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3 years ago

If a carbon atom was broken in half, would it still be carbon?

1 answer:

4 0

If a carbon atom was broken in half, would it still be carbon? 99% of all carbon atoms are the isotope carbon-12m containing 6 protons and 6 neutrons. If you split that evenly, the resulting atom would have 3 protons and 3 neutrons, which would be lithium-6.

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Please help I cannot fail!

alekssr [168]

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4 0

2 years ago

In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. the dish is on a shelf above the poi

Georgia [21]

components of the speed of the coin is given as

$v_x = v cos60$

$v_x = 6.4 cos60 = 3.2 m/s$

$v_y = vsin60$

$v_y = 6.4 sin60 = 5.54 m/s$

now the time taken by the coin to reach the plate is given by

$t = \frac{\delta x}{v_x}$

$t = \frac{2.1}{3.2}$

$t = 0.656 s$

now in order to find the height

$h = vy * t + \frac{1}{2} at^2$

$h = 5.54 * 0.656 - \frac{1}{2}*9.8*(0.656)^2$

$h = 1.52 m$

so it is placed at 1.52 m height

3 0

3 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t

enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

$v_{avg} = \frac{6000}{20*60}$

$v_{avg} = 5m/s$

now since effect of wind is in opposite direction then we can say

$V_{net} = v_{bird} - v_{wind}$

$5 = 9 - v_{wind}$

$v_{wind}= 4 m/s$

Part b)

now if bird travels in the same direction of wind then we will have

$v_{net}= v_{bird} + v_{wind}$

$v_{net} = 9 + 4 = 13 m/s$

now we can find the time to go back

$time = \frac{distance}{speed}$

$time = \frac{6000}{13}$

$time = 7.7 minutes$

Part c)

Total time of round trip when wind is present

$T = t_1 + t_2$

$T = 20 + 7.7 = 27.7 min$

now when there is no wind total time is given by

$T = \frac{6000}{9} + \frac{6000}{9}$

$T = 22.22 min$

So due to wind time will be more

4 0

4 years ago

Pure water has a pH of 7. Pure water _______. A. Is a neutral substance B. Could be either an acid or a base C. Is a base D. Is

Hoochie [10]

A. is a neutral substance a pH of 7 describes something of neutral pH where anything less than 7 is an acid and higher is a base

5 0

3 years ago

Read 2 more answers

A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient

gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

6 0

3 years ago