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3 years ago

A top fuel dragster covers the quarter mile in 4.5 seconds. What is it's speed in miles per hour

Physics

2 answers:

Tju [1.3M]3 years ago

7 0

Answer: The speed of top fuel dragster is 200 miles/hr

Explanation:

We are given:

Speed of top fuel dragster = quarter mile in 4.5 seconds = 0.25 mil/4.5 sec

To calculate the speed of top fuel dragster in miles per hour, we use the conversion factor:

1 hr = 3600 seconds

Converting the speed in miles per hour, we get:

$\Rightarrow (\frac{0.25mil}{4.5s}\times (\frac{3600s}{1hr})\\\\\Rightarrow 200mil/hr$

Hence, the speed of top fuel dragster is 200 miles/hr

gladu [14]3 years ago

6 0

Distance in Miles (.25) x 3600 (seconds in an hour) / time in seconds = 200 MPH. Drag racing calculates the top speed via a speed trap starting 66 feet from the finish line.

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The cheetah is considered the fastest running animal in the world. Cheetahs can accelerate to a speed of 21.7 m/s in 2.50 s and

viktelen [127]

Answer:

1) 64.2 mi/h

2) 3.31 seconds

3) 47.5 m

4) 5.26 seconds

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 0 m/s

v = Final velocity = 21.7 m/s

s = Displacement

a = Acceleration

1) Top speed = 28.7 m/s

1 mile = 1609.344 m

$1\ m=\frac{1}{1609.344}\ miles$

1 hour = 60×60 seconds

$1\ s=\frac{1}{3600}\ hours$

$28.7\ m/s=\frac{\frac{28.7}{1609.344}}{\frac{1}{3600}}=64.2\ mi/h$

Top speed of the cheetah is 64.2 mi/h

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow t=\frac{21.7-0}{2.5}\\\Rightarrow a=8.68\ m/s^2$

Acceleration of the cheetah is 8.68 m/s²

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{28.7-0}{8.68}\\\Rightarrow t=3.31\ s$

It takes a cheetah 3.31 seconds to reach its top speed.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{28.7^2-0^2}{2\times 8.68}\\\Rightarrow s=47.5\ m$

It travels 47.5 m in that time

4) When s = 120 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 120=0\times t+\frac{1}{2}\times 8.68\times t^2\\\Rightarrow t=\sqrt{\frac{120\times 2}{8.68}}\\\Rightarrow t=5.26\ s$

The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds

8 0

3 years ago

A weatherman carried an aneroid barometer from the ground floor to his office atop the Sears Tower in Chicago. On the level grou

Sophie [7]

Answer:

442.36038 m or 1451.31362 ft

Explanation:

$P_1$ = Initial pressure = 30.15 inHg

$P_2$ = Final pressure = 28.607 inHg

$\rho$ = Density of air = 0.075 lb/ft³

$1\ lb/ft^3=16.0185\ kg/m^3$

$1\ in=0.0254\ m$

$1\ m=3.28084\ ft$

Density of mercury = 13560 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Difference in pressure is given by

$P_1-P_2=\rho gh\\\Rightarrow h=\frac{P_1-P_2}{\rho g}\\\Rightarrow h=\frac{(30.15-28.607)\times 13560\times 0.0254\times 9.81}{0.075\times 16.0185\times 9.81}\\\Rightarrow h=442.36038\ m\\\Rightarrow h=442.36038\times 3.28084\\\Rightarrow h=1451.31362\ ft$

The height of the building is 442.36038 m or 1451.31362 ft

4 0

3 years ago

To determine an athlete's body fat, she is weighed first in air and then again while she's completely underwater. it is found th

il63 [147K]

The difference in weight is due to the displacement of water (the buoyancy of water is acting on the athlete thus giving her smaller weight).

The amount of weight displaced or the amount of buoyant force is:

Fb = 690 N - 48 N

Fb = 642 N

From newtons law, F = m*g. Using this formula, we can calculate for the mass of water displaced:

m of water displaced = 642N / 9.8m/s^2

m of water displaced = 65.5 kg

Assuming a water density of 1 kg/L, and using the formula volume = mass/density:

V of water displaced = 65.5kg / 1kg/L = 65.5 L

The volume of water displaced is equal to the volume of athlete. Therefore:

V of athlete = 65.5 L

The mass of athlete can also be calculated using, F = m*g

m of athlete = 690 N/ 9.8m/s^2

m of athlete = 70.41 kg

Knowing the volume and mass of athlete, her average density is therefore:

average density = 70.41 kg / 65.5 L

average density = 1.07 kg/L = 1.07 g/mL

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3 years ago

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An Austin volleyball player bumps a 5 kg ball into the air. It reaches a height of 2.8 meters. How fast was the ball going as it

KATRIN_1 [288]

Answer:

v = 7.4 m/s

Explanation:

Given that,

Mass if a volleyball, m = 5 kg

The ball reaches a height of 2.8 m

We need to find how fast the ball is going as it bumped into the air. Ket the velocity is v. Using the conservation of energy to find it as follows :

$mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 2.8} \\\\=7.4\ m/s$

So, the required speed is 7.4 m/s. Hence, the correct option is (b).

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3 years ago

When the mass of an object decreases, the force of gravity - Remains Unchanged - Decreases - increases - Becomes irregular

Gwar [14]

The force of gravity decreases

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3 years ago

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