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lesya692 [45]
3 years ago
6

Circle the letter of the sentence that tells how Bohr' model of the atom differed from Rutherford's model A. Bohr's model focuse

d on the nucleus B. Bohr's model focused on the protons C. Bohr's model focused on the neutrons D. Bohr's model focused on the electrons
Physics
1 answer:
photoshop1234 [79]3 years ago
4 0

Answer:

A. Bohr's model focused on the nucleus

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AVprozaik [17]
42- C
43- A
44- C
45- A
46- B
47- D
7 0
3 years ago
A mass of 100g stretches a spring 5cm. If the mass is set in motion from itsequilibrium position with a downward velocity of 10
stepladder [879]

Answer:

The required IVP is;

u'' + 196u = 0

where;

u(0) = 0 and

u'(0) = -10

Explanation:

See the attached for explanation

3 0
3 years ago
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ELEN [110]

Answer:

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8 0
2 years ago
Calculate a. The heat that must be supplied to a 500.0 g copper kettle containing 400.0 g of water to raise its temperature from
uysha [10]

Explanation:

<h2>For Copper</h2>

dH copper = mCdT copper

<em>(Specific Heat of copper=0.385 J/g C )   </em>

dH = 500 g (0.385 J/g C) (78 C rise)

dH = 15,015 Joules

<h2> For Water</h2>

dH water =  m C dT water

<em>(Specific Heat of copper=</em>4.184 J/g-C<em>)   </em>

<em />

dH = 400 g (4.184 J/g-C) (78 C rise)

dH = 130,540 Joules

total heat = 15,015 + 130,540 = 145,555 Joules

<h2>Percentage for Water  </h2>

(130,540 Joules  / 145,555 Joules) x 100 = 89.7 %

If we consider that we have 3 Significant Figures,

then, the answers become ,

15.0 KJ must be added for Copper

130.5 KJ must be added for Water

and the total of 145 KJ must be added in the kettle with the water

89.7 %  of heat goes to the Water

7 0
3 years ago
A 3.00-kg mass rests on the ground. It is attached to a string which goes vertically to and over an ideal pulley. A second mass
s344n2d4d5 [400]

Answer:A

Explanation:

mass of object=3 kg

distance moved=50 cm

time t=1 s

s=ut+\frac{at^2}{2}

0.5=0+\frac{a\cdot 1^2}{2}

a=1 m/s^2

Let T is the tension in rope

T-mg=ma

Let M be the other mass

Mg-T=Ma

T=M(g-a)

M=m\frac{g+a}{g-a}

M=3\times \frac{9.8+1}{9.8-1}

M=3\times 1.227

M=3.67 kg

8 0
3 years ago
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