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Hatshy [7]
4 years ago
12

On a standard number line, the reference point is usually _____

Physics
1 answer:
KATRIN_1 [288]4 years ago
5 0
The reference point would be a. 0
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Be sure to answer all parts. Assume the diameter of a neutral helium atom is 1.40 × 102 pm. Suppose that we could line up helium
ss7ja [257]

Answer:

The number of atoms are 1.86\times10^{8}.

Explanation:

Given that,

Diameter D = 1.40\times10^{2}\ pm

D=1.40\times10^{2}\times10^{-12}\ m

Distance = 2.60 cm

We calculate the number of atoms

Using formula of numbers of atoms

Number\ of\ atoms =\dfrac{2.60\times10^{-2}}{1.40\times10^{2}\times10^{-12}}

Number\of\atoms =1.86\times10^{8}

Hence, The number of atoms are 1.86\times10^{8}.

8 0
3 years ago
You are 12 miles north of your base camp when you begin walking north at a speed of 2mph. what is your location, relative to you
Leona [35]
Http://www.calculator.net/pace-calculator.html?ctype=distance&ctime=05%3A00%3A00&cdistance=5&cdistanceunit=Miles&cpace=02%3A00%3A00&cpaceunit=tpm&printit=0&x=87&y=24 a pace calculator
4 0
3 years ago
A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 13 m/s, skate
kolbaska11 [484]

Let us say that:

1 = 1st player notation

2 = 2nd player notation (the opponent)

 

a. First let us establish the distance travelled by the 2nd player:

d2 = 13 m/s * (t + 1.5)

d2 = 13 t + 19.5

 

Then the distance of the 1st player:

d1 = v0 t + 0.5 a t^2         (v0 initial velocity = 0 since he started from rest)

d1 = 0.5 * 4 m/s^2 * t^2

d1 = 2 t^2

 

The two distances must be equal, d1 = d2:

2 t^2 = 13 t + 19.5

t^2 – 6.5 t = 9.75

Completing the square:

(t – 3.25)^2 = 9.75 + (- 3.25)^2

t – 3.25 = ±4.5

t = -1.25, 7.75

Since time cannot be negative, therefore:

t = 7.75 seconds

 

So he catches his opponent after 7.75 seconds.

 

b. Using the equation:

d1 = 2 t^2

d1 = 2 * (7.75)^2

d1 = 120.125 m

 

So he travelled about 120.125 meters when he catches up to his opponent.

3 0
3 years ago
A soccerball is kicked straight out from a hill at 15 m/s and lands 42m away. how tall is the hill
ozzi

Answer:

The height of the hill is, h = 38.42 m

Explanation:

Given,

The horizontal velocity of the soccer ball, Vx = 15 m/s

The range of the soccer ball, s = 42 m

The projectile projected from a height is given by the formula

                             S = Vx [Vy + √(Vy² + 2gh)] / g

Therefore,

                            h = S²g/2Vx²                     (Since Vy = 0)

Substituting the values

                             h = 42² x 9.8/ (2 x 15²)

                                = 38.42 m

Hence, the height of the hill is, h = 38.42 m

4 0
4 years ago
Please help me! I don’t understand how to solve this problem.
tester [92]
To solve these problems first draw the free body diagram:

8 0
4 years ago
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