How do i figure this out ? Anyone help me please

Un movil aumenta su velocidad de 10m/s a 20m/s acelerando uniformemente a razon de 5m/s2 ¿que distancia logro en dicha operacion

Kruka [31]

v₀ = initial velocity of the mobile = 10 m/s

v = final velocity of the mobile = 20 m/s

a = acceleration of the mobile = 5 m/s²

d = distance traveled during this operation = ?

Using the kinematics equation

v² = v²₀ + 2 a d

inserting the above values in the equation

20² = 10² + 2 (5) d

400 = 100 + 10 d

subtracting 100 both side

400 - 100 = 100 - 100 + 10 d

300 = 10 d

dividing both side by 10

300/10 = 10 d/10

d = 30 m

hence mobile travels 30 m.

8 0

3 years ago

The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12. How many grainsare there in the ball

Alina [70]

Answer:

the number of grains in the ball is 274,848

Explanation:

Given that;

diameter = 0.5 mm

so radius r = 0.25 mm

first we determine the volume of the ball using the following equation;

V = 4/3×πr³

we substitute

V = 4/3×π(0.25)³

V = 0.06544 mm³

Now form table 1.1 "Grain sizes" a metal with grain size number of 12 has about 4,200,000 grains/mm³

so;

Number of grains N = 0.06544 × 4,200,000

N = 274,848 grains

Therefore, the number of grains in the ball is 274,848

5 0

3 years ago

A mass is hung from a spring and set in motion so that it oscillates continually up and down. The velocity v of the weight at ti

otez555 [7]

To solve the problem it is necessary to identify the equation in the manner given above.

This equation corresponds to the displacement of a body under the principle of simple harmonic movement.

Where,

$\xi = Acos(\omega t +\phi)$

PART A) Our equation corresponds to

$y = -5cos(4\pi t)$

Therefore the value of omega is equivalent to that of

$\omega = 4\pi$

From the definition we know that the period as a function of angular velocity is equivalent to

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{4\pi}$

$T = \frac{1}{2}$

This same point is the equivalent of the maximum point of the speed that the body can reach, since the internal expression of the $cos\theta$ Is equivalent to . So the maximum speed that the body can reach is,

$y = -5cos(4\pi t)$

$y = -5cos(4\pi (1/2))$

$y = -5*(-1)$

$y = 5$

Therefore the maximum felocity will be 5ft / s

PART B) The period of graph is the time taken to reach from one maximum point to next point maximum point, then

$t = \frac{T}{2} = \frac{1}{2}*\frac{1}{2}$

$t = \frac{1}{4}s$

5 0

3 years ago

A string of length L, fixed at both ends, is capable of vibrating at 309 Hz in its first harmonic. However, when a finger is pla

Scorpion4ik [409]

Answer:

i = 0.3326 L

Explanation:

A fixed string at both ends presents a phenomenon of standing waves, two waves with the same frequency that are added together. The expression to describe these waves is

2 L = n λ n = 1, 2, 3…

The first harmonic or leather for n = 1

Wave speed is related to wavelength and frequency

v = λ f

λ = v / f

Let's replace in the first equation

2 L = 1 (v / f₁)

For the shortest length L = L-l

2 (L- l) = 1 (v / f₂)

These two equations form our equation system, let's eliminate v

v = 2L f₁

v = 2 (L-l) f₂

2L f₁ = 2 (L-l) f₂

L- l = L f₁ / f₂

l = L - L f₁ / f₂

l = L (1- f₁ / f₂)

.

Let's calculate

l / L = (1- 309/463)

i / L = 0.3326

4 0

3 years ago

) Suppose a particle travels along a straight line with velocity v(t) = t 2 e −3t meters per second after t seconds. How far doe

pshichka [43]

Answer:

x(t=3s) = 0.07 m to the nearest hundredth

Explanation:

v(t) = t² e⁻³ᵗ

Find displacement after t = 3 s.

Recall, velocity, v = (dx/dt)

v = (dx/dt) = t² e⁻³ᵗ

dx = t² e⁻³ᵗ dt

∫ dx = ∫ t² e⁻³ᵗ dt

This integration will be done using the integration by parts method.

Integration by parts is done this way...

∫ u dv = uv - ∫ v du

Comparing ∫ t² e⁻³ᵗ dt to ∫ u dv

u = t²

∫ dv = ∫ e⁻³ᵗ dt

u = t²

(du/dt) = 2t

du = 2t dt

∫ dv = ∫ e⁻³ᵗ dt

v = (-e⁻³ᵗ/3)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ (-e⁻³ᵗ/3) 2t dt

= (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt

But the integral (∫ 2t (-e⁻³ᵗ/3) dt) is another integration by parts problem.

∫ u dv = uv - ∫ v du

u = 2t

∫ dv = ∫ (-e⁻³ᵗ/3) dt

u = 2t

(du/dt) = 2

du = 2 dt

∫ dv = ∫ (-e⁻³ᵗ/3) dt

v = (e⁻³ᵗ/9)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ 2t (-e⁻³ᵗ/3) dt = 2t (e⁻³ᵗ/9) - ∫ 2 (e⁻³ᵗ/9) dt = 2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)

Putting this back into the main integration by parts equation

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt = (-t²e⁻³ᵗ/3) - [2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)]

x(t) = ∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k (k = constant of integration)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k

At t = 0 s, v(0) = 0, hence, x(0) = 0

0 = 0 - 0 - (2/27) + k

k = (2/27)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + (2/27)

At t = 3 s

x(3) = (-9e⁻⁹/3) - (6e⁻⁹/9) - (2e⁻⁹/27) + (2/27)

x(3) = -0.0003702294 - 0.0000822732 - 0.0000091415 + 0.0740740741 = 0.07361243 m = 0.07 m to the nearest hundredth.

7 0

3 years ago

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