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stepan [7]
3 years ago
15

Prove the three laws of motion​

Physics
2 answers:
Vaselesa [24]3 years ago
7 0

Answer:

The first law, also called the law of inertia, was pioneered by Galileo. This was quite a conceptual leap because it was not possible in Galileo's time to observe a moving object without at least some frictional forces dragging against the motion. In fact, for over a thousand years before Galileo, educated individuals believed Aristotle's formulation that, wherever there is motion, there is an external force producing that motion.

The second law, $ f(t)=m\,a(t)$ , actually implies the first law, since when $ f(t)=0$ (no applied force), the acceleration $ a(t)$ is zero, implying a constant velocity $ v(t)$ . (The velocity is simply the integral with respect to time of $ a(t)={\dot v}(t)$ .)

Newton's third law implies conservation of momentum [138]. It can also be seen as following from the second law: When one object ``pushes'' a second object at some (massless) point of contact using an applied force, there must be an equal and opposite force from the second object that cancels the applied force. Otherwise, there would be a nonzero net force on a massless point which, by the second law, would accelerate the point of contact by an infinite amount.

Explanation:

RUDIKE [14]3 years ago
7 0
The first law, also called the law of inertia, was pioneered by Galileo. This was quite a conceptual leap because it was not possible in Galileo's time to observe a moving object without at least some frictional forces dragging against the motion. In fact, for over a thousand years before Galileo, educated individuals believed Aristotle's formulation that, wherever there is motion, there is an external force producing that motion.
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I hope you are able to read this question?? Help ASAP this question is on the quiz tommorow
Masteriza [31]

You would be correct.

Because you have only JUST released the arrow, and how close he is to the target, it would have the same amount of energy when it strikes the target. Yes, the kinetic energy would be destroyed when you hit the target but not right away. And yes, the potential energy would also be destroyed once you release the arrow, but it goes straight back once it stops moving, aka when it hits the target, although it has only just stopped moving.

Hope this helps!

8 0
3 years ago
Rosalind is an impulsive child who seldom waits for her turn when playing with other children. Rosalind's father is also an impu
wariber [46]

Answer:

Alburt Bandura

Explanation:

Rosalind has learned to be impulsive and seldom waits for her turn when playing with other children, this trait she has learned  from her father, who is also an impulsive man.

The theorist who would most likely suggest that Rosalind has learned to be impulsive from watching her father is Alburt Bandura. Alburt in his famous Bebo Doll experiment confirmed that children learn from the adults behavior in their life.

3 0
3 years ago
An 80-kgkg quarterback jumps straight up in the air right before throwing a 0.43-kgkg football horizontally at 15 m/sm/s . How f
vladimir1956 [14]

Answer:

V = 0.0806 m/s

Explanation:

given data

mass quarterback = 80 kg

mass football = 0.43 kg

velocity = 15 m/s

solution

we consider here momentum conservation is in horizontal direction.

so that here no initial momentum of the quarterback

so that final momentum of the system will be 0

so we can say

M(quarterback) ×  V = m(football) × v (football)   ........................1

put here value we get

80 ×  V  = 0.43  × 15

V = 0.0806 m/s

5 0
3 years ago
The current in some DC circuits decays according to the function I=I0e−t/τ, where I is the current at some point in time, I0 is
almond37 [142]

Answer: 1.95

Explanation:

You should start off from the decay formula and solve for τ:

I = I_{0}e^{\frac{t}{\tau\\  } }

\frac{I}{I_{0}} = e^{\frac{-t}{\tau} }

Apply inverse logarithmic function:

ln(\frac{0.2 A}{1.2 A} ) = \frac{-t}{\tau}

The final form will be:

\tau=\frac{-3.5s}{ln(\frac{0.2A}{1.2A} )}

Inputing values for I, IO, and t:

\tau=\frac{-3.5S}{ln(\frac{0.2 A}{1.2 A} )} = 1.95

3 0
3 years ago
A balloon is ascending at 12.4m/s at a height of 81.3m above the ground when a package is dropped. a) How long did it take to re
tankabanditka [31]

Answer:

3secs

Explanation:

Given the following parameters

height H= 81.3m

Velocity v = 12.4m/s

Required

Time it take to reach the ground

Using the equation of motion

H = ut+1/2gt²

81.3 = 12.4t + 1/2(9.8)t²

81.3 = 12.4t + 4.9t²

4.9t² + 12.4t - 81.3 = 0

Using the general formula to find t

t = -12.4±√12.4²-4(4.9)(-81.3)/2(4.9)

t = -12.4±√153.76+1593.48/2(4.9)

t = -12.4±√1747.24/9.8

t = -12.4+41.8/9.8

t = 29.4/9.8

t = 3secs

Hence it took 3secs to reach the ground

5 0
3 years ago
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