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hoa [83]
3 years ago
5

If a car races around a 2.3 mile oval track 220 times, what is the displacement of the car in meters when it crosses the finish

line.
Physics
1 answer:
Wewaii [24]3 years ago
5 0

Answer: 0 miles.

Explanation:

First let's define what the displacement is:

The displacement is defined as the difference between the final position and the initial position.

So, if for example, you start at your house, then go to a store, and then go back to your house, your total displacement will be zero, because the initial position and the final position are the same.

Then when the car reaches the final line, it will be in the same place that it started (because this is an oval track, so the finish line connects with the starting point)

Then the total displacement is 0miles.

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A force of 100N is applied to an area of 100mm².what is the pressure exerted on the area in N/m².​
Dmitry_Shevchenko [17]

Answer:

P = 1000000[Pa] = 1000 [kPa]

Explanation:

To solve this problem we must use the definition of pressure, which is equal to the relationship of force over area.

P=F/A

where:

P = pressure [Pa] (units of pascals)

F = force = 100 [N]

A = area = 100 [mm²]

But first we must convert the units from square millimeters to square meters.

A=100[mm^{2}]*\frac{1^{2} m^{2} }{1000^{2}mm^{2}  } =0.0001[m^{2} ]

Now replacing:

P=100/0.0001\\P=1000000[Pa]

3 0
3 years ago
Blue light of wavelength λ passes through a single slit of width d and forms a diffraction pattern on a screen. If we replace th
ololo11 [35]

Answer:

We can retain the original diffraction pattern if we change the slit width to d) 2d.

Explanation:

The diffraction pattern of a single slit has a bright central maximum and dimmer maxima on either side. We will retain the original diffraction pattern on a screen if the relative spacing of the minimum or maximum of intensity remains the same when changing the wavelength and the slit width simultaneously.

Using the following parameters: <em>y</em> for the distance from the center of the bright maximum to a place of minimum intensity, <em>m</em> for the order of the minimum, <em>λ </em>for the wavelength, <em>D </em>for the distance from the slit to the screen where we see the pattern and <em>d </em>for the slit width. The distance from the center to a minimum of intensity can be calculated with:

                                                    y\approx\frac{m\lambda D}{d}

From the above expression we see that if we replace the blue light of wavelength λ by red light of wavelength 2λ in order to retain the original diffraction pattern we need to change the slit width to 2d:

<em>                                                 </em>y\approx\frac{m\lambda D}{d} =\frac{m2\lambda D}{2d}

7 0
3 years ago
Which of the following is responsible for the overall integration of the autonomic nervous system (ANS)?
Dafna1 [17]

Answer:

D. hypothalamus

Explanation:

The autonomic nervous system (ANS) is a component of the peripheral nervous system that is in charge of the involuntary functions, this means it regulates the flight-or-fight response, heart and respiratory rate, sexual arousal, urination, etc.<em> The hypothalamus is the integrator of these functions, it receives the information from the nervous system and responds to it.</em>

I hope you find this information useful and interesting! Good luck!

3 0
3 years ago
1. Is The larger the mass the greater the acceleration is on a smaller object?
zheka24 [161]

Answer:No,a smaller object would have more acceleration because it would take less force to move.

7 0
3 years ago
In Trial II, the same spring is used as in Trial I. Let us use this information to find the suspended mass in Trial II. Use 0.51
Masteriza [31]

Answer:

M_2=0.79kg

Explanation:

From the question we are told that:

Period T=0.517s

Trial 1

Spring constant \mu=117N/m

Period T_1=0.37

Mass m=0.400kg

Trial 2

Period T_2=0.52

Generally the equation for Spring Constant  is mathematically given by

\mu=\frac{4 \pi^2 M}{T^2}

Since

\mu _1=\mu_2

Therefore

\frac{4 \pi^2 M_1}{T_1^2}=\frac{4 \pi^2 M_2}{T_2^2}

M_2=M_1*(\frac{T_2}{T_1})^2

M_2=0.400*(\frac{0.52}{0.37}})^2

M_2=0.79kg

5 0
3 years ago
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