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marishachu [46]
3 years ago
7

A car travels due east 14 miles in 18 minutes. Then, the car turns around and returns to its starting point, taking an additiona

l 15 minutes for the return leg. Calculate (a) the average speed and (b) the average velocity of the vehicle.
Physics
1 answer:
N76 [4]3 years ago
4 0

Answer:

(a) The average speed is 0.85 milles/minute

(b) The average velocity is zero

Explanation:

In order to answer part (a) and (b) you have to apply the formulas for average speed and average velocity which are:

<em>-Average speed formula:</em>

v=\frac{d}{t}

where d is the total distance traveled and t is the tota time

Replacing the given values:

v=\frac{14+14}{18+15}\\v=\frac{28}{33} \\v=0.85 milles/minute

Notice that you have to replace the total distance, which is 14 milles for the go plus 14 milles for the return. The same for the total time.

<em>-Average velocity formula:</em>

V = Δx/Δt

Where V is the velocity vector, Δx is the displacement and Δt is the change in time

V= \frac{X2-X1}{t2-t1}

Where X2 is the final position and X1 is the initial position

In this case X1= 0 i and X2=0 i (i is the unit vector in the x direction). So, the displacement is zero.

Therefore, the average velocity is:

V= 0 i [milles/minute]

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If a runner has a speed of 8.66m/s and runs for 46.2s what distance is covered? tv = d
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3 years ago
Residential building codes typically require the use of 12-gauge copper wire (diameter 0.205 cm) for wiring receptacles. Such ci
Alecsey [184]

Given Information:  

Current = I = 20 A

Diameter = d = 0.205 cm = 0.00205 m

Length of wire = L = 1 m

Required Information:  

Energy produced = P = ?

Answer:  

P = 2.03 J/s

Explanation:  

We know that power required in a wire is

P = I²R

and R = ρL/A

Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m

L is the length of the wire and A is the area of the cross-section and is given by

A = πr²

A = π(d/2)²

A = π(0.00205/2)²

A = 3.3x10⁻⁶ m²

R = ρL/A

R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶

R = 5.09x10⁻³ Ω

P = I²R

P = (20)²*5.09x10⁻³

P = 2.03 Watts or P = 2.03 J/s

Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A

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