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Sergio [31]
3 years ago
5

you (60 kg) are standing in a (500 kg) elevator that is moving upwards from a ground floor on a building what is the power ratin

g of the motor that can lift this elevator if the elevator travels a distance of 20m upwards in 15 s
Physics
1 answer:
PtichkaEL [24]3 years ago
5 0

Explanation:

Power = work / time

Power = force × distance / time

P = (650 kg) (10 m/s²) (20 m) / (15 s)

P = 8667 W

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Your complete chemical equation is CH4+O2=CO3+H2O
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a plane travels 204 km, northeast in 15.0 minutes. It also increases elevation by 1.6 km, upward in the same amount of time. Wha
LUCKY_DIMON [66]

Answer:

230 m/s northeast, 1.8 m/s up

Explanation:

204 kilometres = 204000 metres

15.0 minutes = 900 seconds

Velocity = Distance / Time

= 204000 / 900

= 230 m/s northeast (to 2 sf.)

1.6km = 1600 metres

Velocity = 1600 / 900

= 1.8 m/s up (to 2 sf.)

Read more on Brainly.com - brainly.com/question/13863590#readmore

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3 years ago
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Suppose you exert a 25-N force to lift a ball 0.4 m in 2 s. How much work is done?
Sergio [31]

work is force x distance = 25 x 0.4

= 2.5x4 = 10joules

pwer would be 10j/2s watts .... 5 watts

3 0
3 years ago
Which statement best explains why atoms form chemical bonds with other atoms?
Alex73 [517]
Because atoms is something that pops or has bubbles in it
7 0
3 years ago
A rifle bullet with mass 8.00g strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless, horizontal s
mafiozo [28]

Answer:

block velocity   v = 0.09186 = 9.18 10⁻² m/s  and speed bollet   v₀ = 11.5 m / s

Explanation:

We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.

Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)

Before the crash

     p₀ = m v₀ + 0

After the crash

   p_{f} = (m + M) v

    p₀ = p_{f}

    m v₀ = (m + M) v                    (1)

Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring

Initial

    Em₀ = K = ½ m v2

Final

    E m_{f}= Ke = ½ k x2

   Emo = E m_{f}

   ½ m v² = ½ k x²

   v² = k/m  x²

Let's look for the spring constant (k), with Hook's law

   F = -k x

   k = -F / x

   k = - 0.75 / -0.25

   k = 3 N / m

Let's calculate the speed

  v = √(k/m)   x

  v = √ (3/8.00)   0.15

  v = 0.09186 = 9.18 10⁻² m/s

This is the spped of  the  block  plus bullet rsystem right after the crash

We substitute calculate in equation  (1)

   m v₀ = (m + M) v

  v₀ = v (m + M) / m

  v₀ = 0.09186 (0.008 + 0.992) /0.008

  v₀ = 11.5 m / s

6 0
3 years ago
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