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Pachacha [2.7K]
3 years ago
11

How many atoms of carbon, C, are in 0.020 g of carbon?

Physics
1 answer:
9966 [12]3 years ago
6 0

Answer:

9.6352× 10²⁰ C atoms

Explanation:

From the given information,

The molar mass of Carbon = 12 g/mol

number of moles = 0.020g/ 12 g/mol

number of moles = 0.0016 mol

If 1 mole of C = 6.022 × 10²³ C atoms

∴

0.0016 mol of C = (6.022 × 10²³ C atoms/ 1 mol of C)×0.0016 mol of C

= 9.6352× 10²⁰ C atoms

Hence, the number of carbon atoms present in 0.020 g of carbon = 9.6352× 10²⁰ C atoms

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maw [93]
Counter clockwise torque is 360Nm. 
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40 * 9 = 360
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3 years ago
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Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia
Oksanka [162]

Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km  west

velocity V1 = 8.9 km/h

distance B = 4.5 km  east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7     ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x )  / 8.9      .............2

now equating equation 1 and 2

time A = time B

x / 7  =   ( 10.2 - x )  / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

so B meet A 0.01 km east of flagpole

4 0
3 years ago
A professional racecar driver buys a car that can accelerate at 5.9 m/s2. The racer decides to race against another driver in a
3241004551 [841]

Answer:

(a) Time will be t = 3.56 sec

(b) Distance traveled by car when they are side by side is 37.38712 m

(b) Velocity of race car = 21.004 m/sec

velocity of stock car = 12.816 m/sec            

Explanation:

We have given acceleration of the car a_1=5.9m/sec^2

Acceleration of the stock car a_2=3.6m/sec^2

When 1st car overtakes the second car then distance traveled by both the car will be same

(a) So s_1=s_2

As both car starts from rest so initial velocity of both car will be 0 m/sec

It is given that stock car leaves 1 sec before

So \frac{1}{2}\times 5.9\times t^2=\frac{1}{2}\times (t+1)^2\times 3.6

After solving t = 3.56 sec

(b) From second equation of motion s=ut+\frac{1}{2}at^2=0\times 3.56+\frac{1}{2}\times 5.9\times 3.56^2=37.38712m

(c) From first equation pf motion v = u+at

So velocity of race car v = 0+5.9×3.56 = 21.004 m/sec

Velocity of stock car v = 0+ 3.6×3.56 = 12.816 m/sec

3 0
3 years ago
1.15) What distance will 650 joules of work move a box weighing 50 newtons?​
son4ous [18]

Answer:

The answer is 13 however make sure if they ask for a certain measurement like meter answer it by saying 13 meters.

Explanation:

This basically turns into basic algebra if you know the formula for work. The formula for work is W=F*d  

Here are the variables that you know 650J=50N*d so you need d.

All you do is divide 650J by 50N and you get a total of 13 (meters since I don't know what they want you to put it in).

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galina1969 [7]

Answer:

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Explanation:

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The tangential acceleration instead is given by

a_t = \alpha r

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r is the distance from the centre of the merry go round

Since the girl is near the outer edge and the boy is closer to the centre, the value of r for the girl is larger than for the boy, so the girl has greater tangential acceleration.

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