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k0ka [10]
3 years ago
10

PLEASE HEEEEEEELP

Physics
1 answer:
Zepler [3.9K]3 years ago
8 0

<h2>\large{\underbrace{\underline{\fcolorbox{White}{pink}{\bf{ANSWER♥︎}}}}}</h2><h3>kinetic energy is given as</h3>

KE = (0.5) m v²

given that : v = speed of the bottle in each case = 4 m/s when m = 0.125 kg

KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J

when m = 0.250 kg KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J

when m = 0.375 kg KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J

when m = 0.0.500 kg KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J

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It's a chemical change.

Explanation:

The ingredients will react together once mixed and heated. Also the ingredients can not be pulled apart.

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If a Ferrari, with an initial velocity of 14m/s, accelerates at a rate of 60
ss7ja [257]

Answer:

254

Explanation:

use the formula "final Velocity- initial velocity / time = acceleration"

so "X - 14 /4 = 60"

60 x 4 = X - 14

240 +14 = X

X = 254

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Which of the following is an example of heat being transferred by radiation? Choose all that apply, Fire, Sun, Candle Flame, Sto
OlgaM077 [116]

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Fire, Sun, Candle Flame, Stove

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3 years ago
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What is the mass of a truck in grams of it produces a force of 1500N while accelerating at a rate of 6 m/s²?​
aleksley [76]

Answer:

250,000

Explanation:

<h2> </h2>

<h2>formula = ( F=ma </h2>

  • F=1500N
  • a=6m/s^2
  • F= ma
  • m=?
  • 1500/6 = m
  • m=250 kg
  • 1kg =1000gm so 250kg =250,000gm
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5 0
3 years ago
A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density
Amiraneli [1.4K]

Answer:

a)      K = 2/3 π G m ρ R₁³ / R₂ ,  b) U = - G m M / r

Explanation:

The law of universal gravitation is

     F = G m M / r²

Part A

Let's use Newton's second law

     F = m a

The acceleration is centripetal

     a = v² / R₂

     

      G m M / R₂² = m v² / R₂

      v² = G M / R₂

They give us the density of the planet

    ρ = M / V

    V = 4/3 π R₁³

    M =   ρ V

    M =   ρ 4/3 π R₁³

    v² = 4/3 π G  ρ R₁³ / R₂

    K = ½ m v²

    K = ½ m (4/3 π G ρ R₁³ / R₂)

    K = 2/3 π G m ρ R₁³ / R₂

Part B

Potential energy and strength are related

     F = - dU / dr

     ∫ dU = - ∫ F. dr

The force was directed towards the center and the vector r outwards therefore there is an angle of 180º between the two cos 180 = -1

    U- U₀ = G m M ∫ dr / r²

    U - U₀ = G m M (- r⁻¹)

We evaluate for

    U - U₀ = -G m M (1 / r_{f} -  1 /r_{i})

They indicate that for ri = ∞     U₀ = 0

    U = - G m M / r

6 0
3 years ago
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