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1 year ago

A 110-pound person pulls herself up 4.0 feet. This took her 2.5 seconds. How much power was developed?

Physics

2 answers:

mrs_skeptik [129]1 year ago

5 0

Answer:

0.32 Hp is the answer. right

lesya [120]1 year ago

4 0

Answer: Power developed by her is <u><em>0.32 hp</em></u>

Explanation:

Work done , W= Force(F) X Displacement (d) in the direction of force

Thus work done , $W=Fd=110\times 4.0 ft-lbf = 440 ft-lbf$

Now 1.0 ft-lbf = 1.355 joules of energy

Therefore , work done, W= 440x 1.355 J = 596.2 J

Average Power , $P=\frac{work done(W)}{time (t)}$

=> $P=\frac{596.2}{2.5} watts = 238.5 watts$

We know 746 hp = 1.0 watts

=> $P=\frac{238.5}{746} hp=0.32 hp$

Thus average power developed by her is 0.32 hp

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Answer:

0.018 J

Explanation:

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Answer:

90.77%

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Explanation:

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Total Number of production time = 205hours

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2 years ago

The position function x(t) of a particle moving along an x axis is x = 4.00 - 6.00t2, with x in meters and t in seconds. (a) at

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The position function x(t) of a particle moving along an x axis is $x=4.00 - 6.00t^2$

a) The point at which particle stop, it's velocity = 0 m/s

So dx/dt = 0

0 = 0- 12t = -12t

So when time t= 0, velocity = 0 m/s

So the particle is starting from rest.

At t = 0 the particle is (momentarily) stop

b) When t = 0

$x=4.00 - 6.00*0^2 = 4m$

SO at x = 4m the particle is (momentarily) stop

c) We have $x=4.00 - 6.00t^2$

At origin x = 0

Substituting

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t = 0.816 seconds or t = - 0.816 seconds

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2 years ago

Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te

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Answer:

Please see attachment

Explanation:

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2 years ago

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

$F_{No}= 7771.125 \ N$

$F_p = 2.2*10^{6} N$

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

2 years ago