A car’s velocity as a function of time is given by vx(t) = α + βt2, where α = 3.00 m/s and β = 0.100 m/s3. (a) Calculate the average acceleration for the time interval t = 0 to t = 5.00 s. (b) Calculate the instantaneous acceleration for t = 0 and t = 5.00 s. (c) Draw vx-t and ax-t graphs for the car’s motion between t = 0 and t = 5.00 s.

Answer:

They cannot be equal

Explanation:

Let 2 vector equal in magnitude and opposite direction. One of them is 1 and the other is -1

The sum of their magnitudes is 1 + 1 = 2

The magnitude of their sum of 2 vector is 0 since the 2 are in opposite directions, they cancel out each other and their final magnitude is thus 0.

So the sum of the magnitudes of two vectors cannot be equal to the magnitude of the sum of the same two vectors.

Answer:

(a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

Explanation:

Given that,

Length of tube = 11.0 cm

(a). We need to calculate the frequency of this standing wave

Using formula of fundamental frequency

[tex]f_{1}=\dfrac{v}{4l}[/tex]

Put the value into the formula

[tex]f_{1}=\dfrac{344}{4\times0.11}[/tex]

[tex]f_{1}=781.81\ Hz[/tex]

[tex]f_{1}=0.782\ kHz[/tex]

(b). If the test tube is half filled with water

When the tube is half filled the effective length of the tube is halved

We need to calculate the frequency

Using formula of fundamental frequency of the fundamental standing wave in the air

[tex]f_{1}=\dfrac{v}{4(\dfrac{L}{2})}[/tex]

Put the value into the formula

[tex]f_{1}=\dfrac{344}{4\times\dfrac{0.11}{2}}[/tex]

[tex]f_{1}=1563.63\ Hz[/tex]

[tex]f_{1}=1.563\ kHz[/tex]

Hence, (a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

When measuring weight on a scale that is accurate to the nearest 0.1

pound, the real limits for the weight of 145 pounds is 144.9- 145.1

This is an instrument which is used to measure the weight of objects. There

may be differences in the measurement as a result of air interference and

other factors.

We were told that the accuracy is to the nearest 0.1 pound which means

= 145± 0.1

= (145-0.1) - (145+0.1)

= 144.9 - 145.1

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The real limits for the weight of 145 pounds on a scale accurate to the nearest 0.1 pound are 144.95 to 145.05 pounds.

When measuring weight on a scale that is accurate to the nearest 0.1 pound, the real limits for the weight of 145 pounds would be 144.95 to 145.05 pounds. This is because the scale is accurate to the tenths place, meaning it can measure weights up to one decimal place. In this case, the weight of 145 pounds would be rounded to 145.0 pounds. Therefore, the real limits would be 144.95 pounds (145.0 - 0.05) to 145.05 pounds (145.0 + 0.05).

Final answer:

Using Bernoulli's equation and the continuity equation, we find that as the diameter of a pipe decreases, the velocity of the water increases, and to conserve the total mechanical energy, gauge pressure in the constricted segment decreases.

Explanation:

The situation described in the question can be analyzed using the principle of conservation of energy, specifically Bernoulli's equation for incompressible fluid flow. We are given that water flows at a speed of 5.9 m/s through a horizontal pipe of diameter 3.1 cm, and the gauge pressure is 1.5 atm. With the constriction of the pipe's diameter to 2.1 cm, we want to find the new gauge pressure of the water.

According to Bernoulli's equation, the total mechanical energy per unit volume is the same at all points along the streamline, i.e.,

[tex]P + \(\frac{1}{2}\)\(\rho\)v^2 + \(\rho\)gh = constant[/tex]

where P is the pressure, \(\rho\) is the density of the fluid, v is the flow velocity, g is acceleration due to gravity, and h is the elevation. For the situation described, h remains constant (horizontal pipe), and there is no change in the gravitational potential energy. Therefore, we can simplify the equation to:

[tex]P1 + \(\frac{1}{2}\)\(\rho\)v1^2 = P2 + \(\frac{1}{2}\)\(\rho\)v2^2[/tex]

Since the fluid is incompressible and the flow rate must be conserved, the velocity v2 in the constricted segment is determined by the continuity equation:

[tex]A_1v_1 = A_2v_2[/tex]

To solve this problem, we'll use the principle of continuity, which states that the product of cross-sectional area and velocity is constant for an incompressible fluid flowing through a tube. Mathematically, it can be expressed as:

Step 1: Find  [tex]\( v_2 \)[/tex]

Using the continuity equation:

[tex]\[ A_1 V_1 = A_2 V_2 \][/tex]

We can find [tex]\( v_2 \)[/tex] as:

[tex]\[ V_2 = \frac{\frac{\pi D_1^2}{4}}{\frac{\pi D_2^2}{4}} \times V_1 \]\[ V_2 = \frac{D_1^2}{D_2^2} \times V_1 \][/tex]

Since [tex]\( A = \frac{\pi D^2}{4} \)[/tex], we can rewrite the equation as:

[tex]\[ V_2 = \frac{\frac{\pi D_1^2}{4}}{\frac{\pi D_2^2}{4}} \times V_1 \]\[ V_2 = \frac{D_1^2}{D_2^2} \times V_1 \][/tex]

Now, let's plug in the values:

[tex]\[ V_2 = \frac{(0.031 \, \text{m})^2}{(0.021 \, \text{m})^2} \times 5.9 \, \text{m/s} \]\[ V_2 = \frac{0.000961 \, \text{m}^2}{0.000441 \, \text{m}^2} \times 5.9 \, \text{m/s} \]\[ V_2 = \frac{0.000961}{0.000441} \times 5.9 \, \text{m/s} \]\[ V_2 \approx 12.84 \, \text{m/s} \][/tex]

Step 2: Calculate   [tex]\( P_2 \)[/tex]

We'll use Bernoulli's equation, which states that the sum of pressure energy, kinetic energy, and potential energy per unit volume is constant along any streamline of flow. In this case, we'll ignore changes in height (assuming horizontal flow), and the equation simplifies to:

[tex]\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \][/tex]

Where:

- [tex]\( P_1 \)[/tex] is the initial pressure

-  [tex]\( P_2 \)[/tex] is the pressure in the constricted segment

- [tex]\( \rho \)[/tex]is the density of the fluid (we'll assume water,[tex]\( \rho = 1000 \, \text{kg/m}^3 \))[/tex]

-[tex]\( v_1 \) and \( v_2 \)[/tex] are initial and final velocities, respectively.

We rearrange this equation to solve for  [tex]\( P_2 \)[/tex]:

[tex]\[ P_2 = P_1 + \frac{1}{2} \rho v_1^2 - \frac{1}{2} \rho v_2^2 \][/tex]

Now, let's plug in the values:

[tex]\[ P_2 = P_1 + \frac{1}{2} \times 1000 \times (5.9)^2 - \frac{1}{2} \times 1000 \times (12.84)^2 \]\[ P_2 = P_1 + 17421.5 - 82626.24 \]\[ P_2 = P_1 - 65204.74 \][/tex]

Step 3: Convert  [tex]\( P_2 \)[/tex] to atmospheres (gauge pressure)

To express the pressure in atmospheres and considering atmospheric pressure as the baseline, we need to subtract the atmospheric pressure from [tex]\( P_2 \)[/tex]. Assuming atmospheric pressure is [tex]\( 101.3 \, \text{kPa} \) (or \( 101300 \, \text{Pa} \))[/tex]:

[tex]\[ P_{\text{gauge}} = \frac{P_2 - \text{atmospheric pressure}}{\text{atmospheric pressure}} \]\[ P_{\text{gauge}} = \frac{-65204.74 - 101300}{101300} \]\[ P_{\text{gauge}} \approx -1.64 \][/tex]

So, the gauge pressure in the constricted segment is approximately [tex]\( -1.64 \)[/tex] atmospheres.

Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = [tex]v_{oy}[/tex] t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = [tex]v_{oy}[/tex] t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

The minimum muzzle velocity of a cannon to clear a 25.0 m tall cliff from 60.0 m away involves using projectile motion equations to find the initial velocity components, while ensuring the shell reaches the required height and distance.

To solve for the minimum muzzle velocity for a cannon to shoot a shell past a 25.0 m cliff from 60.0 m away, we can use projectile motion equations. First, we separate the initial velocity into its horizontal (vx) and vertical (vy) components:

The shell must reach a height of at least 25.0 m to clear the cliff. We use the equation of motion in the vertical direction, considering the initial vertical velocity (vy) and the displacement (s = 25 m), to find the time (t) it takes to reach the top of the cliff. Then, using the horizontal velocity (vx), we can calculate how far the shell would travel horizontally in that time, ensuring that it covers at least 60.0 m. With the horizontal distance (d) and time (t) determined, we can calculate the shell's trajectory past the cliff edge.

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Answer:

The sequence is A,B,H,B,F

Explanation:

Answer:

Same momentum but speed is reduced 4 times.

Explanation:

According to the law of momentum conservation, the total momentum of the system before and after the collision is the same

Before the collision, the bigger car is at rest, only the 1st car of mass m is moving at speed v

p = mv

After the collision, both cars are connected and moving at speed V

P = (m + 3m)V = 4mV

These 2 momentum are equal

p = P

mv = 4mV

V = v/4

So after the collision, they have the same momentum but the speed decreased 4 times

Final answer:

After the collision of a moving car with mass m and a stationary car with mass 3m, the final velocity of the connected system is v/4, a quarter of the original car's velocity. The momentum of the system remains unchanged due to the conservation of momentum, but the total mechanical energy is typically not conserved as some is lost to other forms of energy.

Explanation:

Conservation of Momentum and Collision of Railway Cars

When a railroad car of mass m moving with an initial velocity v collides with and connects to another car at rest with a mass of 3m, the conservation of momentum principle applies. Given that neither external forces nor friction are significant, the total system momentum before and after the collision remains constant. The initial momentum of the system, which is mv (since the second car is at rest), must equal the final momentum of the now combined mass (4m) moving at the new velocity v'. To find the final velocity, we use the conservation of momentum equation:

mv = (m + 3m)v'

Which simplifies to:

v' = (mv)/(4m) = v/4

This equation shows that the speed of the combined railroad cars after the collision is a quarter of the initial speed of the moving car. While the speed of the connected car system is reduced compared to the initial speed of the car with mass m, the total momentum remains the same because the mass has increased fourfold.

Energy Conservation After the Collision

As for energy conservation, the total mechanical energy may not be conserved in an inelastic collision (like when cars connect after collision). Although the momentum is conserved, some kinetic energy is usually lost as other forms of energy such as heat, sound, or deformation of the vehicles.

Answer:

Explanation:

a ) Position of the plane with respect to observer ( origin ) is

R = 2.71 i + 1.37 j + 4.65 k

magnitude of R = √ (2.71² + 1.37² + 4.65²)

√(7.344 + 1.8769 + 21.6225)

=√30.8434

= 5.55 km

b ) angle with north

cos Ф = 1.37 / 5.55

= .2468

Ф = 75°

c )

R = 2.71 i + 1.37 j + 4.65 k

=

Answer:

Current, [tex]I=6.78\times 10^{-11}\ A[/tex]

Explanation:

In this case, we need to find the current in amperes if 1400 Na+ ions flow across a cell membrane in 3.3 μs.

Charge, [tex]q=1400\times 1.6\times 10^{-19}=2.24\times 10^{-16}\ C[/tex]

Time taken, [tex]t=3.3\ \mu s=3.3\times 10^{-6}\ s[/tex]

Let I is the current. It is given by total charge per unit time. It is given by :

[tex]I=\dfrac{q}{t}[/tex]

[tex]I=\dfrac{2.24\times 10^{-16}}{3.3\times 10^{-6}}[/tex]

[tex]I=6.78\times 10^{-11}\ A[/tex]

So, the current of [tex]6.78\times 10^{-11}\ A[/tex] is flowing across a cell membrane. Hence, this is the required solution.  

Answer:

Titanium.

Explanation:

Density of a metal is defined as the mass of the metal and its volume.

Volume of the metal = total volume- volume of initial water

= 27.4 - 25.3

= 2.1 ml

Mathematically,

Density = mass/volume

= 0.00945 kg/0.0000021 m3

= 4500 kg/m3.

The metal is Titanium.

Answer:

The question is incomplete.the complete question is giving below "College Physics 10+5 pts

Your bedroom has a rectangular shape, and you want to measure its area. You use a tape that is precise to 0.001 and find that the shortest wall in the room is 3.547 long. The tape, however, is too short to measure the length of the second wall, so you use a second tape, which is longer but only precise to 0.01 . You measure the second wall to be 4.79 long. Which of the following numbers is the most precise estimate that you can obtain from your measurements for the area of your bedroom?

A. 17.0m^2

B. 16.990m^2

C.16.99m^2

D.16.9m^2

E. .16.8m^2

b. If your bedroom has a circular shape, and its diameter measured 6.32 , which of the following numbers would be the most precise value for its area?

a)30 m^2

b) 31.4 m^2

c)31.37 m^2

d)31.371 m^2

Answers:

A. 17.0m²

B. 31.4m²

Explanation:

First, we determine the area of the room using the measured parameters I.e

Length=4.79 long

Breadth= 3.547 long

Hence area is calculated as

A=length *Breadth

A=4.79*3.547

A=16.99013m²

From the question the different measurements have different precision, the answer should match the number of significant figures of the least precisely known number in the calculation which is 3 significant digits.

Hence the correct answer will be 17.0m²

B. The are of a circle is express as

A=πd²/4

A=π(6.32)²/4

A=31.37069m²

The π and 4 are exact number they have no effect on the accuracy on the accuracy of the area. Hence we express the answer to the same significant figure as giving in the question..

The correct answer will be 31.4m²

The most precise estimate for the area of the rectangular bedroom is 17.004613 units. The most precise value for the area of the circular bedroom is 31.373968 units.

To find the most precise estimate for the area of your rectangular bedroom, you need to consider the precision of the measurements. The first wall measurement is 3.547, which has a precision of 0.001. The second wall measurement is 4.79, which has a precision of 0.01. Since the second measurement has a lower precision, it will determine the precision of the final estimate. Therefore, the most precise estimate for the area of your bedroom is obtained by multiplying the two measurements: 3.547 x 4.79 = 17.004613.

To find the most precise value for the area of your circular bedroom, you need to consider the precision of the diameter measurement. The diameter is measured as 6.32, which has a precision of 0.01. The formula to calculate the area of a circle is A = πr^2, where r is the radius (half of the diameter). So, the radius is 6.32/2 = 3.16. Therefore, the most precise value for the area of your circular bedroom is obtained by calculating the area using the radius: A = π(3.16)^2 = 31.373968.

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Answer:f1 = 7.90Hz, f2= 15.811Hz, f3 = 23.71Hz.

Explanation: length of string = 10m, mass of string = 100g = 0.1kg, T= 250N

We need the velocity of sound wave in a string when plucked with a tension T, this is given below as

v = √T/u

Where u = mass /length = 0.1/ 10 = 0.01kg/m

Hence v = √250/0.01, v = √25,000 = 158.11

a) at the lowest frequency.

At the lowest frequency, the length of string is related to the wavelength with the formulae below

L = λ/2, λ= 2L.

λ = 2 * 10

λ = 20m.

But v = fλ where v = 158.11m/s and λ= 20m

f = v/ λ

f = 158.11/ 20

f = 7.90Hz.

b) at the first frequency.

The length of string and wavelength for this case is

L = λ.

Hence λ = 10m

v = 158.11m/s

v= fλ

f = v/λ

f = 158.11/10

f = 15.811Hz

c) at third frequency

The length of string is related to the wavelength of sound with the formulae below

L =3λ/2, hence λ = 2L /3

λ = 2 * 10 / 3

λ = 20/3

λ= 6.67m

v = fλ where v = 158.11m/s, λ= 6.67m

f = v/λ

f = 158.11/6.67

f = 23.71Hz.

Answer:

2.8 cm

Explanation:

[tex]y_1[/tex] = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

[tex]\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm[/tex]

Fringe width is also given by

[tex]\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}[/tex]

For second order

[tex]\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1[/tex]

Distance between two second order minima is given by

[tex]y_2=2\beta_2[/tex]

[tex]\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm[/tex]

The distance between the two second order minima is 2.8 cm

The distance between the two second-order minima is 2.80 cm.

The central maximum's width given is 1.40 cm (this is the distance between the first-order minima, m = ±1). So, the distance from the center to the first-order minimum (m = ±1) is 0.70 cm.

Using the formula for the first-order minimum (m = 1):

a sin(θ₁) = λ

We know that the distance to the first minima (y₁) with the screen distance (L) is given by:

tan(θ₁) ≈ sin(θ₁) = y₁ / L

so, for the first-order minima:

y₁ = Lλ / a

We have y₁ = 0.70 cm, L = 1.20 m. Solving for aλ:

a = Lλ / 0.007

Next, for the second-order minima (m = 2), we use:

y₂ = 2Lλ / a

Thus, the distance between the second-order minima will be:

2y₂ = 2 × 2Lλ / a = 2 × 1.40 cm = 2.80 cm

So, the distance between the two second-order minima is 2.80 cm.

Answer:

Short circuit

Explanation:

In an ideal inductor circuit with constant current and voltage, it implies that the voltage drop in the circuit is zero (0).

Also, In circuit analysis, a short circuit is defined as a connection between two nodes that forces them to be at the same voltage.

In an ideal short circuit, this means there is no resistance and thus no voltage drop across the connection. That is voltage drop is zero (0).

Therefore, the circuit element is short circuit.

For circuits with constant currents and voltages, an ideal inductor is equivalent to a short circuit or a piece of wire with no resistance, as it does not affect the circuit voltage or resistance.

An ideal inductor in a circuit with constant currents and voltages acts as if it is effectively a wire with no resistance, since an ideal inductor does not dissipate energy when the current is constant. However, if we consider the energy transfer in a circuit with an inductor and another circuit element, often referred to as a 'black box', which could be a resistor, we notice energy is exchanged only when there is a change in current. Using the lumped circuit approximation, the inductor's magnetic fields are assumed to be completely internal, meaning it only interacts with other components via the current flowing through the wires, not by overlapping magnetic fields in space.

Given the Kirchhoff's voltage law, which states that the sum of the voltage drops in a closed loop must be zero, an inductor with a constant current will have a voltage drop that is zero, making it equivalent to a short circuit in terms of its impact on the voltage in the circuit. Therefore, for circuits with constant currents and voltages, an ideal inductor is equivalent to a short circuit or a piece of wire with no resistance, as it does not contribute any voltage drop or generate heat like a resistor would.

Answer:

According to the instructions given, only options a and b are correct.

That is,

C = (5/3) - (4D/3)

C = 1 – (5D/2)

D= -4/7

C= 17/7

Explanation:

3C + 4D = 5 and 2C + 5D = 2

So, following the instructions from the question,

1) we'll pick the variable with the smallest coefficient and isolate it.

In eqn 1, C has the smallest coefficient,

3C = 5 - 4D (isolated!)

In eqn 2, C still has the smallest coefficient,

2C = 2 - 5D

2) Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient.

For eqn 1,

3C = 5 - 4D, divide through by the coefficient of C,

C = (5/3) - (4D/3)

This matches option a perfectly.

For eqn 2,

2C = 2 - 5D, divide through by the coefficient of C,

C = (2/2) - (5D/2) = 1 - (5D/2)

This matches option b perfectly!

Further solving the equations now,

Since C = C

(5/3) - (4D/3) = 1 - (5D/2)

(5D/2) - (4D/3) = 1 - (5/3)

(15D - 8D)/6 = -2/3

7D/6 = -2/3

D = -4/7

Substituting this into one of the eqns for C

C = 1 - (5D/2)

C = 1 - (5/2)(-4/7) = 1 - (-10/7) = 1 + (10/7) = 17/7.

QED!

Answer:

[tex]2.1276595745\times 10^{-7}\ cm[/tex]

Explanation:

Volume of oil = 0.1 mL = 0.1 cm³

Area of oil = [tex]47\ m^2[/tex]

Converting to [tex]cm^2[/tex]

[tex]1\ m=10^2\ cm[/tex]

[tex]1\ m^2=10^4\ cm^2[/tex]

[tex]47\ m^2=47\times 10^4\ cm^2[/tex]

Volume is given by

[tex]V=At\\\Rightarrow t=\dfrac{V}{A}\\\Rightarrow t=\dfrac{0.1}{47\times 10^4}\\\Rightarrow t=2.1276595745\times 10^{-7}\ cm[/tex]

The thickness of the layer is [tex]2.1276595745\times 10^{-7}\ cm[/tex]

Hi, there is not much information about what do you need to do, but base on the C++ code you need to complete it to count the number of items in the array, using the instructions already written.

Answer:

#include <iostream>

using namespace std;

int main()

{

   int numList [] = {2,3,4,5,6,7,9,11,12,13,14,15,16};

   int count=0;

   for(int spot=0; spot < (sizeof(numList)/sizeof( numList[ 0 ])); ++spot)

   {

        cout << numList[spot];

        cout << "\n";

        ++count;

   }

   cout << "The number of items in the array is: ";

   cout << count;

   return 0;

}

Explanation:

To complete the program we need to finish the for statement, we want to know the number of items, we can get it by using this expression:  (sizeof(numList)/sizeof( numList[ 0 ])), sizeof() function returns the number of bytes occupied by an  array, therefore, the division between the number of bytes occupied for all the array (sizeof(numList)) by the number of bytes occupied for one item of the array (sizeof( numList[ 0])) equal the length of the array. While iterating for the array we are increasing the variable count that at the end contains the result that we print using the expression cout << "The number of items in the array is: "

Final answer:

The force of water on a rectangular dam is given by the formula F = pgh²L/2, where p is the density of water, h is the depth at the dam, and L is the length of the dam. Using the given values, we can calculate the force of the water on the dam to be approximately 1,130,946 lb.

Explanation:

The force of water on a rectangular dam is given by the formula F = pgh²L/2, where p is the density of water, h is the depth at the dam, and L is the length of the dam. We are given the values p = 62.4 lb/ft³, h = 35 ft, and L = 101 ft.

Substituting these values into the formula, we have F = (62.4 lb/ft³) * (35 ft)² * (101 ft) / 2.

Simplifying the expression, we get F = 1,130,946 lb. Therefore, the force of the water on the dam is approximately 1,130,946 lb.

Answer:

6.38 V

Explanation:

Note: Since the capacitors are identical, The same Charge flows through them.

For the first capacitor,

The Energy stored in a capacitor is given as

E = 1/2qv ............... Equation 1

Where E = Energy stored by the capacitor, q = charge in the capacitor v = Voltage.

make q the subject of the equation

q = 2E/v ................ Equation 2

Given: E = 8 J, v = 51 v.

Substitute into equation 2

q = 8/51

q = 0.3137 C.

For the second capacitor,

v = 2E/q ................... Equation 3

Given: q = 0.3137 C, E = 2 J.

Substitute into equation 3

v = 2/0.3137

v = 6.38 V

Hence the voltage across the second capacitor = 6.38 V

Answer:

trajectory

Explanation:

according to issac newton plants are important

Answer:

No.

Explanation:

Through uniform We can presume you mean constant in both  that is in magnitude as well as in direction ⠀

When you apply a steady, non-zero acceleration to a resting body, the velocity can be   parallel to the acceleration and not always perpendicular at any time later.

We can do the same for an object with an initial velocity in a direction which is different than that of acceleration, and the direction of its velocity will asymptotically approach to that of acceleration over time.

In both cases  the motion of a object undergoing a non-zero persistent acceleration can not be uniform That is the concept of acceleration: the velocity change over time.

The related concepts to solve this problem is the Glide Ratio. This can be defined as the product between the height of fall and the lift-to-drag ratio. Mathematically, this expression can be written as,

[tex]R = h (\frac{L}{D})_{max}[/tex]

Replacing,

[tex]R = 5000ft (7.7)[/tex]

[tex]R = 38500ft[/tex]

Converting this units to miles.

[tex]R = 38500ft (\frac{1mile}{5280ft})[/tex]

[tex]R = 7.2916miles[/tex]

Therefore the glide in terms of distance measured along the ground is 7.2916miles

The far that can it glide in terms of distance measured along the ground is 38,500 ft.

Given that,

Based on the above information, the calculation is as follows:

[tex]\frac{lift}{drag} = \frac{Distance}{height}\\\\7.7 = Distance \div 5500\\\\[/tex]

So, the distance is 38,500 ft.

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Answer:

72.3 MeV

Explanation:

E = Total energy of muon = 178 MeV

[tex]E_0[/tex] = Rest energy of a muon = 105.7 MeV

Kinetic energy of the muon at the surface of the Earth is given by the difference of the total energy and the rest energy of the muon

[tex]K=E-E_0\\\Rightarrow K=178-105.7\\\Rightarrow K=72.3\ MeV[/tex]

The kinetic energy of a muon at the surface is 72.3 MeV

Answer:

The velocity of the object 1 s after it is dropped is -9.8 m/s.

Explanation:

Hi there!

The instantaneous velocity is defined by the change in height over a very small time. Mathematically, it is expressed as the derivative of the function h(t):

instantaneous velocity = dh/dt

dh/dt = h´(t) = -2 · 4.9 · t

h´(t) = -9.8 · t

Now we have to eveluate the function h´(t) at t = 1 s:

h´(1) = -9.8 · (1) = -9.8

The velocity of the object 1 s after it is dropped is -9.8 m/s.

Final answer:

The instantaneous velocity of the object one second after it is dropped is 9.8 m/s, moving downwards.

Explanation:

The question asks for the instantaneous velocity of a dropped object after a specific time has elapsed, which is a physics concept related to mechanics and motion. Given the height equation h(t) = -4.9t2 + 136, the instantaneous velocity at time t can be found by taking the derivative of the height equation with respect to time, which gives us the velocity as a function of time v(t) = dh/dt. At t = 1 second, the derivative of the height equation is v(1) = -9.8(1) = -9.8 m/s. The negative sign indicates that the object is moving downwards. However, when we speak about the speed or magnitude of the velocity (instantaneous velocity), we typically refer to the positive value, which would be 9.8 m/s.

The stuntwoman should place the foam mats approximately 37 meters south of the helicopter's position. Calculations are based on her dropping from a height of 30.0 m with an initial horizontal velocity of 15 m/s, taking into account gravitational acceleration and ignoring air resistance.

To solve this problem, we need to calculate two things: how long the stuntwoman is in the air and how far she will travel horizontally during this time. Given the constant velocity components are 10.0 m/s upward and 15.0 m/s horizontal towards the south and ignoring air resistance, we'll tackle part (a) of the question first.

Part (a) - Determining the landing spot of the stuntwoman

Firstly, we acknowledge that the upward component of the helicopter's velocity will momentarily counteract gravity for the stuntwoman, but since air resistance is ignored, this effect is instantaneously null once she begins her descent. The primary considerations then are the height of 30.0 m and the horizontal component of 15.0 m/s.

To calculate the time (t) it takes for the stuntwoman to hit the ground, we use the equation for vertical motion under gravity:

h = 1/2gt^2

Where h is the height (30.0 m) and g is the acceleration due to gravity (~9.8 m/s^2). Solving for t, we get:

t = sqrt((2*h)/g) = sqrt((2*30)/9.8) ≈ 2.47 s

With the time in air known, we calculate the horizontal displacement (d) using:

d = v*t

Where v is the horizontal velocity (15.0 m/s). Thus, d ≈ 15.0 m/s * 2.47 s ≈ 37.05 m.

This means the stuntwoman should place the foam mats around 37 meters south of the helicopter's position at the moment she drops.

Part (b) - Drawing the graphs

For simplicity, the explanation of graph drawing is summarized: The x-t graph will show a linear increase in displacement over time, illustrating constant velocity in the horizontal direction. The y-t graph will depict a parabola, indicating acceleration (deceleration up then acceleration down) due to gravity in the vertical component. The vx-t graph will be a horizontal line showing constant horizontal velocity, and the vy-t graph will start at a positive value, decrease to zero at the peak of her motion, and then increase negatively as she accelerates downwards.

Answer:

[tex]v=369.27\frac{m}{s}[/tex]

Explanation:

The speed of the waves in a string is related with the tension and mass per unit length of the string, as follows:

[tex]v=\sqrt\frac{T}{\mu}[/tex]

First, we calculate the mass per unit length:

[tex]\mu=\frac{m}{L}\\\mu=\frac{55*10^{-3}kg}{100m}\\\mu=5.5*10^{-4}\frac{kg}{m}[/tex]

Now, we calculate the speed of the waves:

[tex]v=\sqrt\frac{75N}{5.5*10^{-4}\frac{kg}{m}}\\v=369.27\frac{m}{s}[/tex]

Given values,

or,                   [tex]= 55\times 10^{-3} \ kg[/tex]

As we know,

→ [tex]v = \sqrt{\frac{T}{\mu} }[/tex]

or,

→ [tex]\mu = \frac{m}{L}[/tex]

By putting the values,

      [tex]= \frac{55\times 10^{-3}}{100}[/tex]

      [tex]= 5.5\times 10^{-4} \ kg/m[/tex]

hence,

The speed of wave:

→ [tex]v = \sqrt{\frac{75}{5.5\times 10^{-4}} }[/tex]

     [tex]= 369.27 \ m/s[/tex]

Thus the above response is correct.  

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To solve this problem we will apply the concepts related to the deformation of a body and the normal effort, from which we will obtain the area. From this area applied to the geometric concept of a circular bar we will find the diameter.

The deformation equation in a rod tells us that

[tex]\delta = \frac{PL}{AE}[/tex]

Here,

P = Load

L = Length

A = Cross-sectional area

E = Elastic Modulus

Rearranging the Area,

[tex]A = \frac{PL}{\delta E}[/tex]

Replacing we have that the area is,

[tex]A = \frac{(109*10^{3})(6.1)}{(10*10^{-3})(150*10^9)}[/tex]

[tex]A = 0.000443266m^2[/tex]

[tex]A = 44.32*10^{-6}m^2[/tex]

Using the geometric expression for the Area we have,

[tex]A = \frac{\pi}{4} d^2[/tex]

[tex]d = \sqrt{\frac{4A}{\pi}}[/tex]

[tex]d = \sqrt{\frac{4(44.32)}{\pi}}[/tex]

[tex]d = 7.51mm[/tex]

Therefore the smalles diameter rod is 7.51mm

Answer:

The average acceleration during the 6.0 s interval was -27 m/s².

Explanation:

Hi there!

The average acceleration is defined as the change in velocity over time:

a = Δv/t

Where:

a = acceleration.

Δv = change in velocity = final velocity - initial velocity

t = elapsed time

The change in velocity will be:

Δv = final velocity - initial velocity

Δv = -74 m/s - 87 m/s = -161 m/s

(notice the negative sign of the velocity that is in opposite direction to the direction considered positive)

Then the average acceleration will be:

a = Δv/t

a = -161 m/s / 6.0 s

a = -27 m/s²

The average acceleration during the 6.0 s interval was -27 m/s².

Answer:

Explanation:

Given

volume of Tank [tex]V=0.25\ m^3[/tex]

Specific gravity [tex]=2[/tex]

specific gravity is the defined as the ratio of density of fluid to the density of water

Density of water [tex]\rho _w=1000\ kg/m^3[/tex]

Density of Fluid [tex]\rho =2\times 1000=2000\ kg/m^3[/tex]

We know mass of a fluid is given by the product of density and volume

[tex]m=\rho \times V[/tex]

[tex]m=2000\times 0.25[/tex]

[tex]m=500\ kg[/tex]    

Answer:

f'  = 485 Hz

Explanation:

given,

Frequency of whistle,f = 500 Hz

speed of source, v_s = 10 m/s

Speed of observer, v_o - 20 m/s

speed of sound,v = 340 m/s

Apparent frequency heard = ?

Using Doppler's effect formula to find apparent frequency

[tex]f' = (\dfrac{v-v_0}{v-v_s})f[/tex]

[tex]f' = (\dfrac{340-20}{340-10})\times 500[/tex]

[tex]f' = 0.9696\times 500[/tex]

f'  = 485 Hz

Hence, the apparent frequency is equal to 485 Hz.

To determine the apparent frequency heard by the listener is equal to 545.45 Hz.

Given the following data:

To determine the apparent frequency heard by the listener, we would apply Doppler's effect of sound waves:

Mathematically, Doppler's effect of sound waves is given by the formula:

[tex]F_o = \frac{V \;+ \;V_o}{V\; - \;V_s} F[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]F_o = \frac{340 \;+ \;20}{340\; - \;10} \times 500\\\\F_o =\frac{360}{330} \times 500\\\\F_o =1.0909 \times 500[/tex]

Apparent frequency = 545.45 Hz.

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