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3 years ago

15

All are examples of electric forces except _________. A. a neutron pushing on another neutron B. an electron pushing on a proton

C. an electron pushing on another electron

1 answer:

Contact [7]3 years ago

6 0

Answer:

A) a neutron pushing on another neutron.

Explanation:

Neutrons have no charge, therefore, there is no electric force among them. Protons and electrons on the other hand do have electric charge (electrons negative charge and protons positive charge) that generate electric forces between them that can be repelling forces (if the charges are of the same sign), or attractive forces (if the charges are of opposite signs).

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In 1913 Niels Bohr formulated a method of calculating the different energy levels of the hydrogen atom.

Svetlanka [38]

They are right the answer is A true

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3 years ago

Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic

Nuetrik [128]

Answer:

(a) A = $3.90 \AA$

(b) $A = 4.50 \AA$

(c) $A = 5.51 \AA$

(d) $A = 9.02 \AA$

Solution:

As per the question:

Radius of atom, r = 1.95 $\AA = 1.95\times 10^{- 10} m$

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = $2\times 1.95 = 3.90 \AA$

(b) For body centered cubic lattice:

$A = \frac{4}{\sqrt{3}}r$

$A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA$

(c) For face centered cubic lattice:

$A = 2{\sqrt{2}}r$

$A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA$

(d) For diamond lattice:

$A = 2\times \frac{4}{\sqrt{3}}r$

$A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA$

6 0

3 years ago

When making a turn, do not have the steering wheel turned in the direction of the turn before beginning the turning maneuver.a)

RideAnS [48]

Answer:

a) True.

Explanation:

If you turn the wheel in the direction of the turn before beginning the turning maneuver then it's possible that there might be not enough space available for turning and also if you are waiting for the traffic to get clear with rear ended then it will get pushed forward onto the coming traffic.

7 0

3 years ago

A 3250-kg aircraft takes 12.5 min to achieve its cruising

scoundrel [369]

Answer:

effeciency n = = 49%

Explanation:

given data:

mass of aircraft 3250 kg

power P = 1500 hp = 1118549.81 watt

time = 12.5 min

h = 10 km = 10,000 m

v =85 km/h = 236.11 m/s

$n = \frac{P_0}{P}$

$P_o = \frac{total\ energy}{t} = \frac{ kinetic \energy + gravitational\ energy}{t}$

kinetic energy $= \frac{1}{2} mv^2 =\frac{1}{2} 3250* 236 = 90590389.66 kg m^2/s^2$

kinetic energy $= 90590389.66 kg m^2/s^2$

gravitational energy $= mgh = 3250*9.8*10000 = 315500000.00 kg m^2/s^2$

total energy $= 90590389.66 +315500000.00 = 409091242.28 kg m^2/s^2$

$P_o =\frac{409091242.28}{750} = 545454.99 j/s$

$effeciency\ n = \frac{P_o}{P} = \frac{545454.99}{1118549.81} = 0.49$

effeciency n = = 49%

8 0

3 years ago