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tatyana61 [14]
3 years ago
5

True of false all matter requires it’s own space

Physics
1 answer:
aleksley [76]3 years ago
8 0

the answer would be True

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A jet starts at rest at the end of a runway and reaches a speed of 80 m/s in 20 s. What is its acceleration?
ahrayia [7]

Answer:

Acceleration=velocity/time.

=80/2=40m/s^2.

4 0
3 years ago
U. A hockey player takes a slap shot at a puck at rest on
faltersainse [42]

Answer:

a) 7200 ft/s²

b) 140 ft

c) 3.7 s

Explanation:

(a) Average acceleration is the change in velocity over change in time.

a_avg = Δv / Δt

We need to find what velocity the puck reached after it was hit by the hockey player.

We know it reached 40 ft/s after traveling 90 feet over rough ice at an acceleration of -20 ft/s².  Therefore:

v² = v₀² + 2a(x − x₀)

(40 ft/s)² = v₀² + 2(-20 ft/s²)(100 ft − 10 ft)

v₀² = 5200 ft²/s²

v₀ = 20√13 ft/s

So the average acceleration impacted to the puck as it is struck is:

a_avg = (20√13 ft/s − 0 ft/s) / (0.01 s)

a_avg = 2000√13 ft/s²

a_avg ≈ 7200 ft/s²

(b) The distance the puck travels before stopping is:

v² = v₀² + 2a(x − x₀)

(0 ft/s)² = (5200 ft²/s²) + 2(-20 ft/s²)(x − 10 ft)

x = 140 ft

(c) The time the puck takes to travel 10 ft without friction is:

t = (10 ft) / (20√13 ft/s)

t = (√13)/26 s

The time the puck travels over the rough ice is:

v = at + v₀

(0 ft/s) = (-20 ft/s²) t + (20√13 ft/s)

t = √13 s

So the total time is:

t = (√13)/26 s + √13 s

t = (27√13)/26 s

t ≈ 3.7 s

8 0
2 years ago
Fission fusion worksheet answers
fredd [130]
Http://tomschoderbekchem.blogspot.com/2014/09/nuclear-fission-and-fusion-worksheet.html

5 0
3 years ago
What is the threshold velocity vthreshold(water) (i.e., the minimum velocity) for creating Cherenkov light from a charged partic
VladimirAG [237]

Complete Question

The  complete question is shown on the first uploaded image  

Answer:

A

   v_w  =  2.256 *10^{8} \  m/s

B

  v_e  =  2.21 *10^{8} \  m/s

C

The  correct option is  B  

Explanation:

From the question we are told that

      The refractive  index of water is  n_w  =  1.33

      The  refractive  index of ethanol is  n_e  =  1.36

       

Generally the threshold velocity for creating Cherenkov light   from a charged particle as it travels through water is mathematically evaluated as

       v_w  =  \frac{c}{n_w }

Where  c is the speed of light with value  c =  3.0 *10^{8} \  m/s

       v_w  =  \frac{3.0 *10^{8}}{1.33 }

       v_w  =  2.256 *10^{8} \  m/s

Generally the threshold velocity for creating Cherenkov light   from a charged particle as it travels through water is mathematically evaluated as

            v_e  =  \frac{ c}{n_e }

  =>       v_e  =  \frac{3.0 *10^{8}}{1.36 }

=>          v_e  =  2.21 *10^{8} \  m/s

4 0
3 years ago
A bare 4 AWG copper conductor Installed horizontally near the bottom or vertically, and within that portion of a concrete foun d
neonofarm [45]

Answer:

20 ft

Explanation:

This is the length where the direct contact could be used as ground electrode

4 0
3 years ago
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