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2 years ago

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2. An object is moving with an initial velocity of 12 m/s. It accelerates at a rate of 1.5 m/s^2 over a distance of 40 m. What i

s its new velocity?

1 answer:

nikklg [1K]2 years ago

4 0

Answer:

16.2 m/s

Explanation:

2ad=Vf^2-Vi^2

2 (1.5) (40) = Vf^2 -(12)^2

Vf= 16.2 m/s

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Based on the data you found, about how many of the 100 aliens would become thin if the temperature were 35 C?

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A rotating wheel requires 5.00 s to rotate 28.0 revolutions. Its angular velocity at the end of the 5.00-s interval is 96.0 rad/

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Answer:

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Explanation:

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3 years ago

A grasshopper hops along a level road. on each hop, the grasshopper launches itself itself at an angle of 45 degrees. and achiev

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Answer: 1.9 m/s

Explanation:

1) Data:

height, y = 0

range, x = 0.77m

angle, α = 45°

question, horizontal velocity, Vox = ?

2) Physical principles and formulas

Projectile motion.

$y=V_{0y}t-gt^{2} /2\\ \\ x=V_{0x}t\\ \\ V_{0y}=V_0sin\alpha \\ \\ V_{0x}=V_0cos\alpha$

3) Solution

$0.77=V_0cos(45)t\\ t=0.77/(V_0cos(45))$

$y=0=V_0sin(45)t-gt^{2} /2$

Factor

$0=t[V_0sin(45)-gt/2]$

Discard t = 0 and solve for the other factor:

$gt/2=V_0sin(45)\\ \\ V_0= gt/(2sin(45))=9.81[0.77/V_{0}cos(45)]/[2sin(45)]\\ \\ V_0^2=9.81[0.77/(2cos(45)sin(45)=9.81[0.77/sin(90)]=7.55\\ \\ V_0}=\sqrt{7.55} =2.8$

Find Vx:

$Vx=V_{0x}=V_{0}cos(45)=2.75(\sqrt{2} /2]=1.9$

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