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Lelu [443]
3 years ago
6

An 0.08 kg arrow is shot from a bow with a velocity of 25 m/s.

Physics
1 answer:
sukhopar [10]3 years ago
5 0
123846284741741741741741852851 to he power of 46
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fomenos

Explanation:

I'm corona positive and isolated feeling depressed just logged in to talk someone but people ignoring me thanks for this behaviour got disappointed bye everyone logging out had a great time

8 0
2 years ago
What is a living organism? Give a few examples.
zhuklara [117]

Answer:

It's a individual form of life. Examples of this are bacterium , protists and fungus

7 0
2 years ago
Find the change in kinetic energy of a 0.650 kg fish leaping to the right at 15.0 m/s that collides inelastically with a 0.950 k
yaroslaw [1]
The change in the kinetic energy refers to the work done in displacing a body, thus, the change in the kinetic energy of an object refers to the work done on the object.
The correct formula to use is:
W = Initial kinetic energy - Final kinetic energy;
Where, W = change in kinetic energy
Final kinetic energy and initial kinetic energy = 1/2 MV^2
Initial velocity = 15 m/s
Final velocity  = 13.5 m/s
Initial mass = 0.650 kg
Final mass = 0.950 kg
W = 1/2 [0.650* (15 *15)] - 1/2 [0.950 * (13.5 * 13.5)]
W = 146.25 - 173.13 = 26.88
Therefore, the change in kinetic energy is 26.88 J.
 The negative sign has to be ignored, because change in kinetic energy can not be negative.
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4 0
3 years ago
Read 2 more answers
State the law of universal gravitation, and use examples to explain how changes in mass and changes in distance affect gravitati
juin [17]
F(g)= Gm1m2/ r^2 If mass is increased, so will the force of gravity because it is in direct relationship with the gravitational force, but if distance is increased, the force of gravity will decrease because it is indirectly related ( since it is on the bottom of the equation)
6 0
3 years ago
HELP PLEASE 20 POINTS SHOW WORK, ALL EQUATIONS
nataly862011 [7]

Answer:

s = 3 m

Explanation:

Let t be the time the accelerating car starts.

Let's assume the vehicles are point masses so that "passing" takes no time.

the position of the constant velocity and accelerating vehicles are

s = vt = 40(t + 2)  cm

s = ½at² = ½(20)(t)² cm

they pass when their distance is the same

½(20)(t)² = 40(t + 2)

10t² = 40t + 80

0 = 10t² - 40t - 80

0 = t² - 4t - 8

t = (4±√(4² - 4(1)(-8))) / 2(1)

t = (4± 6.928) / 2  ignore the negative time as it has not occurred yet.

t = 5.464 s

s = 40(5.464 + 2) = 298.564 cm

300 cm when rounded to the single significant digit of the question numerals.

7 0
3 years ago
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