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

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.

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: 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.

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

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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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.

The force needed to lift the 1800 kg car with the small piston is 2,649 N or approximately 270 kg (since 1 kg is equal to 9.81 N).  

The hydraulic lift works based on Pascal's principle, which states that the pressure applied to a confined fluid is transmitted equally in all directions throughout the fluid.

Assuming there is no loss of energy due to friction or other factors, the force exerted on the small piston will be equal to the force exerted on the large piston. This can be expressed as:

F1/A1 = F2/A2

where F1 is the force exerted on the large piston, A1 is the area of the large piston, F2 is the force exerted on the small piston (which we want to find), and A2 is the area of the small piston.

We can rearrange this equation to solve for F2:

F2 = (F1/A1) x A2

Given that the area of the large piston is 100 cm², we can calculate the force exerted on the large piston by using the weight of the car and the gravitational acceleration:

F1 = m x g = 1800 kg x 9.81 m/s² = 17,658 N

Substituting the values into the equation, we get:

F2 = (17,658 N / 100 cm2) x 15 cm² = 2,649 N

Therefore, the force needed to lift the 1800 kg car with the small piston is 2,649 N or approximately 270 kg (since 1 kg is equal to 9.81 N).

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The light ray will deviate from its original path with 19.5° after refraction.

Applying Snell's law to calculate the angle of refraction:

n1 sin θ1 = n2 sin θ2

where n1 and θ1 =  the refractive index and the angle of incidence in the first medium (air),

n2 and θ2 =  the refractive index and the angle of refraction in the second medium (glass).

In this example,

n1 = 1.00 (refractive index of air), θ1 = 30°, and

n2 = 1.5 (refractive index of glass).

We then calculate for  θ2:

n1 sin θ1 = n2 sin θ2

1.00 * sin 30° = 1.5 * sin θ2

0.5 = 1.5 * sin θ2

sin θ2 = 0.5 / 1.5 = 1/3

θ2 = sin^-1(1/3)

θ2 = 19.5°

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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.

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

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

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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The two risk factors that can affect the ability of a child to learn in school is poor parenting and malnutrition.

A risk factor can be defined as any predisposing factor that can expose an individual to harm.

A risk factor that affects a child is an aspect of the child or environment that increases the probability of poor outcomes.

The two risk factors that can affect the ability of a child to learn in school include the following:

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

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

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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Answer:

[tex]80\; {\rm m\cdot s^{-1}}[/tex].

Explanation:

The frequency [tex]f[/tex] of a wave is the number of cycles completed in unit time ([tex]1\; {\rm s}[/tex] in this example.) In this question, [tex]f = 40\; {\rm s^{-1}}[/tex] ([tex]1\; {\rm Hz} = 1\; {\rm s^{-1}}[/tex]) means that the wave would complete [tex]40[/tex] cycles in every [tex]1\; {\rm s}[/tex].

The wavelength [tex]\lambda[/tex] of a wave is the distance the wave travels in each cycle. It is given that [tex]\lambda = 2\; {\rm m}[/tex].

The goal is to find the wave speed, which is the distance that this wave travels in unit time ([tex]1\; {\rm s}[/tex].)

In this question, it is given that [tex]\lambda = 2\; {\rm m}[/tex] and [tex]f = 40\; {\rm s^{-1}}[/tex]. Thus, this wave would travel a total of [tex]40\, (2\; {\rm m}) = 80\; {\rm m}[/tex] for the [tex]40[/tex] cycles completed in each unit time of [tex]1\; {\rm s}[/tex] ([tex]\lambda = 2\; {\rm m}[/tex] for each cycle.) The speed of this wave would be [tex]80\; {\rm m\cdot s^{-1}}[/tex].

Formally, the speed [tex]v[/tex] of this wave can be found by multiplying the wavelength [tex]\lambda[/tex] of this wave by its frequency [tex]f[/tex]:

[tex]\begin{aligned}v &= \lambda\, f \\ &= (2\; {\rm m})\, (40\; {\rm s^{-1}) \\ &= 80\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

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

The top surface of the oil on that side of the tube is 0.6 times higher than the surface of the water on the other side of the tube.

The principle of hydrostatics, also known as Pascal's principle, states that when an external pressure is applied to a fluid in a container, that pressure is transmitted uniformly in all directions within the fluid, regardless of the shape or volume of the container. In other words, the pressure applied to a confined fluid will be distributed evenly throughout the fluid and will not change in magnitude at any point within the fluid. This principle is important in a number of applications, such as hydraulic systems, which use fluids to transmit force and pressure from one point to another. It is also used to explain how liquids exert pressure on the walls of their container and how objects can float or sink in fluids.

We can use the principles of hydrostatics to solve this problem. Let's call the height difference between the two water surfaces h. We can assume that the oil completely covers the water on one side of the tube and does not mix with it, so the oil and water form two separate liquid columns with a common interface. Let's call the height difference between the oil and water surfaces on the same side of the tube H.

The pressure at any given point in a fluid depends only on the depth of that point below the surface of the fluid and the density of the fluid. Since the two water columns are at the same height, they experience the same pressure from the atmosphere. Similarly, the two oil columns experience the same pressure from the atmosphere.

Now consider a point on the interface between the oil and water on the same side of the tube. This point is at a depth of h+H below the water surface on the other side of the tube, so the pressure at this point is greater than atmospheric pressure by an amount equal to the product of the density of water, the acceleration due to gravity, and the total depth (h+H):

P = Patm + ρwatergh

where P is the pressure at the interface, Patm is atmospheric pressure, ρwater is the density of water, g is the acceleration due to gravity, and h+H is the total depth.

Similarly, the pressure at this point is less than atmospheric pressure by an amount equal to the product of the density of oil, the acceleration due to gravity, and the depth of the oil column (H):

P = Patm - ρoilgH

Since the interface between the oil and water is at the same pressure, we can equate these two expressions for P:

Patm + ρwatergh = Patm - ρoilgH

Solving for H, we get:

H = h(ρwater/ρoil)

Substituting the given values, we get:

H = 0.6h

Therefore, the top surface of the oil on that side of the tube is 0.6 times higher than the surface of the water on the other side of the tube.

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