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1 month ago

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A car drives off a cliff next to a river at a speed of 30 m/s and lands on the bank on theother side. The road above the cliff i

s horizontal and 8.3 m above the other shore wherethe car lands. The tires on the car all hit at once and the air resistance is insignificant.How long is the car in the air?

1 answer:

Softa [3K]1 month ago

7 0

Answer: 1.301 seconds

Explanation:

Provided values

Initial Velocity(u)=30 m/s

Cliff Height=8.3 m

Time for covering 8.3 m

$h=ut+\frac{at^2}{2}$

The initial vertical velocity is 0.

$8.3=\frac{gt^2}{2}$

$t^2=1.69$

$t=1.301 s$

Horizontal distance

$R=u\times t$

$R=30\times 1.301=39.04 m$

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Compressed air is used to fire a 60 g ball vertically upward from a 0.70-m-tall tube. The air exerts an upward force of 3.0 N on

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Answer:

2.87 m

Explanation:

Given parameters:

Mass of the ball (m) = 60 g = 0.06 kg

Height of the tube (h) = 0.70 m

Force applied on the ball by compressed air (F) = 3.0 N

Initial velocity of the ball (u) = 0 m/s (Assumed)

Final velocity of the ball at the tube's exit (v) =?

Acceleration of the ball (a) =?

The ball's weight is derived from multiplying mass and gravity. Therefore,

Weight (W) = $mg=0.06\times 9.8=0.588\ N$

Thus, the total force acting on the ball equals the net of upward force minus the weight.

Net force = Air force - Weight

$F_{net}=F-mg\\F_{net}=3.0-0.588 = 2.412\ N$

According to Newton's second law, net force equals the mass multiplied by acceleration.

$F_{net}=ma\\\\a=\frac{F_{net}}{m}=\frac{2.412\ N}{0.06\ kg}=40.2\ m/s^2$

Acceleration (a) is calculated as 40.2 m/s².

Using the motion equation, we find:

$v^2=u^2+2ah\\\\v^2=0+2\times 40.2\times 0.7\\\\v=\sqrt{56.28}=7.5\ m/s$

Let’s denote the maximum height achieved as 'H'.

Next, we apply the principle of energy conservation from the pipe's peak to the maximum height.

A decrease in kinetic energy equals an increase in potential energy.

$\frac{1}{2}mv^2=mgH\\\\H=\frac{v^2}{2g}$

Substituting the values, we solve for 'H', yielding:

$H=\frac{56.28}{2\times 9.8}\\\\H=\frac{56.28}{19.6}=2.87\ m$

Hence, the ball ascends to a height of 2.87 m above the top of the tube.

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1 month ago

the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point

serg [3582]

Answer:

The pressure measured at this moment is 0.875 mPa

Explanation:

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Applying the pressure formula

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Here, $A_{1}v_{1}=Q_{1}$

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