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Natasha2012 [34]
3 years ago
14

Why does physics involve math?

Physics
1 answer:
Scrat [10]3 years ago
3 0
Physics involves math because calculations have to be made and u use math to make calculations
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NEED HELP NOW DUE TODAY Which of these statements is most likely correct about Newton's law on gravity? (2 points)
yKpoI14uk [10]
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4 0
3 years ago
What happens to the position of an object as an unbalanced force acts on?
jarptica [38.1K]

Answer:

Unbalanced forces change the motion of an object. If an object is at rest and an unbalanced force pushes or pulls the object, it will move. Unbalanced forces can also change the speed or direction of an object that is already in motion.

5 0
3 years ago
Read 2 more answers
is this true or false? an object with a mass of 9.0kg has an acceleration of 3m/s2. the force acting on it is 3N
ludmilkaskok [199]

The answer is "False". The force acting on the object is 27 N.

According to Newton's second law, when a force <em>F</em> acts on am object of mass <em>m</em>, it produces an acceleration <em>a</em>. The force is given by the expression,

F=ma

Thus, if the body has a mass of 9.0 kg and if it has an acceleration of 3 m/s², then, on substituting the values in the equation for force,

F=ma\\ =(9.0kg)(3m/s^2)\\ =27N

Thus, it can be seen that the force acting on the body is 27 N and not 3 N as is mentioned in the statement. Hence the statement is false.

5 0
3 years ago
Mature salmon swim upstream, returning to spawn at their birthplace. During the arduous trip they leap vertically upward over wa
Katarina [22]

Answer: minimum speed of launch must be 7.45m/s

Explanation:

Given the following:

Height or distance (s) = 2.83m

The final velocity(Vf) at maximum height = 0

Upward motion, acceleration due to gravity(g) us negative = -9.8m/s^2

From the 3rd equation of motion:

V^2 = u^2 - 2gs

Where V = final velocity

u = initial velocity

Therefore, u = Vi

u = √Vf^2 - 2gs

u = √0^2 - 2(-9.8)(2.83)

u = √0 + 55.468

u = √55.468

u = 7.4476 m/s

u = 7.45m/s

7 0
3 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
3 years ago
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