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nataly862011 [7]
2 years ago
13

An electric hoist does 56,447 J of work in raising 115 kg load. How high (in meters) was the load lifted?

Physics
1 answer:
Ugo [173]2 years ago
8 0

\\ \rm\rightarrowtail W=mgh

\\ \rm\rightarrowtail 56447=1150h

\\ \rm\rightarrowtail h=56447/11150

\\ \rm\rightarrowtail h=49.1m

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The equipotential surfaces surrounding a point charge are concentric spheres with the charge at the center. If the electric pote
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Answer:

1.06 m

Explanation:

Since the charge is at the centre of two concentric spheres, we use the formula for electric potential due to a point charge. V = kq/r. Let r₁ be the radius of the sphere with potential, V₁ = 200 V and  r₂ be the radius of the sphere with potential, V₂ = 82.0 V. From V = kq/r, r = kq/V. So that r₁ = kq/V₁ and r₂ = kq/V₂. The magnitude of the difference r₁ - r₂ is the distance between the two surfaces. q the charge equals 1.63 × 10⁻⁸ C

r₂ - r₁ = kq/V₂ - kq/V₁ = kq(1/V₂ - 1/V₁) = 1.63 × 10⁻⁸ × 9 × 10⁹ (1/82 -1/200) m =  1.63 × 10⁻⁸ × 9 × 10⁹ (0.0122 - 0.005) = 1.63 × 10⁻⁸ × 9 × 10⁹(0.0072) m = 1.06 m

The distance between them is 1.06 m

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3 years ago
Which of these people have a stressed induced stomachache
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Molly & Caden have a stressed
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3 years ago
A 5.93 kg ball is attached to the top of a vertical pole with a 2.35 m length of massless string. The ball is struck, causing it
Delvig [45]

Answer

given,

mass of ball = 5.93 kg

length of the string = 2.35 m

revolve with velocity of 4.75 m/s

acceleration due to gravity = 9.81 m/s²

T cos θ = mg

T cos θ = 5.93\times 9.81

T cos θ = 58.17

T sin \theta =\dfrac{mv^2}{r}

T sin \theta =\dfrac{5.93\times 4.75^2}{2.35 sin \theta}

T sin^2 \theta =56.93

sin^2 \theta = 1 - cos^2 \theta

T (1 - cos^2 \theta) =56.93

T (1 - (\dfrac{58.17}{T})^2) =56.93

T² - 56.93T - 3383.75 = 0

T =  93.22 N

cos \theta = \dfrac{58.17}{93.22}

θ = 51.39°

6 0
3 years ago
When the distance between two masses is doubled, the gravitational attraction between
mrs_skeptik [129]

The new gravitational attraction will be 1/4 as much

Explanation:

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

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

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, the original force between the two objects is F, when they are separated by a distance r.

Later, the distance between the two objects is doubled, so the new distance is

r'=2r

Therefore, the new force will be

F'=G\frac{m_1 m_2}{(2r)^2}=\frac{1}{4}(\frac{Gm_1 m_2}{r^2})=\frac{F}{4}

Therefore, the new force will be one-fourth as much.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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