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Masteriza [31]
3 years ago
6

What is Newton's third law ?​

Physics
2 answers:
abruzzese [7]3 years ago
8 0
<h3>Answer:</h3>

Newton's Third Law states "To every action, there is an equal and opposite reaction; action and reaction act on different bodies."

tatyana61 [14]3 years ago
3 0

Answer:

every action has an equal and opposite reaction

plz mark me brainliest

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How much work is done by a person who picks up a 3-kg crate from the floor, raises it 2m and sends it flying with a speed of 4m/
erastovalidia [21]

Answer:

  82.8 J

Explanation:

The work done to raise the crate is ...

  PE = Mgh = (3 kg)(9.8 m/s^2)(2 m) = 58.8 J

The kinetic energy added to send the box flying is ...

  KE = (1/2)Mv^2 = (1/2)(3 kg)(4 m/s)^2 = 24 J

So, the total work involved in this activity is ...

  58.8 J +24 J = 82.8 J

6 0
3 years ago
How are active and passive transport different?
sashaice [31]


Passive transport is when molecules travel from HIGHER concentration to LOWER concentration through a concentration gradient.

Active transport is the OPPOSITE. So it travels from LOWER to HIGHER concentration through a concentration gradient.

There are also other differences, but this is the main difference.


8 0
3 years ago
two identical springs of spring constant 7580 N/m are attached to a block of mass 0.245 kg. What is the frequency of oscillation
ad-work [718]

The frequency of oscillation on the frictionless floor is 28 Hz.

<h3>Frequency of the simple harmonic motion</h3>

The frequency of the oscillation is calculated as follows;

f = (1/2π)(√k/m)

where;

  • k is the spring constant
  • m is mass of the block

f = (1/2π)(√7580/0.245)

f = 28 Hz

Thus, the frequency of oscillation on the frictionless floor is 28 Hz.

Learn more about frequency here: brainly.com/question/10728818

#SPJ1

3 0
2 years ago
7. What is another name for the force of
leonid [27]
Weight because both gravity and weight are pulling an object down
3 0
3 years ago
Read 2 more answers
On an asteroid, the density of dust particles at a height of 3 cm is ~30% of its value just above the surface of the asteroid. A
Anvisha [2.4K]

From the law of atmosphere

N_v(y) = n_0*e^{-\frac{mgy}{Kb*T}}

Where

n_0 = constant and is number density where the height y = 0cm

n_V = Number density at height y=3cm

Kb = Boltzmann constant = 1.38*10^{-23}J/K

T=20K

m = 10^{-19}kg

Re-arranging the equation to have the value of the gravity,

\frac{N_v(y)}{n_0} = e^{-\frac{mghy}{KbT}}

ln(\frac{N_v(y)}{n_0}) = -\frac{mgy}{KbT}

Since it is 30% of value above surface, therefore N_v = 0.3n_0

ln(\frac{0.3n_0}{n_0}) = -\frac{mgy}{KbT}

g = -\frac{KbT ln(0.3)}{my}

g = -\frac{(1.38*10^{-23}J/K)(20K)(Ln(0.3))}{10^{-19}(3*10^{-2})}

g = \frac{1.38*2*ln(0.3)*10^{-22}}{3*10^{-4}}

g = 1.104*10^{-1}m/s^2

g = 0.1m/s^2

Therefore the correct answer is C.

4 0
4 years ago
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