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bixtya [17]
3 years ago
8

First, record some data for this comparison in the table below.

Physics
1 answer:
kondor19780726 [428]3 years ago
5 0

Answer:

Record your measured values of displacement and velocity for times t = 8.0 seconds and t = 10.0 seconds in the columns below.

Next, use the measured displacement and velocity values at t = 7.0 seconds and t = 9.0 seconds to interpolate the values of displacement and velocity at t = 8.0 seconds.

Use the following formula to interpolate and extrapolate. Remember, x and y here represent values on the x and y axes of the graph. The x values will really be time and the y values will be either displacement (x) or velocity (vx).

Explanation:

Record your measured values of displacement and velocity for times t = 8.0 seconds and t = 10.0 seconds in the columns below.

Next, use the measured displacement and velocity values at t = 7.0 seconds and t = 9.0 seconds to interpolate the values of displacement and velocity at t = 8.0 seconds.

Use the following formula to interpolate and extrapolate. Remember, x and y here represent values on the x and y axes of the graph. The x values will really be time and the y values will be either displacement (x) or velocity (vx).

This is the answer

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A car with mass mc = 1490 kg is traveling west through an intersection at a magnitude of velocity of vc = 9.5 m/s when a truck o
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v= - 4.507 i - 2.363 j

Explanation:

 Given that

mc= 1490 kg

vc= 9.5 m/s ( - i)

mt=  1650 kg

vt = 6.4 m/s ( -j)

There is any external force so linear momentum will remain conserve.

Lets take final speed is v.

mc .vc + mt . vt = ( mc+mt) v

1490 x 9.5 ( - i) + 1650 x 6.4 ( -j) = ( 1490+1650) v

14,155 ( -i) + 10,560 ( - j) = 3140 v

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4 years ago
Light with a single wavelength falls on two slits separated by 0.510 mm. In the resulting interference pattern on a screen 2.24
scZoUnD [109]

Answer: 655.7 nm

Explanation:

Given

The slits are separated by d=0.510\ mm

Distance between slits and screen is D=2.24\ m

Adjacent bright fringes are \beta =2.88\ mm apart

Also, the distance between bright fringes is given by  

\Rightarrow \beta =\dfrac{\lambda D}{d}\quad [\lambda=\text{Wavelength of light}]\\\\\text{Insert the values}\\\\\Rightarrow 2.88\times 10^{-3}=\dfrac{\lambda \cdot 2.24}{0.510\times 10^{-3}}\\\\\Rightarrow \lambda =\dfrac{2.88\times 10^{-3}\times 0.510\times 10^{-3}}{2.24}\\\\\Rightarrow \lambda =0.6557\times 10^{-6}\ m\\\Rightarrow \lambda =655.7\ nm

7 0
3 years ago
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