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3 years ago

How much power is needed to lift a 20-kg object to a height of 4.0 m in 2.5 seconds?

Physics

1 answer:

4vir4ik [10]3 years ago

5 0

Answer:

313.92w

Explanation:

Formula for power:

P=W/∆t = Fv

Givens:

m=20kg

∆y=4.0m

∆t=2.5s

a=9.81m/s²

In order to find power, we first need to solve for work.

W=Fd (force*displacement), f=mg

W=mg∆y

W=(20kg)(9.81m/s²)(4.0m)

W=784.8J

P=W/∆t

P=784.8J/2.5s

P=313.92 watts

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3 years ago

Please i need your help w/h/w you make the following measurements if an object:42kg, and 22m

Rufina [12.5K]

Answer:

$1.9 kg/m^3$

Explanation:

The density of an object is given by

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m = 42 kg

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3 years ago

The first-order rearrangement of CH3NC is measured to have a rate constant of 3.61 × 10–15 s–1 at 298 K and a rate constant of 8

netineya [11]

Answer:

The activation energy for this reaction, Ea = 159.98 kJ/mol

Explanation:

Using the Arrhenius equation as:

$ln\frac {K_2}{K_1}=-\frac {E_a}{R}\times (\frac {1}{T_2}-\frac {1}{T_1})$

Where, Ea is the activation energy.

R is the gas constant having value 8.314 J/K.mol

K₂ and K₁ are the rate constants

T₂ and T₁ are the temperature values in kelvin.

Given:

K₂ = 8.66×10⁻⁷ s⁻¹ , T₂ = 425 K

K₁ = 3.61×10⁻¹⁵ s⁻¹ , T₁ = 298 K

Applying in the equation as:

$ln\frac {8.66\times 10^{-7}}{3.61\times 10^{-15}}=-\frac {E_a}{8.314}\times (\frac {1}{425}-\frac {1}{298})$

Solving for Ea as:

Ea = 159982.23 J /mol

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3 years ago

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3 years ago

Read 2 more answers

Two football players with mass 75kg and 100kg run directly toward each other with speeds of 6 m/s and 8 m/s respectively, If the

Rina8888 [55]

Answer:

2 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = total momentum after collision

mu+m'u' = V(m+m') .................................Equation 1

Where m = mass of the first player, u = initial speed of the first player, m' = mass of the second player, u' = initial speed of the second player, V = combined speed of both players.

Making V the subject of the equation,

V = (mu+m'u')/(m+m')................ Equation 2

Note: Taking the direction of the first player as positive.

Given: m = 75 kg, m' = 100 kg, u = 6 m/s, u' = -8 m/s (opposite the first player),

Substituting into equation 2

V = [(75×6)+(100×(--8))]/(75+100)

V = (450-800)/175

V = 350/175

V = - 2 m/s.

Note: The negative signs tells that the combined speed is in the direction of the second player.

Hence the combined speed of the two players = 2 m/s

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4 years ago