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netineya [11]
3 years ago
11

WILL MARK BRAINLIEST IF THE ANSWER IS CORRECT Vinegar and baking soda react to form sodium acetate (NaC2H3O2), water (H2O), and

carbon dioxide (CO2). ? NaC2H3O2 + H2O + CO2 How many total sodium (Na) atoms are expected in these reactants? A. 0 B. 1 C. 2 D. 3
Physics
1 answer:
TEA [102]3 years ago
4 0

D. 3, hope this helps

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A sock stuck to the inside of the clothes dryer spins around the drum once every 2.0 s at a distance of 0.50 m from the center o
Rashid [163]

a) 1.57 m/s

The sock spins once every 2.0 seconds, so its period is

T = 2.0 s

Therefore, the angular velocity of the sock is

\omega=\frac{2\pi}{T}=\frac{2\pi}{2.0}=3.14 rad/s

The linear speed of the sock is given by

v=\omega r

where

\omega is the angular velocity

r = 0.50 m is the radius of the circular path of the sock

Substituting, we find:

v=(3.14)(0.50)=1.57 m/s

B) Faster

In this case, the drum is twice as wide, so the new radius of the circular path of the sock is twice the previous one:

r' = 2r = 1.00 m

At the same time, the drum spins at the same frequency as before, therefore the angular frequency as not changed:

\omega' = \omega = 3.14 rad/s

Therefore, the new linear speed would be:

v'=\omega' r' = \omega (2r)

And substituting,

v'=(3.14)(1.00)=3.14 rad/s = 2v

So, we see that the linear speed has doubled.

8 0
3 years ago
Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

Age\ of\ Universe; t = \frac{1}{H_0}

Now,

Hubble's constant; H_0 = 51km/s/Mly

We know that;

1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
3 years ago
A block of mass m = 1.5 kg is released from rest at a height of H = 0.81 m on a curved frictionless ramp. At the foot of the ram
kogti [31]

Answer:

0.31 m

Explanation:

m = mass of the block = 1.5 kg

H = height from which the block is released on ramp = 0.81 m

k = spring constant of the spring = 250 N/m

x = maximum compression of the spring

using conservation of energy

Spring potential energy gained by spring = Potential energy lost by block

(0.5) k x² = mgH

(0.5) (250) x² = (1.5) (9.8) (0.81)

x = 0.31 m

7 0
3 years ago
Coasting due west on your bicycle at 8 m/s, you encounter a sandy patch of road 7.2m
jeka57 [31]

Answer:

V = (v1 + v2) / 2 = (8 + 6.5) / 2 = 7.25 m/s     average speed

t = 7.2 / 7.25 = .993 sec      time to cross patch

a = (v2 - v1) / t = (6.5 - 8) / .993 = -1.51 m/s^2     or 1.5 m/s^2

8 0
2 years ago
Coherent light with wavelength 610 nm passes through two very narrow slits, and theinterference pattern is observed on a screen
andriy [413]

Answer:

= 1220 nm

= 1.22 μm

Explanation:

given data:

wavelength \lambda = 610 nm = 610\times  10 ^{-9} m

distance of screen from slits D = 3 m

1st order bright fringe is 4.84 mm

condition for 1 st bright is

d sin \theta =\lambda     ---( 1)

andtan \theta = \frac{y}{ D}

\theta = tan^{-1}\frac{(y }{D})

= 0.0924 degrees

plug theta value in equation 1 we get

d sin ( 0.0924) = 610 \times 10 ^{-9}

d = 3.78\times 10^{-4} m

condition for 1 st dark fringe

d sin \theta =\frac{λ'}{2}

\lambda '= 2 d sin\theta

= 2λ    since from eq (1)

= 1220 nm

= 1.22 μm

7 0
3 years ago
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