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3 years ago

6

Fill in the blanks. Beats only occur when the frequency of two objects is _________________ and when those objects are made to _

________________ together.

1 answer:

Iteru [2.4K]3 years ago

3 0

Beats only occur when the frequency of two objects is interfere with one another and when those objects are made to identical amplitudes together.

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An object is dropped from a height of 25 meters. At what velocity will it hit the ground? A. 7.0 meters/second B. 11 meters/seco

Basile [38]

$Final^{2}=Initial^{2}+2ad \\ x^{2}=0^{2}+2(9.8)(25) \\ x^{2}=490 \\ x=22.13 \\ C$

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3 years ago

Which of these atoms is the most electronegative? select one: a. si b. cl c. p d. f e. c

katrin [286]

The atom that is the most electronegative is fluorine (F).

<h3>What is electronegative?</h3>

Electronegativity, is the tendency for an atom of a given chemical element to attract shared electrons when forming a chemical bond.

Electronegativity increases across the groups from left to right of the periodic table and decreases down the group.

Examples of electronegative elements arranged in decreasing order;

fluorine,
oxygen,
nitrogen,
chlorine,
bromine,
iodine,
sulfur,
carbon, and
hydrogen.

Thus, the atom that is the most electronegative is fluorine (F).

Learn more about electronegativity here: brainly.com/question/24977425

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2 years ago

How does paving parking lots and roads with concrete or asphalt affect surface water and groundwater?

lidiya [134]

Parking lots with roads and concrete reduce infiltration. Infiltration is the process by which water penetrates the soil. Reducing the amount of water that enters the soil can eventually impact groundwater levels in some areas by decreasing it over time. Paved roads lead to increased surface runoff which increases the possibility of flooding in periods of heavy rainfall. This is known as urban flooding.

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3 years ago

A parallel plate capacitor with plates of area A and plate separation d is charged so that the potential difference between its

faust18 [17]

Answer:

Explanation:

Given a parallel plate capacitor of

Area=A

Distance apart =d

Potential difference, =V

If the distance is reduce to d/2

What is p.d

We know that

Q=CV

Then,

V=Q/C

Then this shows that the voltage is inversely proportional to the capacitance

Therefore,

V∝1/C

So, VC=K

Now, the capacitance of a parallel plate capacitor is given as

C= εA/d

When the distance apart is d

Then,

C1=εA/d

When the distance is half d/2

C2= εA/(d/2)

C2= 2εA/d

Then, applying

VC=K

V1 is voltage of the full capacitor V1=V

V2 is the required voltage let say V'

Then,

V1C1=V2C2

V × εA/d=V' × 2εA/d

VεA/d = 2V'εA/d

Then the εA/d cancels on both sides and remains

V=2V'

Then, V'=V/2

The potential difference is half when the distance between the parallel plate capacitor was reduce to d/2

6 0

4 years ago

Read 2 more answers

The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t

Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

$PE=\frac{GMm}{r}$

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

$m = \rho V$

$m = (1030Kg/m^3)(7*10^8m^3)$

$m = 7.210*10^{11}Kg$

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

$r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m$

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

$r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m$

PART A) Potential energy when the ocean is at its furthest point to the moon,

$PE_1 = \frac{GMm}{r_1}$

$PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}$

$PE_1 = 9.05*10^{15}J$

PART B) Potential energy when the ocean is at its closest point to the moon

$PE_2 = \frac{GMm}{r_2}$

$PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}$

$PE_2 = 9.361*10^{15}J$

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

$\Delta KE = \Delta PE$

$\frac{1}{2}mv^2 = PE_2-PE_1$

$v=\sqrt{2(PE_2-PE_1)/m}$

$v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}$

$v = 29.4m/s$

8 0

4 years ago