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3 years ago

A force of 1500 pounds drags a truck 40 feet in 20 seconds. How much work is accomplished? How much power was required?

Physics

2 answers:

mixer [17]3 years ago

6 0

Answer:

The work done and power are 60000 Pounds-feet and 3000 Pounds-feet/sec.

Explanation:

Given that,

Force F = 1500 pounds

Distance d = 40 feet

Time t = 40 sec

Work done :

The work done is the product of the force and distance.

The work done is defined as,

$W = F \times d$

Where, W = work

F = force

d = distance

Put the value into the formula of work

$W= 1500\times40$

$W = 60000\ Pounds-feet$

Power:

The power is equal to the work done divided by time.

The power is defined as,

$P = \dfrac{W}{t}$

Where, P = power

W = work

t = time

Put the value into the formula of power

$P = \dfrac{60000}{20}$

$P=3000\ Pounds-feet/s$

Hence, The work done and power are 60000 Pounds-feet and 3000 Pounds-feet/sec.

svlad2 [7]3 years ago

4 0

Work = f x s = 1500 x 40
power = w/t = (1500 x 40) / 20

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3 years ago

In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh

horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

$Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}$

Where,

$C_d$ is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = $\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2$

A₂ = Area at the throat

A₂ = $\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2$

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

$Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}$

$Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}$

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0

3 years ago

An airplane flies eastward and always accelerates at a constant rate. At one position along its path it has a velocity of 34.3 m

Tomtit [17]

Explanation:

We'll need two equations.

v² = v₀² + 2a(x - x₀)

where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.

x = x₀ + ½ (v + v₀)t

where t is time.

Given:

v = 47.5 m/s

v₀ = 34.3 m/s

x - x₀ = 40100 m

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(47.5)² = (34.3)² + 2a(40100)

a = 0.0135 m/s²

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7 0

4 years ago

If a circuit has a power of 50 W with a current of 4.5 A, what is the resistance in the circuit?

lawyer [7]

Answer:

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2 years ago

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fgiga [73]

Answer:

<h2>The answer is planetary motion</h2>

Explanation:

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2. A line segment joining a planet and the Sun sweeps out equal areas

during equal intervals of time.

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Johannes Kepler was a German astronomer, mathematician, and astrologer

Born: 27 December 1571, Weil der Stadt, Germany

Died: 15 November 1630

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