what do the sound waves made by the bell carry to your ear

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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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!                  

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:

[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:

A supersonic aircraft flies at 3 km altitude at a speed of 1000 m/s on a standard day. How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer

Using the formula for calculating speed expressed as;

Speed = Distance/Time

Given;

Distance = 3km = 3000m

Speed = 1000m/s

Required

How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer (Time)

From the formula;

Time = Distance/speed

Time = 3000/1000

Time = 3seconds

Hence the sound of the aircraft is heard after 3 seconds

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)

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:

C

Explanation:

Some of the mass

Answer:

D. Mass was transformed into energy during the process.

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!

Don't forget to mark me as Brainliest.

Answer:

Explanation:

.... ...

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:

W = - 118.24 J (negative sign shows that work is done on piston)

Explanation:

First, we find the change in internal energy of the diatomic gas by using the following formula:

[tex]\Delta\ U = nC_{v}\Delta\ T[/tex]

where,

ΔU = Change in internal energy of gas = ?

n = no. of moles of gas = 0.0884 mole

Cv = Molar Specific Heat at constant volume = 5R/2 (for diatomic gases)

Cv = 5(8.314 J/mol.K)/2 = 20.785 J/mol.K

ΔT = Rise in Temperature = 18.8 K

Therefore,

[tex]\Delta\ U = (0.0884\ moles)(20.785\ J/mol.K)(18.8\ K)\\\Delta\ U = 34.54\ J[/tex]

Now, we can apply First Law of Thermodynamics as follows:

[tex]\Delta\ Q = \Delta\ U + W[/tex]

where,

ΔQ = Heat flow = - 83.7 J (negative sign due to outflow)

W = Work done = ?

Therefore,

[tex]-83.7\ J = 34.54\ J + W\\W = -83.7\ J - 34.54\ J\\[/tex]

W = - 118.24 J (negative sign shows that work is done on piston)

Answer:

-49.2

Explanation:

Trust me bro

Answer:

first is regular reflection and 2nd is irregular

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].

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:

10 Pa

Explanation:

Pressure is defined as force per unit area.

Mathematically, P= F/A where  P is pressure, F is force and A is area.

Force, F = 60 N

Area, A = 2*3 = 6m²

P = 60/6 = 10 Pa

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:

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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a = v² / R = (2 m/s)² / (0.1 m) = 40 m/s²

Answer:

C

Explanation:

GPS satellites in Earth's orbit constitute the most common way that images of space are captured.

I hope this helps! And this is the real answer.

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: 1.67 x 10^-7N

Explanation:

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

Answer:

false

Explanation:

ke sath hi hai j clin oncol biol to u h

Answer:

I don't know if its just me, but I don't see the statements so I'm not sure how to help answer your question.

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.