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vichka [17]
2 years ago
7

A body moves in an x-y plane with a velocity vm/s. The component of v are Vx=5m/s, Vy=7m/s along x and y axis respectively. What

is the magnitude and direction of the velocity
Physics
1 answer:
mariarad [96]2 years ago
6 0

the answer is in the picture ok

good luck don't worry it's a correct

#carry on learning

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Calculate the number of electrons passing a point in the wire in 1 min when the current is 1 A
Vera_Pavlovna [14]

Explanation:

When one coulomb charge passes through any cross section of the wire per second,the current passing is one ampere. Charge of electron ,e=1.6X10^-19C. n=1/(1.6X10^-19)=6.25X10^18.Sep 17, 2017

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Choose all the answers that apply. Which of the following occurs during data analysis? observations are made statistical informa
denis23 [38]

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Both of them.

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They are both because when your analyzing data , that is what happen's.

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C. Cables transmit infrared waves over longer distances.

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A mass of 20 g stretches a spring 5 cm. Suppose that the mass is also attached to a viscous damper with a damping constant of 40
11111nata11111 [884]

Quasi frequency = 4√6

Quasi period = π√6/12

t ≈ 0.4045

<u>Explanation:</u>

Given:

Mass, m = 20g

τ = 400 dyn.s/cm

k = 3920

u(0) = 2

u'(0) = 0

General differential equation:

mu" + τu' + ku = 0

Replacing the variables with the known value:

20u" + 400u' + 3920u = 0

Divide each side by 20

u" + 20u' + 196u = 0

Determining the characteristic equation by replacing y" with r², y' with r and y with 1 in the differential equation.

r² + 20r + 196 = 0

Determining the roots:

r = \frac{-20 +- \sqrt{(20)^2 - 4(1)(196)} }{2(1)}

r = -10 ± 4√6i

The general solution for two complex roots are:

y = c₁ eᵃt cosbt + c₂ eᵃt sinbt

with a the real part of the roots and b be the imaginary part of the roots.

Since, a = -10 and b = 4√6

u(t) = c₁e⁻¹⁰^t cos 4√6t + c₂e⁻¹⁰^t sin 4√6t

u(0) = 2

u'(0) = 0

(b)

Quasi frequency:

μ = \frac{\sqrt{4km - y^2} }{2m}

= \frac{\sqrt{4(3929)(20) - (400)^2} }{2(20)} \\\\= 4\sqrt{6}

(c)

Quasi period:

T = 2π / μ

T = \frac{2\pi }{4\sqrt{6} } \\\\T = \frac{\pi\sqrt{6}  }{12}

(d)

|u(t)| < 0.05 cm

u(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t < 0.05

solving for t:

τ = t ≈ 0.4045

8 0
2 years ago
A block of lead with dimensions 2.0 dm x 8.0 cm x 35 mm, has a mass of 6.356 kg. calculate the density of lead in g/cm
insens350 [35]

First, let us calculate for the volume of the block of lead using the formula:

V = l * w * h

But we have to convert all units in terms of cm:

l = 2.0 dm = 20 cm

w = 8 cm

h = 3.5 cm

 

Therefore the volume is:

V = (20 cm) * (8 cm) * (3.5 cm)
V = 560 cm^3

 

Next we convert the mass in terms of g:

m = 6.356 kg = 6356 g

 

Density is mass over volume, so:

density = 6356 g / 560 cm^3

density = 11.35 g / cm^3     (ANSWER)

8 0
3 years ago
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