0.000236 kg has how many<br> significant figures?

1. If a fast-moving car making a loud noise approaches and moves past the person, what will happen as the distance between the t

PilotLPTM [1.2K]

<span>So we want to know what will happen when the fast moving car that is making loud noise that is initially approaching the person, passes the person and starts to move away. So Doppler effect is a phenomenon where when the source of a sound is approaching a person, the person hears the sound as higher than if the source was standing still with respect to the person because the wavelength is getting shorter, and as the source is moving avay from the person the sound is getting deeper because the wavelength is getting longer. So the correct answer is A. </span>

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

0.000236 kg has how many significant figures?