In the properties of matter case study lab how did the criminals make the coins look gold

A solid disk of mass 2 kg and radius 2 m is given a horizontal push of 20N at a point .3 m above its center. a. What is the mini

Margaret [11]

Answer:

$\mu_s=1.0205$

Explanation:

Given:

mass of solid disk, $m=2\ kg$

radius of disk, $r=2\ m$

force of push applied to disk, $F=20\ N$

distance of application of force from the center, $s=0.3\ m$

<em>For the condition of no slip the force of static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

$\therefore F$

where:

$f_s$ = static frictional force

$\Rightarrow 20$

$\mu_s>1.0204$

7 0

2 years ago

What do scientist do to receive suggestions an criticism of their research from other scientists?

barxatty [35]

They PUBLISH !

They provide complete written, detailed, technical descriptions of their thoughts, hypotheses, experiments, data, and conclusions, in publications that are read by other scientists around the world.

4 0

3 years ago

Read 2 more answers

A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95

madam [21]

Answer:

The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.

Explanation:

As from the given data

the length of the rope is given as l=30 m

the stretched length is given as l'=41m

the stretched length required is give as y=l'-l=41-30=11m

the mass is m=95 kg

the force is F=380 N

the gravitational acceleration is g=9.8 m/s2

The equation of k is given by equating the energy at the equilibrium point which is given as

$U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}$

Here

m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so

$k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m$

Now the force is

$F=kx\\$ or

$x=\dfrac{F}{k}\\$

So here F=380 N, k=630.92 N/m

$x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m$

So the distance is 0.602 m

6 0

2 years ago

suppose you are walking home after school. The distance from school to your home is five kilometers. on foot you can get home in

agasfer [191]

If it takes ten minutes to travel 5 km, then it'll take 60 minutes to travel thirty, making your biking speed 30 kilometers per hour. 25 minutes divided by ten is 2.5. multiply 5 by 2.5 to get 12.5, your average walking speed.

4 0

3 years ago

Read 2 more answers

A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic

barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, $\Delta E=\dfrac{1}{2}m(v^2-u^2)$