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3 years ago

The person in the middle weighs 500 n, and the tension on the rope is 400 n. what is the weight of the scaffolding?

1 answer:

7 0

Typically a scaffolding has two ropes to support it. One on the left most side while the other one is on the right most side. Based on equilibrium rule, the summation of forces must be 0. The downward force is the weight of person and scaffolding, while the upward force are the tensions on the two ropes, hence:

Weight person + Weight scaffolding = 2 * Tension

500 N + Weight scaffolding = 2 * 400 N

Weight scaffolding = 300 N

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Air "breaks down" when the electric field strength reaches 3 × 106 n/c, causing a spark. A parallel-plate capacitor is made from

Lera25 [3.4K]

Electric field between the plates of parallel plate capacitor is given as

$E = \frac{Q}{A\epsilon_0}$

here area of plates of capacitor is given as

$A = 0.055 * 0.055$

$A = 3.025 * 10^{-3}$

also the maximum field strength is given as

$E = 3 * 10^6 N/C$

now we will plug in all data to find the maximum possible charge on capacitor plates

$3 * 10^6 = \frac{Q}{3.025 * 10^{-3}*8.85 * 10^{-12}}$

$Q = 8.03 * 10^{-8} C$

so the maximum charge that plate will hold will be given by above

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3 years ago

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Answer:

B convection MERRY CHRISTMAS

Explanation:

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2 years ago

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Mila [183]

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2 years ago

In a certain two slit diffraction experiment, two slits 0.02mm wide are spaced0.2mm between centers.(a) How many fringes appear

Kamila [148]

Answer:

a) m = 10 and b) λ = 3.119 10⁻⁷ m

Explanation:

In the diffraction experiments the maximums appear due to the interference phenomenon modulated by the envelope of the diffraction phenomenon, for which to find the number of lines within the maximum diffraction center we must relate the equations of the two phenomena.

Interference equation d sin θ = m λ

Diffraction equation a sin θ = n λ

Where d is the width between slits (d = 0.2 mm), a is the width of each slit (a = 0.02 mm). θ is the angle, λ the wavelength, m and n are an integer.

Let's find the relationship of these two equations

d sin θ / a sin θ = m Lam / n Lam

The first maximum diffraction (envelope) occurs for n = 1, let's simplify

d / a = m

Let's calculate

m = 0.2 / 0.02

m = 10

This means that 10 interference lines appear within the first maximum diffraction.

b) let's use the interference equation, remember that the angles must be given in radians

θ = 0.17 ° (π rad / 180 °) = 2.97 10⁻³ rad

d sin θ = m λ

λ = d sin θ / m

λ = 0.2 10⁻³ sin (2.97 10⁻³) / 2

λ = 3.119 10⁻⁷ m

8 0

3 years ago

Problem 8 I estimate that the Gauss gun (a solenoid) is wound with 500 turns over a distance of 15cm with an average radius of 1

stellarik [79]

Answer:

Energy stored, U = 66.6 J

Explanation:

It is given that,

Number of turns in the solenoid, n = 500

Radius of solenoid, r = 1.5 cm = 0.015 m

Distance, d = 15 cm = 0.15 m

Let U is the energy stored in the solenoid. Its formula is given by :

$U=\dfrac{1}{2}LI^2$

L is the self inductance of the solenoid

$L=\mu_o N^2A d$

N is the no of turns per unit length

$L=\mu_o (n/d)^2A d$

$L=\dfrac{4\pi\cdot10^{-7}\cdot500^{2}\cdot\pi\cdot\left(0.015\right)^{2}}{0.15}$

L = 0.00148 Henry

$U=\dfrac{1}{2}\times 0.00148\times 300^2$

U = 66.6 J

Out of given options, the correct option for the energy stored in the solenoid is 70 J. So, the correct option is (a) "70 J".

6 0

3 years ago