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Amiraneli [1.4K]

3 years ago

11

What are the properties of elements based on their placement on the periodic table?

1 answer:

Step2247 [10]3 years ago

4 0

The property are on there hope this helps

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Two identical cars, one on the moon and one on the earth, have the same speed and are rounding banked turns that have the same r

Naily [24]

Answer:

The value of the centripetal forces are same.

Explanation:

Given:

The masses of the cars are same. The radii of the banked paths are same. The weight of an object on the moon is about one sixth of its weight on earth.

The expression for centripetal force is given by,

$F_{c} = \dfrac{mv^{2}}{r}$

where, $m$ is the mass of the object, $v$ is the velocity of the object and $r$ is the radius of the path.

The value of the centripetal force depends on the mass of the object, not on its weight.

As both on moon and earth the velocity of the cars and the radii of the paths are same, so the centripetal forces are the same.

3 0

3 years ago

If we decrease the distance an object moves we will (2 points)

Fantom [35]

Decrease the amount of force applied

8 0

3 years ago

Read 2 more answers

4. Interference is an example of which aspect of electromagnetic radiation?

Lisa [10]

Answer:

D is the answer wave behavior

6 0

3 years ago

Read 2 more answers

Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

ruslelena [56]

Answer:

a

The orbital speed is $v= 2.6*10^{3} m/s$

b

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

7 0

3 years ago

How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33Ã—Ã—10âˆ

AlekseyPX

Answer:

There are 756.25 electrons present on each sphere.

Explanation:

Given that,

The force of repression between electrons, $F=3.33\times 10^{-21}\ N$

Let the distance between charges, d = 0.2 m

The electric force of repulsion between the electrons is given by :

$F=k\dfrac{q^2}{r^2}$

$q=\sqrt{\dfrac{Fr^2}{k}}$

$q=\sqrt{\dfrac{3.33\times 10^{-21}\times (0.2)^2}{9\times 10^9}}$

$q=1.21\times 10^{-16}\ C$

Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :

q = ne

$n=\dfrac{q}{e}$

$n=\dfrac{1.21\times 10^{-16}}{1.6\times 10^{-19}}$

n = 756.25 electrons

So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.

8 0

3 years ago