Water flows through a pipe of
radius 0.0250 m at 1.50 m/s.
What is the Volume Flow Rate?
(Keep 3 sig figs.)
(Unit=m^3/s)

Help please

The electric force between two point charges is given by the equation

[tex]F=k*q_1*q_2/r^2[/tex]

What is force?

The interaction between two things is measured by the physical quantity known as force. It is a vector quantity, and the sign F is frequently used to denote it. When an object interacts with another object, it feels a push or a pull.

where r is the distance between the charges, q1 and q2 are their magnitudes, and k is the Coulomb constant.

When we enter the problem's specified values, we obtain

[tex]2.2N=8.99*10^9\ N*m^2/C^2*q*-2q/(1.4 m)^2[/tex]

which simplifies to

q = -0.500 N/C.

Thus, the magnitude of charge q is 0.500 N/C.

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a) The time after the release of the first stone that the second stone hits the water is 2.0 s.

b) 15.7 m/s is the initial speed of the second stone.

c)  The speed of the first stone as it hits the water is 15.7 m/s.

d) The speed of the second stone as it hits the water is 28.2 m/s.

What is velocity?

Velocity is a vector quantity that measures both the speed and direction of an object's motion. It is equal to the rate of change of an object's position with respect to time. Velocity is usually represented by the symbol v and is measured in meters per second (m/s).

a) The time between first and second stone's release is 1.0 s. Since the time of release of first stone and the time of splash of both stones are same, the time between the release of second stone and the splash of both stones is 1.0 s.

Thus, the time after the release of the first stone that the second stone hits the water is 2.0 s.

b) The initial speed of the second stone can be calculated using the equation of motion,

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.

Substituting the values,

v² = (1.7)² + 2(9.8) * 59

v = 15.7 m/s

c) The speed of the first stone as it hits the water can be calculated using the equation of motion,

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.

Substituting the values,

v² = (1.7)² + 2(9.8) * 59

v = 15.7 m/s

d) The speed of the second stone as it hits the water can be calculated using the equation of motion,

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.

Substituting the values,

v² = (15.7)² + 2(9.8) * 59

v = 28.2 m/s

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Answer: d. increase

Explanation:

If the sun were more massive, the gravitational force between the sun and Earth would increase. This means that Earth's gravity with the sun would also increase. Therefore, the correct answer is (D) increase.

The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. So, if the mass of one of the objects increases, the gravitational force between them will also increase. In this case, if the mass of the sun were to increase, the gravitational force between the sun and Earth would become stronger, and hence, Earth's gravity with the sun would also increase.

A sound that is 7.8x10-4 W/m2 in intensity is equal to (10 dB)log3.2106 W/m21012 W/m2=185 dB.

How can you determine the relative volume of a sound?

The decibel, often known as the db or 0.1 bel, is the standard measurement unit. Hence, b = 10 log10 (I/I0) can be used to express the relationship between relative intensities, or b, in decibels. This equation can be used to determine that one decibel equals a 26 percent intensity variations.

What does physics mean by relative intensity?

The "decibel level" of a sound is a less formal term for relative intensity level. It is not the same as energy; relative intensity level reflects loudness more faithfully by using a logarithmic scale.

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Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

                               P.E = mgh joule

Considering h  = 0 for the vertical position

The h at ∅ = 42° is  h = L (1 - cos∅)

                              P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

                              P.E = 25 x 9.8 x 2.2 (1 - cos42°)

                                     = 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

                 ∴                 K.E = P.E

                                    1/2 mv² = 138.44

                                    1/2 x 25 x v² 138.44

                                           v² = 11.0752

                                            v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

            Tension on string, T = Force acting on the swing, F

                     

                           

                           =

                           = - 2.2 x 25 x 9.8 [cos0 - cos 42°]

                           = - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

We can use conservation of energy to solve this problem. At the initial position, the child has no kinetic energy and all her energy is potential energy due to her height above the lowest point of the swing. At the bottom of the swing, the child has no potential energy and all her energy is kinetic energy due to her speed.

(a) The potential energy of the child just as she is released can be calculated as:
PE = mgh
where m is the mass of the child, g is the acceleration due to gravity, and h is the height of the child above the lowest point of the swing. At the initial position, h = 2.20 m, so the potential energy is:
PE_initial = mgh = (25 kg)(9.81 m/s^2)(2.20 m) = 544 J

At the bottom of the swing, h = 0, so the potential energy is zero:
PE_bottom = 0 J

The potential energy at the initial position is greater than the potential energy at the bottom of the swing, since the child loses potential energy as she swings down.

(b) We can use conservation of energy to find the speed of the child at the bottom of the swing. At the initial position, all the energy is potential energy. At the bottom of the swing, all the energy is kinetic energy. Therefore, the potential energy at the initial position is equal to the kinetic energy at the bottom of the swing:
PE_initial = KE_bottom
mgh = (1/2)mv^2
where v is the speed of the child at the bottom of the swing. Solving for v, we get:
v = sqrt(2gh)
where sqrt means square root. Substituting the values, we get:
v = sqrt(2(9.81 m/s^2)(2.20 m)) = 6.26 m/s

Therefore, the child will be moving at a speed of 6.26 m/s at the bottom of the swing.

(c) The work done by the tension in the ropes as the child swings from the initial position to the bottom can be found as the change in the total mechanical energy of the child:
W = ΔE = KE_bottom - PE_initial
Substituting the values, we get:
W = (1/2)mv^2 - mgh
W = (1/2)(25 kg)(6.26 m/s)^2 - (25 kg)(9.81 m/s^2)(2

The system's final temperature is roughly 16.7°C.

What is a system's final temperature?

You may determine your substance's final heat by multiplying the temperature change by the initial temperature. Your water's final temperature would be 24 + 6, or 30 degrees Celsius, for instance, if it started off at 24 degrees Celsius.

The following is the formula for energy conservation:

Q1 + Q2 = 0

Q = mcΔT

Q1 + Q2 = 0

568.8

Simplifying and solving for

6394.4 - 106768 = 0

= 16.7°C

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A cesium-133 atom's radiation goes through 1.5 million cycles in around 0.1633 microseconds (or 163.3 nanoseconds).

What frequency does one kind of radiation that cesium-133 emits have?

9,192,631,770 hertz (cycles per second) is the frequency of the microwave spectral line that the isotope cesium-133 emits. The basic unit of time is provided by this. Cesium clocks have an accuracy and stability of 1 second in 1.4 million years.

The radiation emitted by cesium-133 has a frequency of 9,192,631,770 cycles per second, or 9.192631770 109 Hz.

The following formula may be used to determine how long 1.5 million radiation cycles take to complete:

Time is equal to the frequency of cycles.

Plugging in the numbers, we get:

time = 1.5 million / 9.192631770 × 10^9 Hz

time = 1.632995101 × 10^-7 seconds

So it takes approximately 0.1633 microseconds (or 163.3 nanoseconds) for radiation from a cesium-133 atom to complete 1.5 million cycles.

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The value of the charge on each particle is [tex]1.05 x 10^-8 C[/tex].

What is Coulomb's law?

Coulomb's law is a fundamental principle of electrostatics that describes the interaction between electric charges. It states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. We can use Coulomb's law to solve this problem. Mathematically,

[tex]F = k(q1q2)/r^2[/tex]

where F is the force of attraction or repulsion between the two charged particles,[tex]q1[/tex] and [tex]q2[/tex] are the magnitudes of the charges on the two particles, r is the distance between them, and k is Coulomb's constant, which has a value of [tex]9.0 x 10^9 Nm^2/C^2.[/tex]

In this problem, we know that the charges are equal and the distance between them is 10.0 cm. We also know that the force between them is 0.5 N. Therefore,

[tex]0.5 N = k(q^2)/(0.1 m)^2[/tex]

Solving for q, we get:

[tex]q = \sqrt{[(0.5 N)(0.1 m)^2/k]}[/tex]

[tex]q = \sqrt{(0.5 N)(0.01 m)/(9.0 x 10^9 Nm^2/C^2)}[/tex]

[tex]q = 1.05 x 10^-8 C[/tex]

Therefore, the value of the charge on each particle is [tex]1.05 x 10^-8 C.[/tex]

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Answer:

Explanation:

Because you have velocity along the y axis and time along the x axis, this is a velocity v time graph which is an acceleration graph. The slope of the line in this graph IS the acceleration. We can use 2 points and the slope formula to solve for the acceleration:

(0, 0) and (1, 3):

[tex]m=\frac{3-0}{1-0}=3[/tex] m/s squared, choice D.