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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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Neutron is commonly used to initiate a fission chain reaction.
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3 years ago
A car starts from rest and accelerates uniformly over a time of 18 seconds for a distance of 390 m. Determine the acceleration o
Sergeeva-Olga [200]

Answer:

a=2.4\ m/s^2

Explanation:

Given that,

The initial speed of a car, u = 0

Time, t = 18 s

Distance, d = 390 m

We need to find the acceleration of the car. Let it is a. Using the second equation of motion to find it.

d=ut+\dfrac{1}{2}at^2

or

d=\dfrac{1}{2}at^2\\\\a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 390}{(18)^2}\\\\a=2.4\ m/s^2

So, the acceleration of the car is 2.4\ m/s^2.

5 0
3 years ago
If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by
fgiga [73]

Answer:

a) v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) v=65-32(2)=1ft/s

Explanation:

From the exercise we got the ball's equation of position:

y=65t-16t^{2}

a) To find the average velocity at the given time we need to use the following formula:

v=\frac{y_{2}-y_{1}  }{t_{2}-t_{1}  }

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

y_{t=2}=65(2)-16(2)^{2} =66ft

y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft

v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

--

y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

--

y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

--

y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) To find the instantaneous velocity we need to derivate the equation

v=\frac{df}{dt}=65-32t

v=65-32(2)=1ft/s

7 0
3 years ago
A 180 g model airplane charged to 18 mC and traveling at 2.2 m/s passes within 8.6 cm of a wire, nearly parallel to its path, ca
viva [34]

Answer:

a=0.2*10^{-5}g

Explanation:

From the question we are told that:

Mass M=180=>0.18kg

Charge Q=18mC=18*10^-^3C

Velocity v=2.2m/s

Length of Wire L=8.6cm=>0.086

Current I=30A

Generally the equation for Magnetic Field of Wire B is mathematically given by

 B=\frac{\mu_0*I}{2\pi*l}

 B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}

 B=6.978*10^{-5}T

Generally the equation for Force on the plane F is mathematically given by

 F=qvB

Therefore

 ma=qvB

 a=\frac{qvB}{m}

 a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}

 a=2.37*10^{-5}

Therefore in Terms of g's

 a=\frac{2.37*10^{-5}}{9.8}

 a=0.2*10^{-5}g

8 0
2 years ago
PLEASE HELP<br> 8th grade honors science
lora16 [44]

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3 years ago
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