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trasher [3.6K]
3 years ago
8

how much force would be required to produce 88 j of work when pushing a box 1.1meters at an angle of 10 degrees?

Physics
1 answer:
ycow [4]3 years ago
5 0

Answer:81.235N

Explanation:

Work=88J

theta=10°

distance=1.1 meters

work=force x cos(theta) x distance

88=force x cos10 x 1.1 cos10=0.9848

88=force x 0.9848 x 1.1

88=force x 1.08328

Divide both sides by 1.08328

88/1.08328=(force x 1.08328)/1.08328

81.235=force

Force=81.235

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The answer for this question should be TRUE
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A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
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Answer:

a) -2.516 × 10⁻⁴ V

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Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

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R = 1.29 cm

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E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

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3 years ago
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Answer:

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Sunspots are the spot that appears on the sun, this spot appears darker than the surrounding surface of the sun.

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The correct answer is

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Answer:

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Pls check the attached file for step by step solution of the question. It is submitted in this way as typing the equation may not be explanatory.

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