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2 years ago

8

A car is driving east at 120. km/h from Toronto to Ottawa. The distance between the two cities is 425.5 km, how long will it tak

e for the driver to reach Ottawa?

1 answer:

Ket [755]2 years ago

4 0

Answer:

The time it will take for the driver to reach Ottawa is 3 hours 32 minutes and 45 seconds

Explanation:

The given parameters are;

Speed of the car = 120 km/h

Distance from Toronto to Ottawa = 425.5 km

The formula for speed is given as follows;

Speed = Distance/Time

Therefore, to find the time duration it takes from Toronto to Ottawa, we have;

Time duration = Distance from Toronto to Ottawa/(Speed of the car)

The time duration = 425.5/120 = 3.54583 hours = 212.75 min = 12765 seconds

The time it takes from Toronto to Ottawa while driving at 425.5 km/h = 12765 seconds.

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Coefficient of static friction = tan(a) = 0.4
r = 740 m
g = 9.8 m/s²
$v \: = \: \sqrt{gr \tan( \alpha ) }$
v = √(9.8 × 740 × 0.4) m/s
v ≈ 53.85908 m/s

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A jogger runs 10.0 blocks do east, 5.0 blocks due South, and another two. Zero blocks do east. Assume all blocks are equal size,

Annette [7]

Jogger moves in three displacements

d1 = 10 blocks East

d2 = 5 blocks South

d3 = 2 blocks East

now we can say

total displacement towards East direction will be

$d_x = 10 + 2= 12 blocks$

Total displacement towards South

$d_y = 5 block$

now to find the net displacement we can use vector addition

$d = \sqrt{d_x^2 + d_y^2}$

$d = \sqrt{12^2 + 5^2}$

$d = 13 blocks$

<em>so magnitude of net displacement will be equal to 13 blocks</em>

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As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road i

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Answer:

<em>B</em><em>.</em><em> </em><em>Kinetic</em><em> </em><em>friction</em><em> </em>

Explanation:

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2 years ago

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f a car is speeding down a road at 40 miles/hour (mph), how long is the stopping distance D40 compared to the stopping distance

Oksana_A [137]

Answer:

D40 = 2.56 × D25

so number is 2.56 multiple of stopping distance @ 25 mph

Explanation:

given data

speed = 40 miles / hour

distance = D40

speed limit = 25 miles / hour

distance = D25

to find out

express number a multiple of stopping distance @ 25 mph

solution

we know here stopping distance is directly proportional to (speed)²

so here speed ratio is

initial speed = $\frac{40}{25}$

so initial speed = 1.6

so

stopping distance increase = (1.6)²

$\frac{D40}{D25}$ = (1.6)²

$\frac{D40}{D25}$ = 2.56

so here

D40 = 2.56 × D25

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2 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

balandron [24]

Answer with Explanation:

We are given that

Radius of solid core wire=r=2.28 mm= $2.28\times 10^{-3} m$

$1mm=10^{-3} m$

Radius of each strand of thin wire=r'=0.456 mm= $0.456\times 10^{-3} m$

Current density of each wire= $J=3750 A/m^2$

a.Area = $\pi r^2$

Where $\pi=3.14$

Using the formula

Cross section area of copper wire has solid core = $3.14\times (2.28\times 10^{-3})^2=16.3\times 10^{-6} m^2$

Current density = $J=\frac{I}{A}$

Using the formula

$3750=\frac{I}{16.3\times 10^{-6}}$

$I=3750\times 16.3\times 10^{-6}=0.061 A$

Total number of strands=19

Area of strand wire= $A'=19\times 3.14\times (0.456\times 10^{-3})^2=12.4\times 10^{-6} m^2$

$J'=\frac{I'}{A'}$

$3750=\frac{I'}{19\times 3.14(0.456\times 10^{-3})^2}$

$I'=3750\times 19\times 3.14(0.456\times 10^{-3})^2$

$I'=0.047 A$

b.Resistivity of copper wire= $\rho=1.69\times 10^{-8}\Omega-m$

Length of each wire =6.25 m

Resistance, R= $\frac{\rho l}{A}$

Using the formula

Resistance of solid core wire= $R=\frac{1.69\times 10^{-8}\times 6.25}{16.3\times 10^{-6}}=6.5\times 10^{-3}\Omega$

Resistance of strand wire= $R'=\frac{1.69\times 10^{-8}\times 6.25}{12.4\times 10^{-6}}=8.5\times 10^{-3}\Omega$

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2 years ago