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maxonik [38]
4 years ago
11

Sitting on the runway, the velocity of an airplane is equal to zero. When it reaches the end of the runway just before it takes

off, the airplane's velocity will be close to 170 miles per hour.
The change in the velocity of the the airplane over time is called its
A.
acceleration.
B.
position.
C.
velocity.
D.
lift off.
Physics
1 answer:
natita [175]4 years ago
4 0

Answer:

A. acceleration

Explanation:

Acceleration is the change in velocity over change in time.

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In need of help!!!!!!
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Answer:

b

Explanation:

because it fly and you need for it to drop at the right spot for them to get it

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3 years ago
A bike travels 4 miles in half an hour at what speed
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5 0
3 years ago
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A 0.5kg bird has a PE of 137.2 J. What height is its perch?
jekas [21]

PE = mgh

137.2 = .5*9.8*h

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3 years ago
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What phylum do animals with backbones belong to?
kykrilka [37]
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5 0
3 years ago
A particle with kinetic energy equal to 282 J has a momentum of magnitude 26.4 kg · m/s. Calculate the speed (in m/s) and the ma
masha68 [24]

Answer:

v=21.36\,\,\frac{m}{s}\\

m=1.2357\,\,kg

Explanation:

Recall the formula for linear momentum (p):

p = m\,v  which in our case equals 26.4 kg m/s

and notice that the kinetic energy can be written in terms of the linear momentum (p) as shown below:

K=\frac{1}{2} m\,v^2=\frac{1}{2} \frac{m^2\,v^2}{m} =\frac{1}{2}\frac{(m\,v)^2}{m} =\frac{p^2}{2\,m}

Then, we can solve for the mass (m) given the information we have on the kinetic energy and momentum of the particle:

K=\frac{p^2}{2\,m}\\282=\frac{26.4^2}{2\,m}\\m=\frac{26.4^2}{2\,(282)}\,kg\\m=1.2357\,\,kg

Now by knowing the particle's mass, we use the momentum formula to find its speed:

p=m\,v\\26.4=1.2357\,v\\v=\frac{26.4}{1.2357} \,\frac{m}{s} \\v=21.36\,\,\frac{m}{s}

4 0
3 years ago
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