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3 years ago

What might be an example of Newton’s First, Second, and Third Law of Motion

Physics

1 answer:

ch4aika [34]3 years ago

7 0

First law: A ball is stationary on the ground. It doesn't want to move, but eventually a person will come and kick it.

Second law: A car will never be constant it will have to stop then go . Objects take time to accelerate and also take time to decelerate.

Third law: If your sitting on a chair at a table, and you want to go, what do you do? Push against the table. The table will push back at you and hence, your chair moves. If this law wouldn't work then you would have to move the chair manually.

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Atomic physicists usually ignore the effect of gravity within an atom. To see why, we may calculate and compare the magnitude of

STatiana [176]

Answer:

$2.27\cdot 10^{49}$

Explanation:

The gravitational force between the proton and the electron is given by

$F_G=G\frac{m_p m_e}{r^2}$

where

G is the gravitational constant

$m_p$ is the proton mass

$m_e$ is the electron mass

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

$F_G=(6.67259\cdot 10^{-11} m^3 kg s^{-2})\frac{(1.67262\cdot 10^{-27}kg) (9.10939\cdot 10^{-31}kg)}{(3 m)^2}=1.13\cdot 10^{-68}N$

The electrical force between the proton and the electron is given by

$F_E=k\frac{q_p q_e}{r^2}$

where

k is the Coulomb constant

$q_p = q_e = q$ is the elementary charge (charge of the proton and of the electron)

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

$F_E=(8.98755\cdot 10^9 Nm^2 C^{-2})\frac{(1.602\cdot 10^{-19}C)^2}{(3 m)^2}=2.56\cdot 10^{-19}N$

So, the ratio of the electrical force to the gravitational force is

$\frac{F_E}{F_G}=\frac{2.56\cdot 10^{-19} N}{1.13\cdot 10^{-68}N}=2.27\cdot 10^{49}$

So, we see that the electrical force is much larger than the gravitational force.

5 0

4 years ago

Plants make food through photosynthesis, and photosynthesis takes place mostly in plant leaves, but when plants lose their leave

Nastasia [14]

Answer:

The answer is not B or D because they need light co2 and water to make their food so B and D are out. and unless someone is walking during winter to feed it. It will not be A so the answer is C.

Explanation:

3 0

3 years ago

A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa

pav-90 [236]

Answer

given,

current (I) = 16 mA

circumference of the circular loop (S)= 1.90 m

Magnetic field (B)= 0.790 T

S = 2 π r

1.9 = 2 π r

r = 0.3024 m

a) magnetic moment of loop

M= I A

M= $16 \times 10^{-3} \times \pi \times r^2$

M= $16 \times 10^{-3} \times \pi \times 0.3024^2$

M= $4.59 \times 10^{-3}\ A m^2$

b) torque exerted in the loop

$\tau = M\ B$

$\tau = 4.59 \times 10^{-3}\times 0.79$

$\tau = 3.63 \times 10^{-3} N.m$

8 0

3 years ago

A subway train accelerates from rest at one station at a rate of 1.30 m/s^2 for half of the distance to the next station, then d

Minchanka [31]

This problems a perfect application for this acceleration formula:

   Distance = (1/2) (acceleration) (time)² .

During the speeding-up half:   1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half: 1,600 meters = (1/2) (1.3 m/s²) T²

Pick either half, and divide each side by 0.65 m/s²:

   T² = (1600 m) / (0.65 m/s²)

   T = square root of (1600 / 0.65) seconds

Time for the total trip between the stations is double that time.

   T = 2 √(1600/0.65) = <em>99.2 seconds</em> (rounded)

7 0

3 years ago

A certain copper wire has a resistance of 10.5 ω. at what fraction of the length l must the wire be cut so that the resistance o

Delvig [45]

The answer for that questions will be 759. Stars itir

6 0

3 years ago