A car is moving along a straight road. The graph shows how its velocity changes from start of the journey to the end of the journey.

Here's the fact that you need to know:

"The AREA under a speed/time graph is the distance traveled during the time covered by the graph."

Now ... What's the area of that nice trapezoid on the graph ?

Hint: Area of a trapezoid is

(1/2) • (height) • (base-1 + base-2)

Answer:

Oxygen has two vacancies in its electron shell, but carbon has four vacancies.

Explanation:

Oxygen has vacant space for two electrons in its outermost shell. Hence, it can accommodate two electrons from two hydrogen atoms. This leads to the formation of the H₂O molecule. On the other hand, the Carbon atom has vacant space for four electrons. Hence, it can accommodate four electrons from four hydrogen atoms. This leads to the formation of the CH₄ molecule.

In this way the explanation can be provided by the following statement:

Oxygen has two vacancies in its electron shell, but carbon has four vacancies.

Answer:

-Atoms may share electrons in order to fill their electron shells.

- Hydrogen has one valence electron.

- Oxygen has two vacancies in its electron shell, but carbon has four vacancies.

Explanation:

just did it

Answer:

Explanation:

The mass of an object given only the force acting on it and it's acceleration can be found by using the formula

\(m = \frac{f}{a} \\ \)

f is the force in N

a is the acceleration in m/s²

We have

\(m = \frac{20}{2} = 10 \\ \)

We have the final answer as

Hope this helps you

To complete the configuration as described, follow the steps below:

1. Assign IP addresses to the serial interfaces:

  - On RouterA's S0/0/0 interface, assign the first address on the subnet: 192.168.2.25 with a 30-bit mask (255.255.255.252).

  - On RouterB's S0/0/1 interface, assign the last address on the subnet: 192.168.2.26 with a 30-bit mask (255.255.255.252).

 Here's an example of how you can configure the IP addresses on both routers:   ```plaintext

  RouterA# configure terminal

  RouterA(config)# interface Serial0/0/0

  RouterA(config-if)# ip address 192.168.2.25 255.255.255.252

  RouterA(config-if)# no shutdown

  RouterA(config-if)# exit

  RouterA(config)# exit

RouterB# configure terminal

  RouterB(config)# interface Serial0/0/1

  RouterB(config-if)# ip address 192.168.2.26 255.255.255.252

  RouterB(config-if)# no shutdown

  RouterB(config-if)# exit

  RouterB(config)# exit

  ```2. Use the `show controllers` command on RouterA to identify the DCE (Data Communications Equipment) device. The DCE device is the one that provides clocking on the serial interface.

```plaintext

  RouterA# show controllers Serial0/0/0

  ```   Look for the line that indicates the DCE status, which should mention something like "Clock rate 9600" or "DCE V.35."

3. Configure the clock rate on the DCE interface:

  - Assuming RouterA is the DCE device based on the previous step, configure the clock rate of 9600 on RouterA's Serial0/0/0 interface.

 ```plaintext

  RouterA# configure terminal

  RouterA(config)# interface Serial0/0/0

  RouterA(config-if)# clock rate 9600

  RouterA(config-if)# exit

  RouterA(config)# exit

  ```4. Bring both interfaces up:

  - On RouterA:

  ```plaintext

  RouterA# configure terminal

  RouterA(config)# interface Serial0/0/0

  RouterA(config-if)# no shutdown

  RouterA(config-if)# exit

  RouterA(config)# exit

  ``` - On RouterB:

```plaintext

  RouterB# configure terminal

  RouterB(config)# interface Serial0/0/1

  RouterB(config-if)# no shutdown

  RouterB(config-if)# exit

  RouterB(config)# exit

  ```5. Save the changes on both routers:

  - On RouterA:

  ```plaintext

  RouterA# write memory

  ``` - On RouterB:

```plaintext

  RouterB# write memory

  ```

By following these steps, you should have completed the configuration of the serial link between the two routers in a back-to-back configuration.

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Answer:

Explanation:

Hell im good dude

Answer:

most likely just 1

Explanation:

I sucked at sig figs in middle school, in fact it was the lowest test grade i ever received, but it is relatively simple.

because there is only one sig fig involved in each number, there can't be multiple sig figs here. 3000/500=6

500/3000=2(round up from 1.66666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666) you get the point.

The nonconservative work done on the surfer is 2845.5 J.

We can use the work-energy theorem to solve this problem. The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy. In this case, we can calculate the initial and final kinetic energies of the surfer and find the difference, which will give us the net work done.

The initial kinetic energy of the surfer is:

\(K_i = (1/2) * m * v_i^2\)

\(K_i = (1/2) * 73.2 kg * (1.44 m/s)^2\)

K_i = 75.7 J

The final kinetic energy of the surfer is:

\(K_f = (1/2) * m * v_f^2\)

\(K_f = (1/2) * 73.2 kg * (8.89 m/s)^2\)

K_f = 2921.2 J

The change in kinetic energy is:

ΔK = K_f - K_i

ΔK = 2921.2 J - 75.7 J

ΔK = 2845.5 J

According to the work-energy theorem, this change in kinetic energy must be equal to the net work done on the surfer. Therefore, the nonconservative work done on the surfer is:

W_nc = ΔK

W_nc = 2845.5 J

So, the nonconservative work done on the surfer is 2845.5 J.

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Answer:

2.8 radians

Explanation:

\(s = r \beta \\ 1.4 = 0.50 \beta \\ \beta = 2.8 \: rad\)

Answer:

asdadsadada

Explanation:

Answer:

Reflected

Explanation:

During the process of reflection whatever that is sent forth e.g light comes back (reflects)

Answer:

transmitted

Explanation:

took quiz

Answer:

d. 0.48

Explanation:

Answer:

An infrared camera – also called IR camera, thermal means heat it can track your heat camera or thermal camera – is a measuring  by instrument it means its a measuring tool

used for non-contact measurements of the surface temperature of objects.

Explanation:

kids are oof

Answer:

34

Explanation:

The corresponding distance traveled by the orbiter is 1,728.05 feet. Given that the deceleration in feet per second squared during the time that the chute is deployed is −0.0003v2 (speed v in feet per second), the corresponding distance traveled by the orbiter can be found using the equation below;

The equation is:ds=∫v₁ᵛ₂dt Let's put in the values we have;v₁ = 200 mi/hr = 200 * 1.467 = 293.4 ft/s, v₂ = 35 mi/hr = 35 * 1.467 = 51.345 ft/s

∫ds= ∫v₁ᵛ₂ dt = ∫v₁ᵛ₂ (-dv/0.0003v²) = [-1/0.0003 ∫(1/v²)dv] from v₁ to v₂= [-1/0.0003 (-1/v)] from v₁ to v₂ = [1/0.0003 (1/v₁ - 1/v₂)]

The distance covered by the shuttle during the time that the chute is deployed is given by 1/0.0003 (1/v₁ - 1/v₂).∴ d = 1/0.0003 (1/293.4 - 1/51.345) = 1,728.05 feet. Therefore, the corresponding distance traveled by the orbiter is 1,728.05 feet.

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The number of seconds in the 40 second interval starting at t = 0 during which the voltage is below 0.21m V is: 19.06 - 7.03 = 12.03 s = 12.97 (approx.) Thus, the correct option is (c) 12.97.

The voltage V, in an electric circuit is measured in millivolts (mV) and is given by the formula V=0.2

sin0.1π(t -0.5)+0.3, where t is the time in seconds from the start of an experiment. We have to use the graph of the function to estimate how many seconds in the 40 second interval starting at t = 0 during which the voltage is below 0.21mV.

Graph of the given function is shown below:

Graph of the given function

As per the graph, it is observed that the voltage is below 0.21 mV from 7.03 s to 19.06 s.

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Answer:

The correct answer is A. Thermal energy (heat) is defined as the sum of all the kinetic energies of all the particles in an object.

Answer:

accelaration= 1.654 m/s²

Explanation:

accelaration= force/mass

a= \(\frac{F}{m}\)

a= \(\frac{1500}{907}\)

a= 1.654 m/s²

Answer:

both

Explanation:

really you don't know something from kindergarden

Answer:

The concentration of hydrogen ion at pH is equal to 2 :\(= [H^+]=0.01 mol/L\)

The concentration of hydrogen ion at pH is equal to 6 : \([H^+]'=0.000001 mol/L\)

There are 0.009999 more moles of  \(H^+\) ions in a solution at a pH = 2 than in a solution at a pH = 6.

Explanation:

The pH of the solution is the negative logarithm of hydrogen ion concentration in an aqueous solution.

\(pH=-\log [H^+]\)

The hydrogen ion concentration at pH is equal to 2 = [H^+]

\(2=-\log [H^+]\\\)

\([H^+]=10^{-2}M= 0.01 M=0.01 mol/L\)

The hydrogen ion concentration at pH is equal to 6 = [H^+]

\(6=-\log [H^+]\\\\\)

\([H^+]=10^{-6}M= 0.000001 M= 0.000001 mol/L\)

Concentration of hydrogen ion at pH is equal to 2 =\([H^+]=0.01 mol/L\)

Concentration of hydrogen ion at pH is equal to 6 = \([H^+]'=0.000001 mol/L\)

The difference between hydrogen ion concentration at pH 2 and pH 6 :

\(= [H^+]-[H^+]' = 0.01 mol/L- 0.000001 mol/L = 0.009999 mol/L\)

Moles of hydrogen ion in 0.009999 mol/L solution :

\(=0.009999 mol/L\times 1 L=0.009999 mol\)

There are 0.009999 more moles of  \(H^+\) ions in a solution at a pH = 2 than in a solution at a pH = 6.

The minimum amount that the spring must be compressed if the car is to stay on the track is 8.373 Xa.

let A be the starting point

Let B be the loop's top.

Given that K=31,000 N/m, m=840 kg, and r=6.3 m

Early Compression

Xa = ?

Final (spring is in natural state then) (spring is in natural state then)

Xb = 0

Please remember to find the minimal compression now. When the roller coaster is almost at separation, which means the normal force is equal to zero, we need to determine the velocity at the top of the hoop.

total force C = Mac

mVb = mVb²/R

We now use the work energy technique.

-(1/2) * k* (Xb²-Xa²) - mg(2r) = m * (Vb²-Va²) * (1/2)

Va = 0\s

Xb = 0\s-(1/2) * k * (-Xa^2) - mg(2r)

= (1/2)*(m*Vb*2) (1/2)*(k*Xa*2) - mg(2r)

= m * Vb * (1/2)

Isolate the deformation

Using algebra, Xa of the spring equals (1/2) * k * (Xa2) = (1/2) * m * Vb2 + mg (2r)

Substituting m/k * (gR) + m/k * g* into Vb²/ Xa²

2.89 m for 4r Xa²

= 8.373 Xa.

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To find the moments and mass of the triangular lamina, we first need to determine the density function rho(x, y). Once we have the density function, we can calculate the moments mx and my and the mass m.

Given that the vertices of the triangular lamina are (0,0), (0,3), and (6,3), we can see that the base of the triangle is 6 units long, and its height is 3 units. Therefore, the area of the triangular lamina is (1/2) * base * height = (1/2) * 6 * 3

= 9 square units.

To find the density function rho(x, y), we need more information. The density function represents how the mass is distributed over the lamina. Without this information, we cannot proceed with finding the moments and mass accurately.


To find the moments mx and my, we need to integrate the product of the density function rho(x, y) and the coordinates x and y, respectively, over the triangular lamina. The moments mx and my can be calculated using the following formulas:

mx = ∫∫ x * rho(x, y) dA
my = ∫∫ y * rho(x, y) dA

where dA represents the infinitesimal area element.

To find the mass m, we need to integrate the density function rho(x, y) over the triangular lamina. The mass m can be calculated using the following formula:

m = ∫∫ rho(x, y) dA

However, since we don't have the density function rho(x, y) given, we cannot determine the moments mx and my or the mass m accurately.


In order to find the moments mx and my and the mass m of the triangular lamina, we need to know the density function rho(x, y). Without this information, we cannot proceed with the calculations. It is important to have all the necessary data and information to accurately solve problems involving moments and mass.

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If the time period of the pendulum in earth is 1 seconds, then the length  L is 0.24 m. If length in mars is L/2 and acceleration due to gravity is 1/3rd of that of earth, the time period of the pendulum on mars is 1.216 seconds.

The time period of oscillation of a pendulum is the time taken to complete oscillation. It is related to the length l, and acceleration due to gravity g as:

T = 2π √l/g

T for earth = 1 s

g for earth = 9.8 m/s²

then l = T/4π² × g

l = 1/4π²× 9.8 m/s²

 = 0.24 m

If in mars L = L/2 = 0.24 /2 = 0.12 m

g = 9.8 m/s²/3 = 3.22m/s²

Then time period T = 2π × √0.12 m/3.22m/s²

T = 1.216 s.

Therefore, the time period of the Rover's pendulum in mars is 1.216 seconds.

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Answer:

The answer is "a, c and b"

Explanation:

Answer: (a) The length of the wire with a diameter of 0.15m and 15-ohm resistance is 15.77 m

(b) There is no voltage drop in the resistor but there is a voltage drop of 2 V in the wire

Explanation:

The expression for resistance is R = xL/A,  where R is resistance, x is Resistivity, L is the length of the wire and A is the area

Rearranging the equation we get, L = RA/x

The area of a cross-section is A = \(\pi\)(D/2)^2

D in our case is equal to 0.15 x 10^-3 m

x in our case is equal to 1.68 x 10^-8 omega m

Putting all these values in the equation we get L to be equal to 15.77 m

For the second part using our equation V = IR, we can figure out there will be no voltage drop across the resistor but there will be a voltage drop across the wire by 2 V

That is due to the new resistor being added in a series connection and using that information we can find out the current in the wire, as the current stays the same across resistors in a series connection we can find out the voltage in each resistor using our equation V = IR

Subtracting it from the initial voltage through the wire will give us the change i.e. of 2 V

To correct the circuit without removing the 7.0µF capacitor, the technician can add an additional 9.0µF capacitor in parallel.

To find the total capacitance when capacitors are connected in parallel, we simply add their individual capacitances.

Given that the circuit was mistakenly constructed with a 7.0µF capacitor instead of the required 16µF capacitor, the technician needs to add a capacitance of 16µF - 7.0µF = 9.0µF to correct the circuit.

By connecting this additional 9.0µF capacitor in parallel with the existing 7.0µF capacitor, the total capacitance of the circuit will be 7.0µF + 9.0µF = 16µF.

To correct the circuit without removing the 7.0µF capacitor, the technician can add an additional 9.0µF capacitor in parallel. This will result in a total capacitance of 16µF, meeting the required value for the circuit.

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Answer:

636.4 J

Explanation:

The potential energy between one of the charges at the corner of the square and the fifth identical charge is U = kq²/r where q = charge = +50 × 10⁻⁶ C  and r = distance from center of square. = √2 m (since the midpoint of the sides = 1 m, so the distance from the charge at the corner to the center is thus √(1² + 1²) = √2)

Since we have four charges, the additional potential energy to move the charge to the centre of the square is U' = 4U = 4kq²/r

U' = 4kq²/r

= 4 × 9 × 10⁹ Nm²/C² (+50 × 10⁻⁶ C)²/√2 m

= 900 Nm²/√2 m

= 636.4 J

The total external energy required is 636.4 J.

According to the question, a square of side a = 2m has 4 identical charges on the corners with charge Q = 50×10⁻⁶C.

A fifth identical charge is brought at the geometric center of the square. The geometric center is at the center of the diagonal:

\(r=\frac{a}{\sqrt{2} }=\sqrt{2}\;m\)

The potential energy is a state function which means that it depends on the initial and final position.

Now the energy required is equal to the change in potential energy

\(\Delta U=\frac{1}{4\pi\epsilon_o}\frac{4Q^2}{r}\\\\\Delta U=\frac{4\times9\times10^9\times(50\times10^{-6})^2}{\sqrt{2} } \\\\\Delta U=636.4\;J\)

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Answer:

yes   1000 j   as I posted on the same Q elsewhere

Explanation: