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Marrrta [24]
3 years ago
10

0.000236 kg has how many significant figures?

Physics
1 answer:
SCORPION-xisa [38]3 years ago
4 0

Answer:

6

Explanation:

everything after the decimal is counted

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4 0
3 years ago
Read 2 more answers
Bernice draws an oxygen atom. She draws a small circle for the nucleus. Inside of the circle, she draws plus signs for protons a
Lelu [443]

Answer

It should be A and C

Explanation:

because oxygen is number 8 in the periodic table of elements and has a atomic weight of 15.999 you use those numbers to figure out what is true between those.

The 8 for oxygen goes for the number of electrons and proton and to find neutrons u round the 15.999 up which now make it 16 and subtract it by the 8 now you have 8 protons, 8 neutrons, and 8 electrons

5 0
3 years ago
A 295-kg object and a 595-kg object are separated by 4.10 m.
kodGreya [7K]

Answer:

a)F=3 x 10⁻⁷ N

b)x=2.405 m

Explanation:

Given that

m₁=295 kg

m₂=595 kg

d= 4.1 m

a)

m₃=63 kg

r=d/2 = 2.05 m

The force between the mass m₁ and m₃

F_{13}=\dfrac{Gm_1m_3}{r^2}

by putting the values

F_{13}=\dfrac{Gm_1m_3}{r^2}

F_{13}=\dfrac{6.67\times 10^{-11}\times 295\times 63 }{2.05^2}

F₁₃=2.94 x 10⁻⁷ N

The force  between the mass m₂ and m₃

by putting the values

F_{23}=\dfrac{Gm_2m_3}{r^2}

F_{23}=\dfrac{6.67\times 10^{-11}\times 595\times 63 }{2.05^2}

F₂₃=5.94 x 10⁻⁷ N

The net force F

F= F₂₃- F₁₃

F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N

F=3 x 10⁻⁷ N

b)

Lest take at distance x from mass m₂ net force is zero.

F_{23}=\dfrac{Gm_2m_3}{x^2}

F_{13}=\dfrac{Gm_1m_3}{(4.1-x)^2}

Form above two equation

\dfrac{Gm_1m_3}{(4.1-x)^2}=\dfrac{Gm_2m_3}{x^2}

\dfrac{m_1}{(4.1-x)^2}=\dfrac{m_2}{x^2}

\dfrac{295}{(4.1-x)^2}=\dfrac{595}{x^2}

x²=2.01(4.1-x)²

x=1.42 (4.1-x)

x=5.82 - 1.42x

x=2.405 m

4 0
3 years ago
Number 2<br> Is the answer is right?
garik1379 [7]
In question 2
The second line of equation would be 54 = 108 + 10a
get the rest from that
6 0
3 years ago
A Carnot air conditioner takes energy from the thermal energy of a room at 70°F and transfers it as heat to the outdoors, which
IrinaK [193]

Answer:

Explanation:

Given that,

Hot temperature

T_H = 96°F

From Fahrenheit to kelvin

°K = (°F - 32) × 5/9 + 273

°K = (96 - 32) × 5/9 + 273

K = 64 × 5/9 + 273 = 35.56 + 273

K = 308.56 K

T_H = 308.56 K

Low temperature

T_L = 70°F

Same procedure to Levine

T_L = (70-32) × 5/9 + 273

T_L = 294.11 K

A carnot refrigerator working between a hot reservoir and at temperature T_H and a cold reservoir and at temperature T_L has a coefficient of performance K given by

K = T_L / (T_H - T_L)

K = 294.11 / (308.56 - 294.11)

K = 294.11 / 14.45

K = 20.36

Then, the coefficient of performance is the energy Q_L drawn from the cold reservoir as heat divided by work done,

So, for each joules W = 1J

K = Q_L / W

Then,

Q_L = K•W

Q_L = 20.36 × 1

Q_L = 20.36 J

Q_L ≈ 20J

So, approximately 20J of heats are removed from the room

4 0
3 years ago
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