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3 years ago

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David drove the first 6 hours of his journey at 65 km/hour and the last 3 hours of his journey at 80 km/hour. how far is the who

le journey in Km?

1 answer:

Sidana [21]3 years ago

6 0

2434+4678-4567+56788

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A cyclist is riding a bicycle at a speed of 22 mph on a horizontal road. The distance between the axles is 42 in., and the mass

stealth61 [152]

Answer:

The shortest distance is $S = 24.86 ft$

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

The speed of the bicycle is $v_b = 22\ mph = 22 * \frac{5280}{3600} = 32.26 ft/s$

The distance between the axial is $d = 42 \ in$

The mass center of the cyclist and the bicycle is $m_c = 26 \ in$ behind the front axle

The mass center of the cyclist and the bicycle is $m_h = 40 \ in$ above the ground

For the bicycle not to be thrown over the

Momentum about the back wheel must be zero so

$\sum _B = 0$

=> $mg (26) = ma(40)$

=> $a = \frac{26}{40} g$

Here $g = 32.2 ft/s^2$

So $a = \frac{26}{40} (32.2)$

$a = 20.93 ft/s^2$

Apply the equation of motion to this motion we have

$v^2 = u^2 + 2as$

Where $u = 32.26 ft /s$

and $v = 0$ since the bicycle is coming to a stop

$v^2 = (32.26)^2 - 2(20.93) S$

=> $S = 24.86 ft$

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3 years ago

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jeyben [28]

B. insulator

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3 years ago

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The speed of sound through diamond is about 12,000 m/s. The speed of sound through wood is about 3,300 m/s. Which statement expl

Damm [24]

Answer: They have different rigidities.

Explanation:

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2 years ago

A battery has a terminal voltage of 12.0 V when no current flows. Its internal resistance is 2.0 Ω. If a 4.6 Ω resistor is conne

rosijanka [135]

Answer:

Check attachment for solution

Explanation:

Given that 12V battery

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3 years ago

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A professional golfer hits a golf ball of mass 46 g with her 5-iron, and the ball first strikes the ground 155 m away. The ball

RideAnS [48]

Answer:

$C=2.32\times 10^{-4}\ Ns^2/m^2$

Explanation:

It is given that,

Mass of the golf ball, m = 46 g = 0.046 kg

Terminal speed of the ball, v = 44 m/s

The drag force, $F_r=Cv^2$

Where, C is the drag coefficient. At terminal speed, the weight of the ball is balanced by the drag force.

$Cv^2=mg$

$C=\dfrac{mg}{v^2}$

$C=\dfrac{0.046\times 9.8}{(44)^2}$

$C=2.32\times 10^{-4}\ Ns^2/m^2$

Hence, this is the required solution.

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3 years ago