An electric current will always follow?

What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular

monitta

Answer:

a_total = 2 √ (α² + w⁴) , a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

a_total² = a_T²2 + $a_{c}$ ²

where

the centripetal acceleration is

a_{c} = v² / r = w r²

tangential acceleration

a_T = dv / dt

angular and linear acceleration are related

a_T = α r

we substitute in the first equation

a_total = √ [(α r)² + (w r² )²]

a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

w = w₀ + α t

w = 0 + 1.0 2

w = 2.0rad / s

we substitute

a_total = r √(1² + 2²) = r √5

a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

a_total = 2,236 m

7 0

4 years ago

Given the initial wavefunction Ψ (x, 0) = Axexp (-k x) withx> 0 andk> 0, and Ψ (x, 0) = 0 forx <0, what value must A ta

madam [21]

Answer with explanation:

The Normalization Principle states that

$\int_{-\infty }^{+\infty }f(x)dx=1$

Given

$f(x)=xe^{-kx}(x>0\\\\0(x$

Thus solving the integral we get

$\int_{0 }^{+\infty }A\cdot xe^{-kx}dx=1\\\\A\int_{0 }^{+\infty }\cdot xe^{-kx}dx=1$

The integral shall be solved using chain rule initially and finally we shall apply the limits as shown below

$I=\int xe^{-kx}dx\\\\x\int e^{-kx}dx-\int \frac{d(x)}{dx}\int e^{-kx}dx\\\\-\frac{xe^{-kx}}{k}-\int 1\cdot \frac{-e^{-kx}}{k}\\\\\therefore I=\frac{e^{-kx}}{k}-\frac{xe^{-kx}}{k}$

Applying the limits and solving for A we get

$I=\frac{1}{k}[\frac{1}{e^{kx}}-\frac{x}{e^{kx}}]_{0}^{+\infty }\\\\I=-\frac{1}{k}\\\\\therefore A=-k$

3 0

3 years ago

A 0.30 kg softball has a velocity of 15 m/s at an angle of 35 degrees below the horizontal just before making contact with the b

jarptica [38.1K]

Answer:

a) 5.03 kg m/s

b) 10.03 kg m/s

Explanation:

Hi!

Let us consider the origin of coordinates at the pitcher, and pointing directly towards the initial direction of the ball. Therefore, the angle of the velocity with respect to the x axis is -35° (below the horizontal).

The components of the initial momentum are:

px = (0.3 kg)(15 m/s) cos(-35° ) = 4.5 cos(-35° ) kg m/s = 3.69 kg m/s

py = 4.5 sin(-35° ) kg m/s = -2.581 kg m/s

The final momentum will be:

a)

pfx = 0

pfy = - (20 m/s) (0.3 kg) = -6 m/s

And the difference in momentum is:

dpy = pfy - py = -3.419 kg m/s

dpx = pfx - px = -3.69 kg m/s

And its magnitude:

dp = √(dpy^2 + dpx^2) = 5.03 kg m/s

b)

pfx = -6 kg m/s

pfy = 0

And the difference in momentum is:

dpx = pfx - px = -9.69 kg m/s

dpy = pfy - py = 2.581 kg m/s

And its magnitude:

dp = √(dpy^2 + dpx^2) = 10.03 kg m/s

6 0

3 years ago

A flat circular plate of copper has a radius of 0.262 m and a mass of 61.5 kg.

Evgesh-ka [11]

Answer:

0.031 m

Explanation:

Density of copper = ρ = 8960 kg/m³

r = Radius = 0.262 m

m = Mass of plate = 61.5 kg

v = Volume of plate = Volume of cylinder = πr²h

$\rho=\frac{m}{v}\\\Rightarrow 8940=\frac{61.5}{\pi r^2h}\\\Rightarrow h=\frac{61.5}{\pi 0.262^2\times 8940}\\\Rightarrow h=0.031\ m$

So, thickness of plate is 0.031 m

5 0

4 years ago

The hydronium ion concentration of your gastric fluid is 1.0 x 10^-2. you drink a tablespoon of milk of magnesia, which raises t

slamgirl [31]

Answer:

[OH⁻] = 1 × 10⁻¹¹ mol/dm³

Explanation:

Since the hydronium ion concentration of the gastric fluid is [H₃O]⁺ = 1.0 × 10⁻².

The pH at this point is pH = -log₁₀[H₃O]⁺ = -log₁₀[1.0 × 10⁻²] = 2

When milk of magnesia is added, the pH increases by one unit, so the new pH is 1 + 2 = 3

Since pH + pOH = 14, then pOH = 14 - pH = 14 - 3 = 11

The hydroxide ion concentration of the fluid is gotten from

pOH = -log₁₀[OH⁻]

11 = -log₁₀[OH⁻]

-11 = log₁₀[OH⁻]

taking antilog

[OH⁻] = 1 × 10⁻¹¹ mol/dm³

7 0

4 years ago

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