the net electric field by summing up the electric fields:

E_net = E1 + E2

Substituting the values and performing the calculations:

\(E1 = (9 × 10^9 Nm^2/C^2) * (5 × 10^(-9) C) / (0.07 m)^2\)

E1 ≈ 9188571.43 N/C

\(E2 = (9 × 10^9 Nm^2/C^2) * (2 × 10^(-9) C) / (0.03 m)^2\)

E2 ≈ 74000000 N/C

E_net = E1 + E2

E_net ≈ 83188571.43 N/C

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In this particular question, the task is to reformulate the given initial value problem in terms of a new dependent variable, using the constants and variables provided. The initial value problem is related to the study of an antioxidant, Hsian-tsao leaf gum, and its antioxidation activity, which depends on concentration in a certain way.

The dependent variable is a quantitative measure of antioxidant activity at concentration, while the constant represents a limiting or equilibrium value of this activity, and a positive rate constant. The goal is to use this information to reformulate the problem in terms of a new dependent variable.

The specific details of this reformulation will depend on the constants and variables provided, but it is important to focus on the key concepts and to provide a clear and concise answer that directly addresses the question at hand.

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Answer:

but the answer should be no as if 3 bulbs are connected together if one burns out the whole circuit will stop .hope you understood

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The work done by the force of 9 Newtons over a distance of 10 meters is 90 Joules.

To calculate the work done by the force, we need to use the formula:

Work = Force x Distance x cos(theta)

where theta is the angle between the force vector and the displacement vector.

In this case, the body is moving horizontally, so the angle between the force vector and the displacement vector is 0 degrees. Therefore, cos(theta) = cos(0) = 1.

We are given that the force acting on the body is 9 Newtons, and the distance moved by the body is 10 meters.

Substituting these values into the formula, we get:

Work = 9 N x 10 m x cos(0)

Work = 90 Joules

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Answer:

m = 15 kg

Explanation:

Applied force to the cart, F' = 18 N

Force of friction, f = 15 N

Net force acting on the cart,

F = F' - f

F = 18 - 15

F = 3 N

The change in velocity of the cart is 1 m/s every 5 s.

We need to find the mass of the cart.

The formula of net force is given by :

F = ma, where a is the acceleration of the cart

\(F=\dfrac{m\Delta v}{t}\\\\m=\dfrac{Ft}{\Delta v}\\\\m=\dfrac{3\times 5}{1}\\\\m=15\ kg\)

So, the cart's mass is 15 kg.

Answer:

A first-class lever has the fulcrum between the load and the effort. A second-class lever has the load between the effort and the fulcrum. A third-class lever has the effort between the load and the fulcrum. A see-saw is an example of a first-class lever.

Explanation:

To find:

Which of the planets has the highest density.

Explanation:

Neptune is known as the ice planet. It has a higher density than gas giants but a lower density than rocky planets. The density of the Neptune is 1638 kg/m³.

Jupiter is known as a gas giant. Gas giants has the lowest densities. The density of Jupiter is 1326 kg/m³.

The Earth and the Mercury are rocky planets. The densities of rocky planets are more than that of the gas giants and ice giants. The density of the Earth is 5514 kg/m³ and the density of the Mercury is 5429 kg/m³.

Final answer:

On comparing the densities of the planets, the Earth has the highest density.

Thus the correct answer is option C.

Answer:

a large elliptical galaxy

Explanation:

When we arrange the objects in increasing strength of gravitational attraction, the following results:

To arrange the objects in increasing strength of gravitational attraction, we shall determine the gravitational force between each pair oif the objects. Details below:

For electron-dog:

F = GM₁M₂ / r²

F = (6.67×10¯¹¹ × 9.11×10⁻³¹ × 20) / 1²

F = 1.22×10⁻³⁹ N

For butterfly-rock:

F = GM₁M₂ / r²

F = (6.67×10¯¹¹ × 5.0×10⁻⁴ × 8) / 1²

F = 2.67×10⁻¹³ N

For electron-butterfly:

F = GM₁M₂ / r²

F = (6.67×10¯¹¹ × 9.11×10⁻³¹ × 5.0×10⁻⁴) / 1²

F = 3.04×10⁻⁴⁴ N

For rock-dog:

F = GM₁M₂ / r²

F = (6.67×10¯¹¹ × 8 × 20) / 1²

F = 1.07×10⁻⁸ N

From the above calculations, we have the gravitational attraction as:

Thus, arranging in increasing order, we have:

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Answer:

The magnetic field through the coil must change.

Explanation:

To convert gr 1 (grain) to mg, you would multiply the number of grains by 64.8. Therefore, 1 grain is equal to 64.8 mg.

To convert a measurement from one unit to another, we need to use a conversion factor that relates the two units. In this case, we want to convert from grains (gr) to milligrams (mg).

The conversion factor between grains and milligrams is 1 gr = 64.8 mg. This means that for every 1 grain, there are 64.8 milligrams. We can use this conversion factor to convert any given number of grains to milligrams.

For example, if we have 5 grains and want to know how many milligrams that is, we would multiply 5 grains by 64.8 mg/gr. This gives us:

5 gr * 64.8 mg/gr = 324 mg

So, 5 grains is equal to 324 milligrams.

Similarly, if we want to convert 1 grain to milligrams, we just need to multiply 1 grain by the conversion factor of 64.8 mg/gr. This gives us:

1 gr * 64.8 mg/gr = 64.8 mg

So, 1 grain is equal to 64.8 milligrams.

Therefore, to convert any number of grains to milligrams, we simply multiply the number of grains by 64.8.

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Answer:

amplitude

Explanation:

sound travel and it amplifies

Answer:

A

Explanation:

Answer: A

Explanation:

Answer:

Release

Explanation:

The water is heated until

The direction of the angular momentum vector will be upward (d), as that

is the direction in which your fingers will curl using the right-hand rule,

since the front wheel of a bicycle rotates clockwise when viewed from

the rider's perspective. Therefore option d) upward is correct.

To use the right-hand-rule to find the direction of the angular

momentum vector of the front wheel of a bicycle when

riding at a constant speed, we need to follow these steps:

Extend your right hand with your thumb pointing in the direction of the

velocity of the front wheel (forward).

Curl your fingers towards the direction of rotation of the wheel

(clockwise).

The direction in which your fingers curl gives the direction of the angular

momentum vector.

Since the front wheel of a bicycle rotates clockwise when viewed from

the rider's viewpoint, the direction of the  angular momentum vector will

be upward (d), as that is the direction in which your fingers will curl using

the right-hand- rule.

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Answer:

When a net force acts on a body it produces acceleration in body the magnitude of tries acceleration is directly proportional to the net force acting and inversely proportional to the mass of the body

Mathematical form

F=ma

Explanation:

I hope it will help you

Answer:

A. like poles repel or push apart

Answer:

63.8454kgm/s

Explanation:

According to Newton's second law;

Force = mass ×acceleration

Fore = m(v-u)/t

Force = mv-mu/t

MV is the final momentum

mu is the initial momentum = 38kgm/s

Force = 88.3N

Time = 0.338s

Substitute and get mv

88.3 = mv-38/0.338

88.3×0.338 = mv-38

29.8454 = mv-38

MV = 29.8454+34

mv = 63.8454kgm/s

Hence the final momentum is 63.8454kgm/s

Answer:

67.8

Explanation:

im an acellus student, and this is the answer that i got when doing the problem.

Answer:

a. Winter.

Explanation:

Answer:

Winter

Explanation:

Answer:

F/6.

Explanation:

Let 'M' represent the mass of the block.

Let 'F' represent the force on earth.

Let 'a' represent the acceleration on earth.

Let 'Fm' represent the force on the moon.

Let 'am' represent the acceleration on the moon.

Recall:

Force (F) = mass (M) x acceleration (a)

F = M•a

Making M the subject

M = F/a

But,

Acceleration on the moon (am) = 1/6 acceleration (a) of earth.

am = 1/6a

Fm = M•am

M = Fm/am

M = Fm/(1/6a)

Since the mass of an object is constant,

Therefore,

Mass on earth = mass on moon

Mass on earth (M) = F/a

Mass on moon (M = Fm/(1/6a)

F/a = Fm/(1/6a)

Cross multiply

Fm × a = F × 1/6a

Divide both side by a

Fm = (F × 1/6a) /a

Fm = F/6

Therefore, the force necessary to give the moon bound block the same acceleration is F/6.

Answer:

a = 4.125 [m/s^2]

Explanation:

In order to calculate the value of acceleration, we must use Newton's second law, which tells us that the sum of forces must be equal to the product of mass by acceleration.

ΣF = m*a

where:

ΣF = sum of forces [N]

m = mass [kg]

a = acceleration [m/s^2]

Performing a summation of forces on the x-axis, we have:

4000 - 700 = m*a

4000 - 700 = 800*a

a = 4.125 [m/s^2]

Answer:

      k = \(\frac{g \ m_1}{L}\)

Explanation:

In this exercise the two cases presented are a simple harmonic motion, with angular velocity

spring - mass                                      w² = k / m₁

simple pendulum (string- mass m₂)  w² = g / L

angular velocity and period are related

              w = 2π/ T

since they indicate that the two periods are equal, the angular velocities are also equal, therefore we can equal the two equations

             \(\frac{k}{m1} = \frac{g}{L}\)

             k = \(\frac{g \ m_1}{L}\)