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2 years ago

How do the nuclei of covalently bonded atoms help keep the bond together.

Physics

2 answers:

Mrac [35]2 years ago

7 0

Answer:

Explanation:

How do the nuclei of covalently bonded atoms help keep the bond together? ... -Positive particles in the nucleus are attracted to shared electrons, so the atoms stay close together. Positive particles in the nucleus are attracted to shared electrons, so the atoms stay close together.

jolli1 [7]2 years ago

6 0

Answer:

oop

Explanation:

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Which of the following tools is used to observe objects that are very far away? A. a anemometer B. a hand lens C. a microscope D

inn [45]

Answer:

D. a telescope

Explanation:

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3 years ago

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Light from two lasers is incident on an opaque barrier with a single slit of width 4.0 x 10^-4 m. One laser emits light of wavel

Sergio [31]

Answer:

a) y = 2.4 x 10⁻³ m = 0.24 cm

b) y = 3.2 x 10⁻³ m = 0.32 cm

Explanation:

The formula of Young's Double Slit experiment will be used here:

$y = \frac{\lambda L}{d}\\\\$

where,

y = distance between dark spots = ?

λ = wavelength

L = distance of screen = 2 m

d = slit width = 4 x 10⁻⁴ m

a) FOR λ = 480 nm = 4.8 x 10⁻⁷ m:

$y = \frac{(4.8\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}$

y = 2.4 x 10⁻³ m = 0.24 cm

a) FOR λ = 640 nm = 6.4 x 10⁻⁷ m:

$y = \frac{(6.4\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}$

y = 3.2 x 10⁻³ m = 0.32 cm

7 0

2 years ago

A 21.7kg child descends a slide 3.5 m high and reaches the bottom with a speed of 2.2m/s. How much thermal energy due to frictio

nadya68 [22]

Answer:

Subtract the kinetic energy at the bottom from the potential energy loss. The remainder becomes frictional heat.

Potential energy loss:

M g H = 21.7*9.81*3.5 = 745.1 J

Kinetic energy at bottom of slide:

= (1/2) M v^2 = 52.5 J

5 0

3 years ago

HELPP ILL GIVE BRIANLISG

STatiana [176]

Answer:

31. Respiratory System

32. Excretory System

33. Nervous System

34. Reproductive system

35. Skeleton System

Explanation:

Hope it helps you.^_^

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2 years ago

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A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in

koban [17]

Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$ , then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$ , then the electric force is:

$F_{e} = E\cdot q$

$F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)$

$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

7 0

2 years ago