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4 years ago

% = Wo/Wi x 100 Solve for Wo % = Wo/Wi x 100 Solve for Wi

1 answer:

8 0

1) % = (Wo /Wi) * 100

Solve for Wo => Wo = (% / 100) * Wi

For example, % =30% and Wi = 250 => Wo = (30 /100) * 250 = 0.30 * 250 = 75

Wo = 75

2) % = (Wo / Wi) * 100

Solve for Wi

=> Wi = Wo * (%/100)

For example, Wo = 125 and % = 40%

=> Wi = 125 * (40 / 100) = 125 * 0.40 = 50

Wi = 50

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Which of the the input Shelly does on a rake is 80J the output work the rate does on the leaves that 70 J what is the efficiency

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Answer: The efficiency Shelly does is 87.5%.

(70J/80J) x 100% = 87.5%

Answer 2: Gravitational potential energy would be considered an object 10 meters above the ground.

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3 years ago

An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t

Gelneren [198K]

Answer:

$1650\:\mathrm{m}$

Explanation:

We can use the following kinematics equations to solve this problem:

$v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x$ .

Using the first one to solve for acceleration:

$132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}$ .

Now we can use the second equation to solve for the distance travelled by the airplane:

$132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}$ (three significant figures).

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3 years ago

A spaceship is traveling through deep space towards a space station and needs to make a course correction to go around a nebula.

expeople1 [14]

Answer:

Magnitude = 3.64 × $10^6$

စ = 43.9°

Explanation:

given data

ship to travel = 1.7 × $10^6$ kilometers

turn = 70°

travel an additional = 2.7 × $10^6$ kilometers

solution

we will consider here

Px = 1.7 × $10^6$

Py = 0

Qx =2.7 × $10^6$ cos(70)

Qy= 2.7 × $10^6$ sin(70)

so that

Hx = Px + Qx ............1

Hx = 2.62 × $10^6$

and

Hy = Py + Qy ..........2

Hy = 2.53 × $10^6$

so Magnitude = $\sqrt{((2.62 \times 10^6 )^2+(2.53 \times 10^6)^2)}$

Magnitude = 3.64 ×

so direction will be

tan စ = Hy ÷ Hx ......................3

tan စ = $\frac{2.53}{2.62}$

tan စ = 0.9656

စ = 43.9°

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3 years ago

A satellite moves in a circular orbit at a constant speed v0 around Earth at a distance R from its center. The force exerted on

postnew [5]

Answer:

vT = v0/3

Explanation:

The gravitational force on the satellite with speed v0 at distance R is F = GMm/R². This is also equal to the centripetal force on the satellite F' = m(v0)²/R

Since F = F0 = F'

GMm/R² = m(v0)²/R

GM = (v0)²R (1)

Also, he gravitational force on the satellite with speed vT at distance 3R is F1 = GMm/(3R)² = GMm/27R². This is also equal to the centripetal force on the satellite at 3R is F" = m(vT)²/3R

Since F1 = F'

GMm/27R² = m(vT)²/3R

GM = 27(vT)²R/3

GM = 9(vT)²R (2)

Equating (1) and (2),

(v0)²R = 9(vT)²R

dividing through by R, we have

9(vT)² = (v0)²

dividing through by 9, we have

(vT)² = (v0)²/9

taking square-root of both sides,

vT = v0/3

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3 years ago

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3 years ago

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