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Ksenya-84 [330]

3 years ago

12

Araon is nervous about his ap biology quiz. While studying he paces around the room. over a 120 second period, he moves 3m forwa

rd, 2.5 back, 7m forward, 10m back and 2m forward what was the average speed

1 answer:

-Dominant- [34]3 years ago

6 0

Average speed = total distance travelled / total time taken
total distance = 3 + 2.5 + 7 + 10 + 2 = 24.5
total time = 120 seconds
therefore 24.5/120 is
0.2m per second.

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Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is

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Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is $c = 28853.78 \ m^2 /s^2$

Explanation:

From the question we are told that

The acceleration is $a = \frac{c}{s}\ m/s^2$

The initial position of the projectile is s= 1.5m

The final position of the projectile is $s_f = 3 \ m$

The velocity is $v = 200 \ m/s$

Generally $time = \frac{ds}{dv}$

and acceleration is $a = \frac{v}{time }$

so

$a = v \frac{dv}{ds}$

=> $vdv = a ds$

$vdv = \frac{c}{s} ds$

integrating both sides

$\int\limits^a_b vdv = \int\limits^c_d \frac{c}{s} ds$

Now for the limit

a = 200 m/s

b = 0 m/s

c = s= 3 m

d = $s_f$ = 1.5 m

So we have

$\int\limits^{200}_{0} vdv = \int\limits^{3}_{1.5} \frac{c}{s} ds$

$[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.$

$\frac{200^2}{2} = c ln[\frac{3}{1.5} ]$

=> $c = \frac{20000}{0.69315}$

$c = 28853.78 \ m^2 /s^2$

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they absorb sunlight and turn it into glucose I think.

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3 years ago

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In his motorboat, Bill Ruhberg travels upstream at top speed to his favorite fishing spot, a distance of 120120 mi, in 33 hr.

photoshop1234 [79]

Answer:

The rate of the boat in still water is 44 mph and the rate of the current is 4 mph

Explanation:

x = the rate of the boat in still water

y = the rate of the current.

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Time taken upstream = 3 hr

Time taken downstream = 2.5 hr

Speed = Distance / Time

Speed upstream

$\frac{120}{3}=x-y\\\Rightarrow 40=x-y$

Speed downstream

$\frac{120}{2.5}=x+y\\\Rightarrow 48=x+y$

Adding both the equations

$48+40=x-y+x+y\\\Rightarrow 88=2x\\\Rightarrow 44=x$

$40=44-y\\\Rightarrow 40-44=-y\\\Rightarrow y=4$

The rate of the boat in still water is <u>44 mph</u> and the rate of the current is <u>4 mph</u>

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C-14 is an isotope of the element carbon. How does it differ from the carbon atom seen here? A) C-14 has two more protons. B) C-

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Answer:

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Explanation:

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3 years ago

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A constant force of magnitude F acts on an object of mass 0.04kg initially at rest at a point O. If the speed of the object when

vampirchik [111]

Answer:

F = 100 Newtons

Explanation:

F = ?

m = 0.04kg

u = 0m/s ==> u is just an abbreviation for initial velocity, it is conventional.

s = 50m ==> s is just an abbreviation for distance, it is conventional.

v = 500m/s ==> v is just an abbreviation for final velocity, it is conventional.

$v^{2} = u^{2} + 2as\\\\=> a = \frac{v^{2} - u^{2}}{2s}\\a = \frac{500^{2} }{2*50}\\a = 2500ms^{-2}$

Then F = ma = 0.04 x 2500 = 100N

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3 years ago