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3 years ago

7

How do I calculate speed, velocity, and acceleration? I need the formulas too

1 answer:

vodomira [7]3 years ago

5 0

Speed is the secular unit, which means it only measures "how fast it goes."
$speed= \frac{distance}{time}$
Velocity is a victory unit, which means it measures "both how fast it goes AND the direction."
$velocity= displacement/time$
Acceleration is the change of velocity over time.
$acceleration = \frac{change in velocity}{time}$

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A 0.5kg bird has a PE of 137.2 J. What height is its perch?

jekas [21]

PE = mgh

137.2 = .5*9.8*h

h=137.2/(.5*9.8)

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An electron is released from rest at a distance of 0.570 m from a large insulating sheet of charge that has uniform surface char

Lelu [443]

Explanation:

Formula to calculate the electric field of the sheet is as follows.

E = $\frac{\sigma}{2 \epsilon_{o}}$

And, expression for magnitude of force exerted on the electron is as follows.

F = Eq

So, work done by the force on electron is as follows.

W = Fs

where, s = distance of electron from its initial position

= (0.570 - 0.06) m

= 0.51 m

First, we will calculate the electric field as follows.

E = $\frac{\sigma}{2 \epsilon_{o}}$

= $\frac{4.60 \times 10^{-12}C/m^{2}}{2 \times 8.854 \times 10^{-12}C^{2}/N m^{2}}$

= 0.259 N/C

Now, force will be calculated as follows.

F = Eq

= $0.259 N/C \times 1.6 \times 10^{-19} C$

= $0.415 \times 10^{-19} N$

Now, work done will be as follows.

W = Fs

= $0.415 \times 10^{-19} N \times 0.51 m$

= $2.12 \times 10^{-20} J$

Thus, we can conclude that work done on the electron by the electric field of the sheet is $2.12 \times 10^{-20} J$ .

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During skidding of a vehicle, wheel moves ___________ rotation.

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Explanation:

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Assume that you have 0.480 mol of N2 in a volume of 0.700 L at 300 K .

Svetach [21]

Answer:

1) 16.88 atm

2) 34.47 atm

Explanation:

Data:

Volume=0.700L

Temperature = 300K

Number of moles=0.480 mol

Ideal gas constant=0.082057 L*atm/K·mol

1) The ideal gas law is:

$PV=nRT$ (1)

with P the pressure, T the temperature, n the number of moles, V the volume and R the ideal gas constant , so solvig (1) for P:

$P=\frac{nRT}{V}$

$P=\frac{(0.480)(0.082057)(300)}{0.700}=16.88 atm$

2) The vander Walls equation is:

$(P+\frac{a}{V^{2}})(V-b)=RT$

solving for P

$P=\frac{RT}{V-b}-\frac{a}{V^2}=\frac{(0.082057)(300)}{0.700-0.0387}-\frac{1.35}{0.700^2}=34.47 atm$

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