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Westkost [7]
3 years ago
12

According to Newton’s first law of motion, when will an object at rest begin to move?

Physics
1 answer:
soldi70 [24.7K]3 years ago
6 0

2. When an unbalanced force acts upon it

Think of a glass of milk resting on a table. The glass weighs a certain amount more due to the load it carries. It would be unaffected until and unbalanced force (such as a hand) carelessly knocks it over spilling the contents.

Hope this helps :)

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Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1
hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

\sum F=ma

We know that the force between two bodies due to gravity is given by the following equation:

F_{g} = G\frac{m_{1}m_{2}}{r^{2}}

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

F_{g} =G\frac{Mm}{r^{2}}

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

F_{c}=ma_{c}

where the centripetal acceleration is given by:

a_{c}=\omega ^{2}r

where

\omega = \frac{2\pi}{T}

Where T is the period, and \omega is the angular speed of the planet, so:

a_{c} = ( \frac{2\pi}{T})^{2}r

or:

a_{c}=\frac{4\pi^{2}r}{T^{2}}

so:

F_{c}=m(\frac{4\pi^{2}r}{T^{2}})

so now we can do the sum of forces:

\sum F=ma

F_{g}=ma_{c}

G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})

in this case we can get rid of the mass of the planet, so we get:

G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})

we can now solve this for T^{2} so we get:

T^{2} = \frac{4\pi ^{2}r^{3}}{GM}

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

T_{1}^{2}>T_{2}^{2}

So let's see what's going on there, we'll call:

M_{1}= mass of Star 1

M_{2}= mass of Star 2

So:

\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}

we can get rid of all the constants so we end up with:

\frac{1}{M_{1}}>\frac{1}{M_{2}}

and let's flip the inequality, so we get:

M_{2}>M_{1}

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0
3 years ago
A person walks due south from point A for 500 yards and then due west for 300 yards, arriving at point B. Answer the following q
Pepsi [2]

The total displacement of the person walking from point A to point B is 300 yards.

As shown in the figure we can conclude that the required method to calculate the total displacement is the Pythagoras theorem.

<h3>Pythagoras theorem in brief :</h3>

According to the Pythagorean Theorem, the square that represents the hypotenuse, or side of a right triangle that faces the right angle, is equal to the total of the squares on the triangle's legs.(or, in popular algebraic notation, a^2 + b^2 = c^2).

<h3>Calculation: </h3>

Let,

a = 500

b=  300

Hence by using Pythagoras' theorem

Total displacement of the person = \sqrt{500^{2}  + 300^{2} } = \sqrt{900000} = 300

Thus the total displacement of the person from starting point is 300 yards.

Learn more about the displacement examples here:

brainly.com/question/11188852

#SPJ4

5 0
2 years ago
A ball is dropped from a tower that is 206 m high, How long does it take to reach the ground?
Slav-nsk [51]

It will take 6.42 s for the ball that is dropped from a height of 206 m to reach the ground.

From the question given above, the following data were obtained:

Height (H) = 206 m

<h3>Time (t) =? </h3>

NOTE: Acceleration due to gravity (g) = 10 m/s²

The time taken for the ball to get to the ground can be obtained as follow:

H = ½gt²

206 = ½ × 10 × t²

206 = 5 × t²

Divide both side by 5

t^{2}  = \frac{206}{5}\\\\ t^{2} = 41.2

Take the square root of both side

t = \sqrt{41.2}

<h3>t = 6.42 s</h3>

Therefore, it will take 6.42 s for the ball to get to the ground.

Learn more: brainly.com/question/24903556

7 0
3 years ago
HELP ME PLEASEEEEEEEEEEEEEE
ozzi

Answer: The correct statements are:

  • The atoms are very attracted to one another.
  • The atoms are held tightly together.

Explanation:

Solid state: In this state, the molecules are closely packed and cannot move freely from one place to another that means no space between them and the intermolecular force of attraction between the molecules are strong.

In solid substance, the particles are very close to each other due to this the intermolecular forces of attraction are strongest.

The key point about solid are:

  • The atoms are very attracted to one another.
  • The atoms are not moving freely.
  • It will not spread out evenly to fill any container.
  • The atoms are held tightly together.
  • The forces of attraction are strong to bring molecules together.
  • The atoms are close and in fixed positions.
7 0
3 years ago
Two long parallel wires carry currents of 20 a and 5.0 a in opposite directions. the wires are separated by 0.20 m. what is the
Alex777 [14]
The two wires carry current in opposite directions: this means that if we see them from above, the magnetic field generated by one wire is clock-wise, while the magnetic field generated by the other wire is anti-clockwise. Therefore, if we take a point midway between the two wires, the resultant magnetic field at this point is just the sum of the two magnetic fields, since they act in the same direction.

Therefore, we should calculate the magnetic field generated by each wire and then calculate their sum. We are located at a distance r=0.10 m from each wire. 

The magnetic field generated by wire 1 is:
B_1= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(20 A)}{2 \pi (0.10 m)}=  4 \cdot 10^{-5}T

The magnetic field generated by wire 2 is:
B_2= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} NA^{-2} )(5.0 A)}{2 \pi (0.10 m)}= 1 \cdot 10^{-5}T

And so, the resultant magnetic field at the point midway between the two wires is
B=B_1 + B_2 = 4 \cdot 10^{-5} T + 1 \cdot 10^{-5}T=5 \cdot 10^{-5} T
8 0
3 years ago
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