Of the following nonmetals which is the least reactive A.S B. Cl C. Br D. Se

How does the distance traveled by the coin compare to its displacement after ten flips?

Mkey [24]

Answer:

To calculate displacement, simply draw a vector from your starting point to your final position and solve for the length of this line. If your starting and ending position are the same, like your circular 5K route, then your displacement is 0. In physics, displacement is represented by Δs.

Explanation:

hope helps

5 0

3 years ago

A coin of mass m rests on a turntable a distance r from the axis of rotation. The turntable rotates with a frequency of f. What

Crank

Answer:

$\mu = \frac{r (2\pi f)^{2}}{g}$

Explanation:

$N$ = normal force acting on the coin

Normal force in the upward direction balances the weight of the coin, hence

$N = mg$

$f$ = frequency of rotation

Angular velocity of turntable is hence given as

$w = 2\pi f$

$r$ = distance from the axis of rotation

$\mu$ = minimum coefficient of static friction

static frictional force is given as

$f = \mu N\\f = \mu mg$

The static frictional force provides the necessary centripetal force , hence

Centripetal force = Static frictional force

$m r w^{2} = \mu mg\\r w^{2} = \mu g\\\\\mu = \frac{r w^{2}}{g} \\\mu = \frac{r (2\pi f)^{2}}{g}$

3 0

3 years ago

Misconception about falling objects

Yakvenalex [24]

A simple rule to bear in mind is that all objects (regardless of their mass) experience the same acceleration when in a state of free fall. When the only force is gravity, the acceleration is the same value for all objects. On Earth, this acceleration value is 9.8 m/s/s.

8 0

3 years ago

What is the distance from axis about which a uniform, balsa-wood sphere will have the same moment of inertia as does a thin-wall

andrey2020 [161]

Answer:

$D_{s}$ ≈ 2.1 R

Explanation:

The moment of inertia of the bodies can be calculated by the equation

I = ∫ r² dm

For bodies with symmetry this tabulated, the moment of inertia of the center of mass

Sphere $Is_{cm}$ = 2/5 M R²

Spherical shell $Ic_{cm}$ = 2/3 M R²

The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass

I = $I_{cm}$ + M D²

Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation

Let's start with the spherical shell, axis is along a diameter

D = 2R

Ic = $Ic_{cm}$ + M D²

Ic = 2/3 MR² + M (2R)²

Ic = M R² (2/3 + 4)

Ic = 14/3 M R²

The sphere

Is = $Is_{cm}$ + M [ $D_{s}$ ²

Is = Ic

2/5 MR² + M $D_{s}$ ² = 14/3 MR²

$D_{s}$ ² = R² (14/3 - 2/5)

$D_{s}$ = √ (R² (64/15)

$D_{s}$ = 2,066 R

3 0

3 years ago

The light from the Sun heats a metal surface. This is an example of what?

Sedaia [141]

Radiation. That's perhaps the only way that energy gets from the Sun to the earth. The space between the two is a vacuum, so conduction is more or less out as is convection. There's no obvious sign of mirrors, so reflection seems to be out as well.

8 0

3 years ago

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