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lord [1]
4 years ago
9

What is Newton's 3rd Law of Motion? *

Physics
2 answers:
drek231 [11]4 years ago
4 0

Answer:

Newton’s Third Law of Motion says that if a body A exerts a force on body B, then the body B exerts a force of equal magnitude, in the opposite direction, on the body A.

Explanation:

Alexandra [31]4 years ago
3 0

Answer:

For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects.

Explanation:

Every force sent into an object will sent a force in return.

If you smack a table, you can feel the table "push back"

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(25%) Problem 3: A bullet is fired horizontally into an initially stationary block of wood suspended by a string and remains emb
Rus_ich [418]

Answer:

Thus, the initial velocity of the bullet is 507.5 m/s.

Explanation:

mass of bullet, m = 0.0085 kg

mass of block, M = 0.99 kg

Height raised, h = 0.95 m

Let the initial velocity of bullet is u and the final velocity of block and bullet system is v.

Use conservation of energy

Potential energy of bullet bock system = kinetic energy of bullet block system

(M+m)\times g\times h = \frac{1}{2}\times \left ( M+m \right )v^{2}

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 0.95}=4.32 m/s

Now use conservation of linear momentum

mu = (M+m) v

0.0085 x u = (0.99 + 0.0085) x 4.32

0.0085 u = 4.314

u = 507.5 m/s

Thus, the initial velocity of the bullet is 507.5 m/s.  

4 0
3 years ago
Why does a charged balloon stick to a neutral wall?
Jobisdone [24]
<span>When you bring a charged object, such as your balloon, near a neutral object that is classified as an insulator, than a temporary charge is induced in the neutral object. If the charged object is positive, then electrons in the neutral object will be attracted toward the charged object, creating a temporary imbalance of charges in the neutral object.</span>



7 0
3 years ago
A boy throws a ball upward with a velocity of 4.50 m/s at 60.0o. What is the maximum height reached by the ball?
Cloud [144]

Answer:

3.1m

Explanation:

Since we only care about the y direction we only need to find vy. Once u draw your vector you will realize that vy= 4.5sin60=3.897m/s.

use vf²=v²+2a(y)

At the maximum height the velocity is 0 and since the object is in freefall, a=-g

Plug in all values

0=15.1875-2*9.8(y)

solve for y

-15.1875*2/-9.8=y

y=3.1m

3 0
3 years ago
Read 2 more answers
A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters​ [cm]. The rover picks up the rock and lifts it
Makovka662 [10]

Answer: 5166.347

Explanation:

The specific gravity of a solid SG (also called relative density) is the ratio of the density of that solid \rho_{rock} to the density of water \rho_{water}=1 kg/m^{3} (normally at 4\°C):

SG=\frac{\rho_{rock}}{\rho_{water}} (1)

On the other hand, the density of the rock is calculated by:

\rho_{rock}=\frac{m_{rock}}{V_{rock}} (2)

Where:

m_{rock} is the mass of the rock

V_{rock}=\frac{4}{3} \pi r^{3} is the volume of the rock, since is spherical

Well, we already know the value of \rho_{water}, but we need to find \rho_{rock} in order to find the rock's specific gravity; and in order to do this, we firsly have to find m_{rock} and then calculate V_{rock}:

In the case of the mass of the rock, we can calclate it by the following equation:

W_{rock}=m_{rock}g_{mars} (3)

Where:

W_{rock} is the weight if the rock in mars

g_{mars}=3.7 m/s^{2} is the acceleration due gravity in Mars

Isolating m_{rock}:

m_{rock}=\frac{W_{rock}}{g_{mars}} (4)

m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}} (5)

To find W_{rock} we can use the following equation of the potential gravitational energy U:

U=W_{rock}H (6)

Where:

U=2 J=2 Nm is the potential energy

H=20 cm \frac{1m}{100 cm}=0.2 m is the height at which the rock has the mentioned potential energy

Isolating W_{rock}:

W_{rock}=\frac{U}{H} (7)

W_{rock}=\frac{2 Nm}{0.2 m} (8)

W_{rock}=10 N (9)

Substituting (9) in (5):

m_{rock}=\frac{10 N}{3.7 m/s^{2}} (10)

m_{rock}=2.702 kg (11)

Substituting (11) in (2):

\rho_{rock}=\frac{2.702 kg}{V_{rock}} (12) At this point we only need to find the volume of the rock, knowing its diameter is d=10 cm, hence its radius is r=\frac{d}{2}=5 cm

V_{rock}=\frac{4}{3} \pi (5 cm)^{3} (13)

V_{rock}=523.59 cm^{3} \frac{1 m^{3}}{(100 cm)^{3}}=0.000523 m^{3} (14)

Substituting (14) in (12):

\rho_{rock}=\frac{2.702 kg}{0.000523 m^{3}} (15)

\rho_{rock}=5166.34 kg/m^{3} (16)

Substituting (16) in (1):

SG=\frac{5166.34 kg/m^{3}}{1 kg/m^{3}} (17)

Finally we obtain the specific gravity of the​ rock:

SG=5166.347

7 0
3 years ago
Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars
zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

T^2 \propto R^3

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

T^2 \propto R^3

T^2 = kR^3

Substituting the values, we get the value of constant k for mars.

687^2 = k\times (2.279 \times 10^{11})^3

k = 3.92 \times 10^{-29}

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

365^3 = 3.92\times 10^{-29} R^3

R^3 = 3.39 \times 10^{33}

R= 1.50 \times 10^{11}\;\rm m

Hence we can conclude that the distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

6 0
3 years ago
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