The attractive or repulsive force between objects is called

Consider the element carbon. How many electrons can be found in a carbon atom? How many are valence electrons?

Finger [1]

The answer is B: 6; 4. Carbon has four electrons in its outermost shell, which are its valence electrons.

8 0

4 years ago

A 3-kg rock falls for 5 sec. What is its speed after falling for 5 sec? Reminder - Acceleration due to gravity, on Earth = 9.8 m

GaryK [48]

Here too when an object is dropped, we know the initial velocity is zero when in free fall. So V =g*t this implies V= 9.8 *5=49m/s..... do you get it?

3 0

3 years ago

On an essentially frictionless, horizontal ice rink, a skater moving at 5.0 m/s encounters a rough patch that reduces her speed

madreJ [45]

Answer:

The length of the rough patch is 4.345 meters.

Explanation:

According to the Work-Energy Theorem, change in kinetic energy is equal to the dissipated work due to friction. That is:

$K_{1} = K_{2} + W_{loss}$

Where:

$K_{1}$ , $K_{2}$ - Initial and final kinetic energy, measured in joules.

$W_{loss}$ - Work losses due to friction.

By applying definitions of kinetic energy and work, the expression described above is expanded:

$\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + f \cdot \Delta s$

Where:

$v_{1}$ , $v_{2}$ - Initial and final speed of the skater, measured in meters per second.

$m$ - Mass of the skater, measured in kilograms.

$\Delta s$ - Length of the rough patch, measured in meters.

$f$ - Friction force, measured in newtons.

According to the statement, friction force is represented by the following expression:

$f = r \cdot m \cdot g$

Where:

$r$ - Ratio of friction force to weight, dimensionless.

$g$ - Gravitational constant, measured in meters per square second.

Then,

$\frac{1}{2}\cdot m \cdot v_{1}^{2} = \frac{1}{2}\cdot m \cdot v_{2}^{2} + r \cdot m \cdot g \cdot \Delta s$

The equation is simplified algebraically and patch length is cleared afterwards:

$\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2}) = r \cdot g \cdot \Delta s$

$\Delta s = \frac{v_{1}^{2}-v_{2}^{2}}{2 \cdot r \cdot g }$

Given that $v_{1} = 5\,\frac{m}{s}$ , $v_{2} = 2.5\,\frac{m}{s}$ , $r = 0.22$ and $g = 9.807 \,\frac{m}{s^{2}}$ , the length of the rough patch is:

$\Delta s = \frac{\left(5\,\frac{m}{s} \right)^{2}-\left(2.5\,\frac{m}{s} \right)^{2}}{2\cdot (0.22)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta s = 4.345\,m$

The length of the rough patch is 4.345 meters.

6 0

3 years ago

Light bending as it passes through a rain drop is an example of...

Dovator [93]

The answer would be C. Refraction

6 0

3 years ago

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What happen to the bulb when it is in series connection?

coldgirl [10]

If the bulb is in series with something else, then . . .

-- The brightness of the bulb depends on the other device in the circuit.

-- If the other device is designed to use less power than the bulb, then the
other device gets more power than the bulb gets.

-- If the other device is designed to use more power than the bulb, then the
other device gets less power than the bulb gets.

-- If the other device is removed from the circuit, then the bulb doesn't light at all.

This description of the often-screwy behavior of a series circuit may partly explain
why the electric service in your home is not a series circuit.

8 0

3 years ago

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