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oksian1 [2.3K]
2 years ago
7

Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally

100m north What distance did he cover? What was his displacement
Also every Kilometer is = to .6 cm
Find Displacement and Distance
Physics
1 answer:
Bess [88]2 years ago
3 0

Find Displacement and Distance

displacement ...

north is 700+400+100 =1200m n

south=1200m

1200-1200=0


east is 300+300=600m

west is 600m

600-600=0

back at dtart. displ zero


distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m  = 3600m


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gizmo_the_mogwai [7]

Answer:

longitudinal engineering strain = 624.16

true strain is 6.44

Explanation:

given data

diameter d1 = 0.5 mm

diameter d2 = 25 mm

to find out

longitudinal engineering and true strains

solution

we know both the volume is same

so

volume 1 = volume 2

A×L(1) = A×L(2)

( π/4 × d1² )×L(1) = ( π/4 × d2² )×L(2)

( π/4 × 0.5² )×L(1) = ( π/4 × 25² )×L(2)

0.1963 ×L(1) = 122.71 ×L(2)

L(1) / L(2) = 122.71 / 0.1963 = 625.16

and we know longitudinal engineering strain is

longitudinal engineering strain = L(1) / L(2)  - 1

longitudinal engineering strain = 625.16  - 1

longitudinal engineering strain = 624.16

and

true strain is

true strain = ln ( L(1) / L(2))

true strain = ln ( 625.16)

true strain is 6.44

3 0
2 years ago
A motorist inflates the tires of her car to a pressure of 180 kPa on a day when the temperature is -8.0° C. When she arrives at
GuDViN [60]

Apply Gay-Lussac's law:

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P = 180kPa, T = -8.0°C = 265.15K

Final P and T values:

P = 245kPa, T = ?

Set the initial and final P/T values equal to each other and solve for the final T:

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If Siobhan hits a 0.25 kg volleyball with 0.5 N of force, what is the acceleration of the ball?
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8 0
2 years ago
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An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s
max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

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You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

4 0
2 years ago
If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed?
sweet [91]

Answer:

Speed of the electron will be v=5.896\times 10^7m/sec

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Potential difference = V=9.9KV=9.9\times 10^3volt

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1.6\times 10^{-19}\times 9.9\times 10^3=\frac{1}{2}\times 9.11\times 10^{-31}v^2

v=5.896\times 10^7m/sec

4 0
1 year ago
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