Which scientist was the first one to use a telescope

What is the relationship between the kinetic energy of molecules in an object and the object temperature

andrew11 [14]

I think it might be friction

7 0

3 years ago

You have a neutral balloon. What is its charge after 12000 electrons have been removed from it? The elemental charge is 1.6 × 10

AleksandrR [38]

Answer:

1.92×10^-9 microCoulomb

Explanation:

Elemental charge = 1.6×10^-19 Coulomb

Charge of balloon after 12,000 electrons have been removed from it = 12,000 × 1.6×10^-19 Coulomb = 1.92×10^-15 Coulomb = 1.92×10^-15/10^-6 = 1.92×10^-9 microCoulomb

6 0

3 years ago

The law of reflection is quite useful for mirrors and other flat, shiny surfaces. (This sort of reflection is called specular re

Sever21 [200]

Answer:

Explanation:

Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .

Now the reflecting surface is twisted so that is becomes inclined at angle alpha .

The reflected light will be deviated from its original direction by angle

2 x alpha .

Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle

2 x beta

Hence angle between these two reflected beam

= 2 beta - 2 alpha

= 2 ( β - α )

So, angular separation between the rays reflected from the two surfaces

= 2 ( β - α ) .

5 0

3 years ago

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0. The wheel has

irga5000 [103]

Answer:

The total linear acceleration is approximately 0.246 meters per square second.

Explanation:

The total linear acceleration ( $a$ ) consist in two components, <em>radial</em> ( $a_{r}$ ) and <em>tangential</em> ( $a_{t}$ ), in meters per square second:

$a_{r} = \omega^{2}\cdot r$ (1)

$a_{t} = \alpha \cdot r$ (2)

Since both components are orthogonal to each other, the total linear acceleration is determined by Pythagorean Theorem:

$a = \sqrt{a_{r}^{2}+a_{t}^{2}}$ (3)

Where:

$r$ - Radius of the wheel, in meters.

$\omega$ - Angular speed, in radians per second.

$\alpha$ - Angular acceleration, in radians per square second.

Given that wheel accelerates uniformly, we use the following kinematic equation:

$\omega = \omega_{o}+ \alpha\cdot t$ (4)

Where:

$\omega_{o}$ - Initial angular speed, in radians per second.

$t$ - Time, in seconds.

If we know that $r = 0.1\,m$ , $\alpha = 2\,\frac{rad}{s^{2}}$ , $\omega_{o} = 0\,\frac{rad}{s}$ and $t = 0.60\,s$ , then the total linear acceleration is: