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3 years ago

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On a road trip, a driver achieved an average speed of (48.0+A) km/h for the first 86.0 km and an average speed of (43.0-B) km/h

for the remaining 54.0 km. What was her average speed (in km/h) for the entire trip? Round your final answer to three significant figures. A = 15 B = 1

1 answer:

Bond [772]3 years ago

4 0

A = 15, B = 1
48 + 15 = 63 for the first 86km
43 - 1 = 42 for the remaining 54 km
time = distance/speed
t1 = 86/63 = 1.3651hours
t2 = 54/42 = 1.2857hours
speed = distance/time
s = 140/2.6508
average speed = 52.8km/h

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Answer:

The angle's rate of change is: -0.125 (degree/feet).

Explanation:

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The first thing you should know for this case is that density is defined as the quotient between mass and volume.
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Then, the density in state two will be:
d2 = m / v2
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Answer:

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Direction of 2nd team from 1st team will be: 14.90 deg North of east

Explanation:

ASSUME Vector is R and makes angle A with +x-axis,

therefore component of vector R is

$R_x = Rcos A$

$R_y = Rsin A$

From above relation

Assuming base camp as the origin, location of 1st team is

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location of 2nd team is at

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$R_3 = R_2 - R_1$

$R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j$

Using above values:

$R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j$

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distance of 2nd team from 1st team will be:

$\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)$

$\left | R_3 \right | = 53.28 km = 58.2 km$

Direction of 2nd team from 1st team will be:

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Answer:

a) <em>2.278 x 10^-5 volts</em>

b) <em>1.139 x 10^-6 Ampere</em>

c) <em>2.59 x 10^-11 W</em>

Explanation:

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the number of turns N = 200 turns

direction of the magnetic field ∅ = 25°

magnetic field strength B = 0.02 T

varying time = 2 sec

The cross sectional area of the wire = $\pi r^{2}$

==> A = 3.142 x $0.002^{2}$ = 1.257 x 10^-5 m^2

Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°

==> Φ = 2.278 x 10^-7 Wb

The induced EMF is given as

E = NdΦ/dt

where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = <em>2.278 x 10^-5 volts</em>

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b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

$I$ = E/R

where R is the resistor

$I$ = (2.278 x 10^-5)/20 = <em>1.139 x 10^-6 Ampere</em>

<em></em>

<em></em>

<em> </em>c) power delivered to the resistor is given as

P = $I$ E

P = (1.139 x 10^-6) x (2.278 x 10^-5) = <em>2.59 x 10^-11 W</em>

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