At 3.00 m from a source that is emitting sound uniformly in all directions, the sound level (b) is 60.0 dB.
How many meters from the source would the sound level be one-fourth the sound level at 3.00 m?

Answer:

1) 50°

Explanation:

We need to find the angle of incidence first before finding the angle of reflection.

Angle of incidence = 90° - 40°

                                = 50°

Since the angle of incidence is the same as the angle of reflection, the angle of reflection here would be 50°.

Answer:

12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor are equivalent to 6 drinks.

Explanation:

In the United States, a standard "drink" of beer has 12 ounces, a standard "drink" of wine has 5 ounces and standard drink of liquor has 1.5 ounces. Then, we obtain the quantity of drinks by dividing the total volume of each drink by its respective unit volume and summing each term. That is:

[tex]N = \frac{12\,oz}{12\,\frac{oz}{dr} }+\frac{12\,oz}{5\,\frac{oz}{dr} }+\frac{3\,oz}{1.5\,\frac{oz}{dr} }[/tex]

[tex]N = 1\,dr+2.4\,dr+2\,dr[/tex]

[tex]N = 5.4\,dr[/tex]

[tex]N = 6\,dr[/tex]

12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor are equivalent to 6 drinks.

Answer:

[tex]Fr_k=152.88\ N[/tex]

Explanation:

Net Force

The net force is defined as the vector sum of all the forces acting on a body at a certain moment.

We should recall some basic concepts and equations to solve the problem.

If no external forces are applied in the vertical direction, the weight of the object and the normal force have the same magnitude and point to opposite directions.

The friction force is defined as:

[tex]Fr_k=\mu_k N[/tex]

[tex]Fr_s=\mu_s N[/tex]

Where the subindices k and s are referred as to the kinetic and static friction forces respectively.

The condition for the object to move is that the applied force is greater than the friction force.

The crate of oranges has a mass of 30 Kg, thus its weight is:

W = m.g = 30 * 9.8 = 294 N

The normal force is:

N = W = 294 N

The kinetic friction is calculated as:

[tex]Fr_k=0.52* 294[/tex]

[tex]\mathbf{Fr_k=152.88\ N}[/tex]

Answer:

Explanation:

.... ...

Given that,

The equation is "F=kx"

To find,

The derivative equation.

Solution,

We are given with the given equation which is as follows :

F = kx ...(1)

Where, F is force, k is spring constant and x is distance covered by the spring

Differentiate equation (1) wrt x.

[tex]\dfrac{dF}{dx}=\dfrac{d}{dx}(kx)\\\\\dfrac{dF}{dx}=k[/tex]

As k is constant.

Hence, this is the required solution.

Answer:

The weight of the brick is 300 lb.

Explanation:

Given that,

The work done in lifting a load of bricks = 15000 ft lb

It is lifted to a height of 50 ft

We need to find the weight of the brick. The work done by an object is given by :

[tex]W=F\times d[/tex]

Here, F = W (weight of the bricks)

[tex]F=\dfrac{W}{d}\\\\F=\dfrac{15000\ ft-lb}{50\ ft}\\\\F=300\ lb[/tex]

So, the weight of the brick is 300 lb.

The weight of the brick will be "300 lb".

Given:

Work done,

Height,

As we know the formula,

→ [tex]W = F\times d[/tex]

or,

→ [tex]F = \frac{W}{d}[/tex]

By substituting the values, we get

→     [tex]= \frac{15000}{50}[/tex]

→     [tex]= 300 \ lb[/tex]

Thus the response above is right.  

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Answer:

B. 2 Hz

Explanation:

Given parameters:

Period of the wave  = 0.5Hz

Unknown:

Frequency of the wave  = ?

Solution:

Frequency is the number of waves that passes through a point.

Period is the time taken for a number of waves to pass through a point.

  Period is the inverse of frequency

  Frequency  = [tex]\frac{1}{t}[/tex]  

 t is the period

  Frequency  = [tex]\frac{1}{0.5}[/tex]  = 2Hz

Answer:2

Explanation:

Answer:

first is regular reflection and 2nd is irregular

Answer:

C

Explanation:

Some of the mass

Answer:

D. Mass was transformed into energy during the process.

Answer:

C

Explanation:

If the arrows represent light rays, then Rachel sees a candle flame when the light released by the flame is received by her eyes.

Diagram C shows how Rachel can see a candle flame.

This is the visual representation of a substance and the organ responsible for this is the eye.

Images are formed through the eyes receiving light sensations and and relaying it to the brain for processing. The line is the light sensation which shows it being received by the eye therefore making option C the most appropriate option.

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The direction of the current from the positive to the negative battery terminals and the fact that the bulbs are connected in parallel to each other suggests that this circuit is a Closed Parallel Circuit. The correct option is D.

An open circuit and a closed circuit are two types of electrical circuits that describe the flow of electrical current through a circuit.

An open circuit is a circuit in which there is a break in the path of the current, meaning that the current cannot flow through the circuit. In an open circuit, the switch is in the "off" position or there is a broken wire, and so no electrical current can flow through the circuit. This means that there is no voltage or electrical energy being transferred from the source to the load.

On the other hand, a closed circuit is a circuit in which there is a complete path for the current to flow through, meaning that the current can flow through the circuit. In a closed circuit, the switch is in the "on" position, and there is a continuous path for the current to flow from the source to the load and back to the source. This means that there is voltage and electrical energy being transferred from the source to the load.

Here in this question,

The direction of the current from the positive to the negative battery and the fact that the bulbs are connected in parallel to each other suggests that this circuit is a Closed Parallel Circuit.

In a closed parallel circuit, the components are connected in parallel to each other, meaning that they are connected to the same two points in the circuit and the current has multiple paths to flow through. The voltage across each component is the same, and the total current flowing through the circuit is divided among the components according to their resistance.

In contrast, an open series circuit is a circuit in which the components are connected in series, meaning that they are connected end to end in a single path, with no other branch points for the current to follow. If one component fails, the circuit becomes open and the current stops flowing.

Therefore, based on the information provided, the circuit described in the question is a Closed Parallel Circuit.

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Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = [tex]\pi r^2[/tex]

V = Potential applied = 2 mV

k = Dielectric constant = 40

[tex]\epsilon_0[/tex] = Electric constant = [tex]8.854\times 10^{-12}\ \text{F/m}[/tex]

Capacitance is given by

[tex]C=\dfrac{k\epsilon_0A}{d}[/tex]

Charge is given by

[tex]Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}[/tex]

Number of electron is given by

[tex]n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}[/tex]

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

Answer:

x =  0.0734 m = 7.34 cm

Explanation:

First we shall calculate the area of the piston:

[tex]Area = \pi radius^2\\Area = \pi (0.028\ m)^2\\Area = 0.00246\ m^2[/tex]

Now, we will calculate the force on the piston due to atmospheric pressure:

[tex]Atmospheric\ Pressure = \frac{Force}{Area}\\\\Force = (Atmospheric\ Pressure)(Area)\\Force = (101325\ N/m^2)(0.00246\ m^2) \\Force = F = 249.56\ N[/tex]

Now, for the compression of the spring we will use Hooke's Law as follows:

[tex]F = kx\\[/tex]

where,

k = spring constant = 3400 N/m

x = compression = ?

Therefore,

x =  0.0734 m = 7.34 cm

Answer:

Explanation:

Given the initial velocity U = 670m/s

Horizontal velocity Ux = Ucos theta

Vertical component of the cannon velocity Uy = Usin theta

Given

U = 670m/s

theta = 45°

horizontal component of the cannonball’s velocity = 670 cos 45

horizontal component of the cannonball’s velocity = 670(0.7071)

horizontal component of the cannonball’s velocity = 473.757m/s

Vertical component of the cannonball’s velocity = 670 sin 45

Vertical component of the cannonball’s velocity  = 670 (0.7071)

Vertical component of the cannonball’s velocity  = 473.757m/s

Hence pair of answer is 473.8 m/s; 473.8 m/s

Answer:

8

Explanation:

The energy levels is given as

E(n) = n² * h² / ( 8 * m * L²), where

n = 1, 2, 3, 4,... etc

At ground state energy (n = 1), therefore is E(g) = h² / (8 * m * L²).

We can then say that

E(n) = n² * Eg

Therefore, to have E = 64 Eg, we must have n² = 64. And for n² to be equal to 64, we find the square root of n

n = √64

n = 8

Essentially, the needed quantum number is 8

Answer:

The acceleration acting on the train during this

time of travel is 0.85m/s²

HOW TO CALCULATE ACCELERATION:

• The acceleration of a moving body can be calculated by using the formula below:

a = (v - u)/t

Where;

1. a = acceleration (m/s²)

2. v = final velocity (m/s) 3. u = initial velocity (m/s)

4. t = time (s)

According to this question, a = ?, v = 48m/s, u = 14m/s, t = 40s.

= (48 - 14)/40 a=

• a = 34/40

• a = 0.85m/s²

Therefore, the acceleration acting on the train during this time of travel is 0.85m/s².

Hope this helps!

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Answer:

I got this question on Ap3x. The answer is Car C...I got it correct

Explanation:

This is because Car C is at the lowest point with the lowest amount of potential energy. Potential energy is stored at it's highest when it has the "potential" to fall, move, or etc. Car C seems to have gone farther down from the high point of the slope, meaning that most of the potential energy transformed into kinetic energy. All in all, Car C has the least potential energy. (Please give Brainliest, or not...your choice but this is the first question that I answered)

Answer:

A Living things can only come from living things.

B Cells come from pre-existing cells.

Explanation:

sorry if wrong e d g e 2022

Answer: D,E

Explanation:

I JUST TOOK IT

The distance that automobile travel in 1.5 seconds is (S)= 34.275 m.

Distance is a scaler quantity that refers to "how much ground an object has covered" during its motion of time.

To calculate the distance that automobile travel in 1.5 seconds, we are using the formula,

S= ut+(1/2)at²

Here we are given,

u= The initial velocity of the automobile = 20 m/s

t = The amount of time that the automobile travel = 1.5 Seconds.

a=The amount of acceleration of the automobile = 3.8 m/s².

We have to calculate the distance that automobile travel in 1.5 seconds = S

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

S= ut+(1/2)at²

Or, S= 20*1.5+(1/2)*3.8*(1.5)²

Or, S= 30+4.275

Or, S= 34.275m

Thus we can conclude that, the distance that automobile travel in 1.5 seconds is (S)= 34.275 m.

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Answer:

The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

Explanation:

We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:

[tex] P_{1} = P_{2} [/tex]

[tex] m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}} [/tex]

Where:

m₁: is the mass of the ball 1 = 100 g = 0.1 kg

m₂: is the mass of the ball 2 = 300 g = 0.3 kg

[tex]v_{1_{i}}[/tex]: is the initial velocity of the ball 1 = 6.20 m/s

[tex]v_{2_{i}}[/tex]: is the initial velocity of the ball 2 = 0 (it is at rest)

[tex]v_{1_{f}}[/tex]: is the final velocity of the ball 1 =?

[tex]v_{2_{f}}[/tex]: is the initial velocity of the ball 2 =?

[tex] m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}} [/tex]

[tex] v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}} [/tex] (1)        

Now, by conservation of kinetic energy (since they collide elastically):

[tex] \frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2} [/tex]          

[tex] m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2} [/tex]  (2)

By entering equation (1) into (2) we have:

[tex] m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2} [/tex]    

[tex] 0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2} [/tex]            

By solving the above equation for [tex]v_{2_{f}}[/tex]:

[tex]v_{2_{f}} = 3.1 m/s [/tex]

Now, [tex]v_{1_{f}}[/tex] can be calculated with equation (1):

[tex] v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s [/tex]

The minus sign of [tex] v_{1_{f}}[/tex] means that the ball 1 (100g) is moving in the negative x-direction after the collision.

Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

I hope it helps you!                  

The amount of time the ball stay in the air is (t)=3.04 Seconds

The best example that would help us understand and know what time are the clock and the calendar. This clock gives us the exact hour, minutes and seconds. The calendar tells us the exact day, month and year.

To calculate the amount of time the ball stay in the air, we are using the formula,

T=2Vsinθ/g

Here we are given,

V= The velocity of the football.=26 m/s.

θ= The angle that the football makes in the air= 35°

g= The acceleration due to gravity = 9.8 m/s².

We have to calculate, the amount of time the ball stay in the air = T

Now, we put the known values in the above equation,

T=2Vsinθ/g

Or, T= 2*26*sin (35°)/9.80

Or, T= 3.04 Seconds.

Thus, from the above calculation we can conclude that the amount of time the ball stay in the air is (t)=3.04 Seconds

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Explanation:

Objects at rest and in motion respond to the presence of an external unbalanced force by simple changing their magnitude of motion or position.

We have this knowledge from Newtons first law of motion "a body will remain in a state of rest or continue with uniform motion unless if it is acted upon by an external force".

Answer:

The force of friction acts in the direction opposite to the direction of motion. If friction would have been applied to the skier it would have resulted in a lower velocity and less kinetic energy.

Explanation:

Answer:

Image B

Explanation:

although I'm not exactly sure, i've recently gotten this question as well. but model B demonstrates the force- distance trade off because you can see how in that image them distance is increased in the force is decreased with the object being shorter. hopefully this helps in some way

Answer:

3400 m

Explanation:

Both lightning and thunder happen at the same time but one is faster than the other. The distance traveled by a sound can be calculated from its speed such that;

 speed = distance/time, hence, distance = speed x time.

For a thunder with 340 m/s speed and 10 seconds away from lightning, the distance between the thunder and the lightning can be calculated as;

distance = 340 m/s x 10 s = 3400 m

Answer:

Approximately [tex]1.69 \times 10^{-23}\; {\rm N}[/tex].

Explanation:

Look up the value of Coulomb's Constant: [tex]k \approx 8.988 \times 10^{-9}\; {\rm N \cdot m^{2} \cdot C^{-2}}[/tex].

Consider point charges of magnitude [tex]q_{1}[/tex] and [tex]q_{2}[/tex]. If the distance between these charges is [tex]r[/tex], the magnitude of the electrostatic force between them would be [tex](k\, q_{1}\, q_{2}) / (r^{2})[/tex].

In this question, the two [tex](+q)[/tex] charges are [tex]5\; {\rm cm}[/tex] and [tex]3\; {\rm cm}[/tex] away from the center [tex](-2\, q)[/tex] charge, respectively. Convert units to standard unit of distance (meters, [tex]{\rm m}[/tex]) and charge (coulombs, [tex]{\rm C}[/tex]):

[tex]q = 1.15 \; {\rm nC} = 1.15 \times 10^{-9}\; {\rm C}[/tex].

[tex]\begin{aligned} 5\; {\rm cm} = 5\; {\rm cm} \times \frac{1\; {\rm m}}{100\; {\rm cm}} = 0.05\; {\rm m} \end{aligned}[/tex].

[tex]\begin{aligned} 3\; {\rm cm} = 3\; {\rm cm} \times \frac{1\; {\rm m}}{100\; {\rm cm}} = 0.03\; {\rm m} \end{aligned}[/tex].

The magnitude of the electrostatic forces on the [tex](-2\, q)[/tex] charge would be:

[tex]\begin{aligned}\frac{k\, q_{1}\, q_{2}}{r^{2}} &\approx \frac{1}{(0.05\; {\rm m})^{2}} \\ &\quad \times (8.988 \times 10^{-9}\; {\rm N \cdot m^{2} \cdot C^{-2}})\\ &\quad \times ((-2) \, (1.15\times 10^{-9}\; {\rm C}))\, (1.15\times 10^{-9}\; {\rm C})) \\ &\approx 9.509\times 10^{-24}\; {\rm N}\end{aligned}[/tex].

[tex]\begin{aligned}\frac{k\, q_{1}\, q_{2}}{r^{2}} &\approx \frac{1}{(0.03\; {\rm m})^{2}} \\ &\quad \times (8.988 \times 10^{-9}\; {\rm N \cdot m^{2} \cdot C^{-2}})\\ &\quad \times ((-2) \, (1.15\times 10^{-9}\; {\rm C}))\, (1.15\times 10^{-9}\; {\rm C})) \\ &\approx 2.641\times 10^{-23}\; {\rm N}\end{aligned}[/tex].

Since the charges are of opposite sign, the [tex](-2\, q)[/tex] charge would attract both of the [tex](+q)[/tex] charges. In particular, the (approximately) [tex]9.509\times 10^{-24}\; {\rm N}[/tex] force would point to the left. The (approximately) [tex]2.641 \times 10^{-23}\; {\rm N}[/tex] force would point to the right.

As a result, the net force on the [tex](-2\, q)[/tex] charge would point to the right. The magnitude of the net force on this charge would be approximately [tex]2.641 \times 10^{-23}\; {\rm N} - 9.509\times 10^{-24}\; {\rm N} \approx 1.69 \times 10^{-23}\; {\rm N}[/tex].