Ask question

Ask question

All categories

Scorpion4ik [409]

4 years ago

8

The weight of a synthetic ball varies directly with the cube of its radius. a ball with a radius of 2 inches weighs 5.605.60 pou

nds. find the weight of a ball of the same material with a 3-inch radius.

1 answer:

Umnica [9.8K]4 years ago

3 0

The weight of the ball m is directly proportional to the cube of the radius, $r^3$ . This means that the ratio between the weight of the ball with radius 3-in and 2-in should be equal to the ratio between the cubes of the two radii:
$\frac{m_3}{m_2}= \frac{(3 in)^3}{(2 in)^3}$
So we can find the weight of the ball when its radius is 3 inches:
$m_3 = m_2 \frac{(3 in)^3}{(2 in)^3}= (5605.60 p) \frac{3^3}{2^3}=18919 pounds$

You might be interested in

How does the position of an object relate to the energy stored in an object?

Kamila [148]

Answer:

Potential Energy

Explanation:

Potential energy is the energy stored in an object due to it's position relative to some zero position. An object possesses gravitational potential energy if it is positioned at a height above (or below) the zero height.

4 0

3 years ago

An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at tha

LenaWriter [7]

Answer:

$t=6.4534 s$

Explanation:

This is an exercise where you need to use the concepts of <em>free fall objects</em>

Our <u>knowable variables</u> are initial high, initial velocity and the acceleration due to gravity:

$y_{0}=75m$

$v_{oy} =20m/s$

$g=9.8 m/s^{2}$

At the end of the motion, the <u><em>rock hits the ground</em></u> making the final high y=0m

$y=y_{o}+v_{oy}*t-\frac{1}{2}gt^{2}$

If we <em>evaluate the equation</em>:

$0=75m+(20m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$

This is a classic form of <u><em>Quadratic Formula</em></u>, we can solve it using:

$t=\frac{-b ± \sqrt{b^{2}-4ac } }{2a}$

$a=-4.9\\b=20\\c=75$

$t=\frac{-(20) + \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=-2.37s$

$t=\frac{-(20) - \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=6.4534s$

Since the <u><em>time can not be negative</em></u>, the <em>reasonable answer</em> is

$t=6.4534s$

8 0

3 years ago

I have a question regarding friction in rolling without slipping.

Solnce55 [7]

Explanation:

They probably put "rolls without slipping" in there to indicate that there is no loss in friction; or that the friction is constant throughout the movement of the disk. So it's more of a contingency part of the explanation of the problem.

(Remember how earlier on in Physics lessons, we see "ignore friction" written into problems; it just removes the "What about [ ]?" question for anyone who might ask.)

In this case, you can't ignore friction because the disk wouldn't roll without it.

As far as friction producing a torque... I would say that friction is a result of the torque in this case. And because the point of contact is, presumably, the ground, the friction is tangential to the disk. Meaning the friction is linear and has no angular component.

(You could probably argue that by Newton's 3rd Law there should be some opposing torque, but I think that's outside of the scope of this problem.)

Hopefully this helps clear up the misunderstanding for you.

4 0

3 years ago

A 4.0 kg object will have a weight of approximately 14.8 N on Mars. What is the gravitational field strength on Mars?

zhannawk [14.2K]

Field strength =force/unit mass
= 14.8N/ 4kg = 3.7N/kg

8 0

3 years ago

Read 2 more answers

The Chernobyl reactor accident in what is now Ukraine was the worst nuclear disaster of all time. Fission products from the reac

Anna35 [415]

Answer:

The correct answer is "53.15 days".

Explanation:

Given that:

Half life of $131_{I}$ ,

$T_{\frac{1}{2} }= 8 \ days$

Let the initial activity be " $R_o$ ".
and, activity to time t be "R".

To find t when R will be "1%" of $R_o$ , then

⇒ $R=\frac{1}{100}R_o$

As we know,

⇒ $R=R_o e^{-\lambda t}$

or,

∴ $e^{\lambda t}=\frac{R_o}{R}$

By putting the values, we get

$=\frac{R_o}{\frac{R}{100} }$

$=100$

We know that,

Decay constant, $\lambda = \frac{ln2}{T_{\frac{1}{2} }}$

hence,

⇒ $\lambda t=ln100$

$t=\frac{ln100}{\lambda}$

$=\frac{ln100}{\frac{ln2}{8} }$

$=53.15 \ days$

5 0

3 years ago