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Anestetic [448]
3 years ago
10

How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.​

Physics
1 answer:
Rasek [7]3 years ago
6 0

Answer:

t = 4.17 [s]

Explanation:

We know that work is defined as the product of force by distance.

W = F*d

where:

F = force [N] (units of Newtons)

d = distance = 6.34 x 10⁴ [mm] = 63.4 [m]

In order to find the force, we must determine the weight of the box, the weight can be determined by means of the product of mass by gravitational acceleration.

w = m*g

where:

m = mass = 1.47 x 10⁴ [g] = 14.7 [kg]

g = gravity acceleration = 9.81 [m/s²]

w = 14.7*9.81

w = 144.2 [N]

Therefore the work can be calculated.

W = w*d

W = 144.2*63.4

W = 9142.72 [J] (units of Joules)

Power is now defined in physics as the relationship of work at a given time

P = W/t

where:

P = power = 2190 [W]

t = time [s]

Now clearing t, we have.

t = W/P

t = 9142.72/2190

t = 4.17 [s]

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6 0
3 years ago
Determine the mass of Ar in 1.00 liter of a gas mixture at 25oC which contains 0.300 atm of Ne and has a total pressure of 4.00
sergejj [24]

Answer:

The mass of Ar is 36.91g

Explanation:

The gas mixture consist of Neon(Ne) and Argon(Ar)

Partial pressure of Ar = total pressure of mixture - partial pressure = 4 - 0.3 = 3.7 atm

Mole fraction of Ar = partial pressure of Ar ÷ total pressure of mixture = 3.7/4 = 0.925

Mass of Ar = 0.925 × molecular weight of Ar = 0.925 × 39.9 = 36.91g

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3 years ago
What two quantities must stay the same in order for an object to have a constant velocity? A) The speed and kinetic energy must
GarryVolchara [31]

the answer is c) the speed and direction of travel must be constant


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3 years ago
Un globo de aire caliente tiene un volumen de 560 mL a la presión atmosférica normal y una temperatura del aire de 400 ºC. Cuand
dalvyx [7]

Explanation:

charles law V1/T1 =V2/T2

560 x 673 =V2/973

376880 = V2/973

V2 = 376880 x 973 = 366704240mL

4 0
3 years ago
A rock climber hangs freely from a nylon rope that is 15 m long and has a diameter of 8.3 mm. If the rope stretches 5.1 cm, what
irinina [24]

Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

        \Delta L=\frac{PL}{AE}

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

       A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2

Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

       0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg  

Mass of the climber = 69.38 kg

6 0
3 years ago
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