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3 years ago

8

A sphere moves in simple harmonic motion with a frequency of 4.40 Hz and an amplitude of 3.50 cm. (a) Through what total distanc

e (in cm) does the sphere move during one cycle of its motion

1 answer:

Anna [14]3 years ago

6 0

Answer:

It will move a distance of 4 * 3.5 = 4 cm in one cycle.

zero to max A, back to zero, zero to min A, back to zero (4 * A)

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Circular coil 1 has N turns and circular coil 2 has 6N. Coil 1 has area A and coil 2 has area A/4. If Coil 1 has current i runni

solmaris [256]

Answer:

$\frac{L_1}{L_2} =\frac{1}{3}$

Explanation:

Recall that the formula for an inductance (L) for coil on N turns, are A and current I is given by:

$L=\frac{\mu_0\,N^2\,A}{I}$

Then, for the first coil we have;

$L_1=\frac{\mu_0\,N^2\,A}{I}$

and for coil 2 we have:

$L_2=\frac{\mu_0\,(6\,N)^2\,(A/4)}{3\,I}$

then, the quotient L1/L2 can be written as:

$\frac{L_1}{L_2} =\frac{\mu_0\,N^2\,A\,3\,I}{\mu_0\,(6\,N)^2\,(A/4)\,I}=\frac{12}{36} =\frac{1}{3}$

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3 years ago

Where do incident rays that are parallel to the principal axis converge to after reflecting off a concave mirror?

morpeh [17]

Answer:

At focus

Explanation:

A concave mirror is converging in nature. In a mirror, concave in nature, the rays which are parallel to the principal axis are supposed to be coming from very large distances or we assume the source to be placed at infinity for such rays which are parallel to the principal axis.

These rays, parallel to the principal axis, coming from infinity, converges at the focus of the mirror concave in nature after reflecting from the concave mirror

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3 years ago

A horizontal insulating rod of length 11.8-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge.

SCORPION-xisa [38]

Answer:

11.962337 × 10^-4 N

Explanation:

Given the following :

Length L = 11.8

Charge = 29nC = 29 × 10^-9 C

Linear charge density λ = 1.4 × 10^-7 C/m

Radius (r) = 2cm = 2/100 = 0.02 m

Using the relation:

E = 2kλ/r ; F =qE

F = 2kλq/L × ∫dr/r

F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))

2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4

In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214

Hence,

(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N

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4 years ago

The diagram shows a handle with three forces, each 100 N, applied to it. The handle is free to

wolverine [178]

Answer:

с The handle will turn anticlockwise (to the left).

Explanation:

6 0

3 years ago

One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren

katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

$\sigma = \frac{F}{A}$

Where,

$\sigma =$ Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

$\sigma_y = \frac{F_y}{A}$

where,

$\sigma_y =$ Shear strength

$F_y =$ Shear Force

$A_0 =$ Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

$311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}$

Resolving for F,

$F= 1399.5N$

PART B) We need here to apply the shear strength equation, then

$\sigma_y = \frac{F_y}{A}$

$210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}$

$F_y = 945N$

In such a way that the material is more resistant to tensile strength than shear force.

6 0

3 years ago