The statement “Heavy objects fall faster than light objects” is an example of a(n) ______

A golf ball is rolling in the grass. What must happen to stop the ball from continuing to roll?

slamgirl [31]

The net force must be zero

This is in accordance to Newton's first law, which states that any object in motion will remain in motion and any object at rest will remain at rest unless acted upon by an unbalanced force. An unbalanced force is one where the net force is not zero. If no unbalanced force is applied to a moving object, it will keep moving forever. The reason that we do not observe this in our daily lives is due to friction acting as the unbalanced force.

4 0

3 years ago

A toroidal coil of N turns has a central radius b and a square cross section of side a. Find its self-inductance.

Xelga [282]

Answer:

$L = \frac{\mu_0 N^2 (a^2)}{2\pi b}$

Explanation:

As we know that magnetic field due to torroid is given as

$B = \frac{\mu_0 N i}{2\pi b}$

this is approximately constant magnetic field along the axis of the torroid

now the flux linked with one coil of the torroid is given as

$\phi = B.A$

$\phi = \frac{\mu_0 N i}{2\pi b}(a^2)$

now total flux of N number of coils is given as

$\phi_{total} = \frac{\mu_0 N^2 i(a^2)}{2\pi b}$

now we know that self inductance is the property of coil in which flux of the coil will link with the current in the coil

So we know that

$L = \frac{\phi}{i}$

$L = \frac{\mu_0 N^2 (a^2)}{2\pi b}$

3 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

2 years ago

Radioactive plutonium−239 (t1/2 = 2.44 × 105 yr) is used in nuclear reactors and atomic bombs. If there are 3.50 × 102 g of the

madam [21]

Answer:

t = 4.41 10⁻⁴ years

Explanation:

For this exercise we must use the concept of average life time, which is the time in which the quantity and substance decays in half

$T_{1/2}$ = ln2 / λ

Let's calculate the decay constant of plutonium

λ = ln2 / $T_{1/2}$

λ = ln 2 / 2.44 10⁵

λ = 2.84 10⁻⁶ s⁻¹

Radioactive decay is a first order process

N = No e (-λ t)

Where N is the number of nuclei, the mass is this by molecular weight

m = N PM

m / PM = m₀ / PM e (- λ t)

m / m₀ = e (- λ t)

-λ t = ln (m / m₀)

t = -1 /λ ln (m/m₀)

t = - 1 / 2.84 10⁻⁶ ln (0.1 / 0.35)

t = 4.41 10⁻⁴ years

7 0

3 years ago

slader A jet is circling an airport control tower at a distance of 15.9 km. An observer in the tower watches the jet cross in fr

liubo4ka [24]

Answer:

y = 138.96 m

Explanation:

The angle subtended by the moon is the mean of the angle of the arc between the two most extreme points of the moon, we can see that the angle is very small, so we can approximate this arc to a straight line and then use the trigonometric relationships

sin θ = y / L

where L = 15.9 10³ m and θ = 8.74 10⁻³ rad

y = L sin θ

y = 15.9 10³ sin (8.74 10⁻³)

y = 15.9 10³ 0.0087399

y = 138.96 m

4 0

2 years ago

The statement “Heavy objects fall faster than light objects” is an example of a(n) _______.