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4 years ago

8

An electric drag racer is much like its piston engine counterpart, but instead it is powered by an electric motor running off of

onboard batteries. These vehicles are capable of covering a 1/4 mile straight-line track in 12 s .
A) Determine the acceleration of the drag racer in units of m/s2. (Assume that the acceleration is constant throughout the race.)
B) How does the value you get compare with the acceleration of gravity?
C) Calculate the final speed of the drag racer in mi/h.

1 answer:

Lapatulllka [165]4 years ago

6 0

Answer:

5.5879 m/s²

0.56961g

149.9976 mph

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 0.25 miles

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 0.25\times 1609.34=0\times t+\frac{1}{2}\times a\times 12^2\\\Rightarrow a=\dfrac{0.25\times 1609.34\times 2}{12^2}\\\Rightarrow a=5.5879\ m/s^2$

The acceleration is 5.5879 m/s²

Dividing by g

$\dfrac{a}{g}=\dfrac{5.5879}{9.81}\\\Rightarrow a=0.56961g$

The acceleration of these cars is 0.56961 times g

$v=u+at\\\Rightarrow v=0+5.5879\times 12\\\Rightarrow v=67.0548\ m/s$

Converting mph

$67.0548\times \dfrac{3600}{1609.34}=149.9976\ mph$

The speed is 149.9976 mph

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Answer:

The orbital speed of Dactyl is $5.55m/s$

Explanation:

The orbital speed can be determined by the combination of the universal law of gravity and Newton's second law:

$F = G\frac{M \cdot m}{r^{2}}$ (1)

Where G is gravitational constant, M is the mass of the asteroid, m is the mass of the moon and r is the distance between them

In the other hand, Newton's second law can be defined as:

$F = ma$ (2)

Where m is the mass and a is the acceleration

Then, equation 2 can be replaced in equation 1

$m\cdot a = G\frac{M \cdot m}{r^{2}}$ (2)

However, a will be the centripetal acceleration since the moon Dactyl describe a circular motion around the asteroid

$a = \frac{v^{2}}{r}$ (3)

$m\frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}}$ (4)

Therefore, v can be isolated from equation 4:

$m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r$

$m \cdot v^{2} = G \frac{M \cdot m}{r}$

$v^{2} = G \frac{M \cdot m}{rm}$

$v^{2} = G \frac{M}{r}$

$v = \sqrt{\frac{G M}{r}}$ (5)

Finally, the orbital speed can be found from equation 5:

Notice, that it is necessary to express r in units of meters.

$r = 95km \cdot \frac{1000m}{1km}$ ⇒ $95000m$

$v = \sqrt{\frac{(6.672x10^{-11}N.m^{2}/kg^{2})(4.4x10^{16}kg)}{95000m}}$

$v = 5.55m/s$

Hence, the orbital speed of Dactyl is $5.55m/s$

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3 years ago

I’ll give brainliest!! please help and answer correctly! plsss answer quick

Rashid [163]

Answer: The motion of the object will remain the same

Explanation:

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3 years ago

6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri

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Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a = - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

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3 years ago

A college student holds a pail full of water by the handle and whirls it around in a vertical circle at a constant speed. The ra

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The minimum speed of the water must be 3.4 m/s

Explanation:

There are two forces acting on the water in the pail when it is at the top of its circular motion:

The force of gravity, mg, acting downward (where m is the mass of the water and g the acceleration of gravity)
The normal reaction, N also acting downward

Since the water is in circular motion, the net force must be equal to the centripetal force, so:

$N+mg=m\frac{v^2}{r}$

Where:

$g=9.8 m/s^2$

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When this occurs, the equation becomes:

$mg=m\frac{v^2}{r}\\v=\sqrt{gr}$

And substitutin the values of g and r, we find the minimum speed that the water must have in order not to spill out:

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Learn more about circular motion:

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3 years ago

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kow [346]

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3 years ago