Can someone complete this assignment will give brainliest!<br> The assignment is attached

A penny is placed on a rotating turntable. where on the turntable does the penny require the largest centripetal force to remain

Mama L [17]

m = mass of the penny

r = distance of the penny from the center of the turntable or axis of rotation

w = angular speed of rotation of turntable

F = centripetal force experienced by the penny

centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as

F = m r w²

in the above equation , mass of penny "m" and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .

hence greater the distance from center , greater will be the centripetal force to remain in place.

So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.

4 0

3 years ago

What is the meaning of smooth?

damaskus [11]

A surface in which is flat or very soft to the touch and reduces splinters or anything sticking out, having an surface which does not have lumps, or indentations.

5 0

3 years ago

Very large forces are produced in joints when a person jumps from some height to the ground. Calculate the force produced if an

krok68 [10]

Answer:

A) 31 kJ

B) 1.92 KJ

C) 40 , 2.48

Explanation:

weight of person ( m ) = 79 kg

height of jump ( h ) = 0.510 m

Compression of joint material ( d ) = 1.30 cm ≈ 0.013 m

A) calculate the force

Fd = mgh

F = mgh / d

W = mg

F(net) = W + F = mg ( 1 + $\frac{h}{d} )$

= 79 * 9.81 ( 1 + (0.51 / 0.013) )

= 774.99 ( 40.231 ) ≈ 31 KJ

B) calculate the force when the stopping distance = 0.345 m

d = 0.345 m

Fd = mgh hence F = mgh / d

F(net) = W + F = mg ( 1 + $\frac{h}{d} )$

= 79 * 9.81 ( 1 + (0.51 / 0.345) )

= 774.99 ( 2.478 ) = 1.92 KJ

C) Ratio of force in part a with weight of person

= 31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40

Ratio of force in part b with weight of person

= 1920 / 774.99 = 2.48

4 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.

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