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Andreas93 [3]
4 years ago
15

A typical male sprinter can maintain his maximum acceleration for 2.0 s, and his maximum speed is 10 m/s. After he reaches this

maximum speed, his acceleration becomes zero, and then he runs at constant speed. Assume that his acceleration is constant during the first 2.0 s of the race, that he starts from rest, and that he runs in a straight line. (a) How far has the sprinter run when he reaches his maximum speed? (b) What is the magnitude of his average velocity for a race of these lengths: (i) 50.0 m; (ii) 100.0 m; (iii) 200.0 m?
Physics
1 answer:
baherus [9]4 years ago
5 0
20 characters longer later and woah
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Kamal said the distance from the top of the balloon to the ground in the Example is Ï···· 353 ft. What mistake might Kamal have
snow_lady [41]

Complete question is;

Kamal said the distance from the top of the balloon to the ground in the Example image attached is √353 ft. What mistake might Kamal have made?

Answer:

the mistake Kamal made is that she probably used 17 ft as the perpendicular side of the triangle with b as the hypotenuse instead of using 17ft as the hypotenuse

Explanation:

From the image attached, we can see that the distance from the top of the balloon which is blue in color to the ground is denoted by "b".

Now the triangle is a right angle triangle with hypotenuse = 15ft + 2ft = 17 ft; the adjacent side = 8 ft, while the opposite side is "b".

Thus, we can use pythagoras theorem to solve this as;

b = √(17² - 8²)

b = √(289 - 64)

b = √225

b = 15ft

However,we are told Kamal got b as √353 ft.

From inspection of the calculations we just did, if we had used addition instead of subtraction, we would have gotten b = √353 ft.

Thus, we can under that the mistake Kamal made is that she probably used 17 ft as the perpendicular side of the triangle with b as the hypotenuse instead of using 17ft as the hypotenuse.

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The type of tectonic plate boundary that occurs when two plates are colliding into each other is known as a
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The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t
DENIUS [597]

Answer:

terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr

Explanation:

Given the data in the question;

we know that, the force on a body due to gravity is;

F_g = mg

where m is mass and g is acceleration due to gravity

Force of drag is;

F_d = \frac{1}{2}pCAv²

where p is the density of fluid, C is the drag coefficient, A is the area and v is the terminal velocity.

Terminal velocity is reach when the force of gravity is equal to the force of drag.

F_g = F_d

mg =  \frac{1}{2}pCAv²

we solve for v

v = √( 2mg / pCA )

so we substitute in our values

v = √( [2×(86 kg)×9.8 m/s² ] / [ 1.21 kg/m³ × 0.7 × 0.145 m²] )

v = √( 1685.6 / 0.122015 )

v = √( 13814.6949 )

v  = 117.54 m/s

v = ( 117.54 m/s × 3.6 ) = 423.144 km/hr

Therefore terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr

5 0
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