What type of energy is stored in the nucleus of an atom?

a radio antenna broadcasts a 1.0 MHz radio wav e with 21 kW of power. Assume that the radiation is emitted uniformly in all dire

My name is Ann [436]

Answer:

$I=2.67\times 10^{-6}\ W/m$

Explanation:

Given that,

Frequency of a radio antenna is 1 MHz

Power, P = 21 kW

We need to find the the waves intensity 25 km from the antenna . The object emits intenisty evenly in all direction. It can be given by :

$I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{21\times 10^3}{4\pi (25000)^2}\\\\I=2.67\times 10^{-6}\ W/m$

So, the wave intensity 25 km from the antenna is $2.67\times 10^{-6}\ W/m$ .

5 0

3 years ago

A 62 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 12 m. She falls a total of 3

Andrew [12]

Answer:

k = 104.46 N/m

Explanation:

Here we can use energy conservation

so we will have

initial gravitational potential energy = final total spring potential energy

as we know that she falls a total distance of 31 m

while the unstretched length of the string is 12 m

so the extension in the string is given as

$x = L - L_o$

$x = 31 - 12 = 19 m$

so we have

$mgH = \frac{1}{2}kx^2$

$62(9.81)(31) = \frac{1}{2}k (19^2)$

$k = 104.46 N/m$

5 0

3 years ago

What is energy?

Korvikt [17]

Answer:

c

Explanation:

c. the ability to do work or to produce heat

4 0

3 years ago

*PHYSICS HELP*

sveta [45]

My calculator is about 1cm thick, 7cm wide, and 13cm long.

Its volume is (length) (width) (thick) = (13 x 7 x 1) = 91 cm³ .

The question wants me to assume that the density of my calculator
is about the same as the density of water. That doesn't seem right
to me. I could check it easily. All I have to do is put my calculator
into water, watch to see if sinks or floats, and how enthusiastically.
I won't do that. I'll accept the assumption.

If its density is actually 1 g/cm³, then its mass is about 91 grams.

The choices of answers confused me at first, until I realized that
the choices are actually 1g, 10² g, 10⁴ g, and 10⁶ g.

My result of 91 grams is about 100 grams ... about 10² grams.

Your results could be different.

3 0

3 years ago

Determine the number of bonding electrons and the number of nonbonding electrons in the structure of of2.

Gnoma [55]

OF2 -
O has 6 electrons in outer shell and F has 7 in its outer shell 
Therefore, you have to account for 20 electrons total in the 
structure (7+7+6 = 20) 
therefore draw it linear first. F ---- O-----F 
The two bonds take care of 4 electrons now you have to add another 16. 
Therefore 3 lone pairs on each F and 2 lone pair on O. 
If you check for formal charges, all the atoms are neutral 
F will have 3 lone pairs + 1 bond = 7 electrons (bond = 1/2 electron for formal charge distribution) therefore both the F's are neutral 
Now look at the O: it should have 6.. it has two lone pair and 2 bonds = 4 electrons and 2 bonds = 1 electron each = 2 electrons from bonds = 6 total electrons for formal charge which is exactly the # it should have. There is no need for any double bond in this as there are no charges to be separated. 
Now if u look at the # of domains around O you will see if you include the lone pairs it has a sp3 hybridization (4 domains) therefore a tetrahedron which has 2 lone pairs and 2 bonds.. since there are two lone pairs, the lone pair/bond pair repulsion is so high it is going to repel the two Fluorines and form a bent structure, looks a lot like H2O.

4 0

3 years ago

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