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1 year ago

10

A refrigerator with a weight of 1,127 newtons needs to be moved into a house using a ramp. The length of the ramp is 2.1 meters,

and its height is 0.85 meters. What is the percentage efficiency if a force of 496 newtons is applied to the refrigerator?
A. 44 percent
B. 58 percent
C. 64 percent
D. 76 percent
E. 92 percent

2 answers:

irinina [24]1 year ago

7 0

The answer will be D. 76 percent since 496 is out of 1,127N

lina2011 [118]1 year ago

3 0

496/1127 = 0.44 = 44%

<span>sin A = 0.85/2.1. </span>
<span>A = 23.9o. </span>

<span>Fp = 1127 sin23.9 = 457 N. = Force parallel to the ramp. </span>

<span>Fn = 1127 Cos23.9 = 1,030 N. = Force </span>
<span>perpendicular to the ramp = Normal force. </span>

<span>Eff. = Fp/Fap = 457/496 = 0.92 = 92%
Correct answer: 92%</span>

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Answer:

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Given;

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Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much l

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a) 4F0

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Explanation:

a)

The magnitude of the gravitational force between two objects is given by the formula

$F=G\frac{m_1 m_2}{r^2}$

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G is the gravitational constant

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For the system planet A - Star A, we have:

$m_1=M_p\\m_2 = M_s\\r=R$

So the force is

$F_A=G\frac{M_p M_s}{R^2}=F_0$

For the system planet B - Star B, we have:

$m_1 = 4 M_p\\m_2 = M_s\\r=R$

So the force is

$F=G\frac{4M_p M_s}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.

For the system planet C - Star C, we have:

$m_1 = M_p\\m_2 = 4M_s\\r=R$

So the force is

$F=G\frac{M_p (4M_s)}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.

b)

The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:

$G\frac{mM}{r^2}=m\frac{v^2}{r}$

where

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M is the mass of the star

v is the tangential speed

We can re-arrange the equation solving for v, and we find an expression for the speed:

$v=\sqrt{\frac{GM}{r}}$

For System A,

$M=M_s\\r=R$

So the tangential speed is

$v_A=\sqrt{\frac{GM_s}{R}}$

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$M=M_s\\r=R$

So the tangential speed is

$v_B=\sqrt{\frac{GM_s}{R}}=v_A$

So, the speed of planet B is the same as planet A.

For system C,

$M=4M_s\\r=R$

So the tangential speed is

$v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A$

So, the speed of planet C is twice the speed of planet A.

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1 year ago

An airplane travels horizontally at a constant velocity v. An object is dropped from the plane and one second later another obje

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Answer:

the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

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so we can say that two object when dropped after some interval of time then they always lie in same vertical line

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So the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

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2 years ago

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Answer:

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v = Final velocity

$s=ut+\frac{1}{2}at^2\\\Rightarrow 3.7=0\times 6.1+\frac{1}{2}\times a\times 6.1^2\\\Rightarrow a=\frac{3.7\times 2}{6.1^2}\\\Rightarrow a=0.198871271164\ m/s^2$

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$F=ma+f\\\Rightarrow F=74000\times 0.198871271164+14000\\\Rightarrow F=28716.4740661\ N$

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Work done is given by

$W=Fs\\\Rightarrow W=28716.4740661\times 3.7\\\Rightarrow W=106250.954045\ J$

The fraction is given by

$\dfrac{54450.9540448}{106250.954045}=0.512474965841$

The teams 51.2474965841% of the work goes to kinetic energy of the plane.

3 0

2 years ago

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Hoochie [10]

Answer:

A. 2.57 K

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From specific heat capacity,

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Given: Q = 300 J, m = 500 g = 0.5 kg

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ΔT = 2.57 K

Hence the right option is A. 2.57 K

3 0

2 years ago