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3 years ago

In the following problems, assume the following parameters:

Physics

1 answer:

topjm [15]3 years ago

6 0

Answer:

1) a) I₁ = 0.2941 kg m², b) I₂ = 0.2963 kg m², c) I_{total} = 0.5904 kg m²

3) α = 6.31 10⁶ rad / s²

Explanation:

1) The moment of inertia for bodies with high symmetry is tabulated, for a divo with an axis passing through its center is

I = ½ m r²

a) moment of inertia of the upper disk

I₁ = ½ m₁ r₁²

I₁ = ½ 1,468 0.633²

I₁ = 0.2941 kg m²

b) upper aluminum disc moment of inertia

I₂ = ½ m₂ r₂²

I₂ = ½ 1.479 0.633²

I₂ = 0.2963 kg m²

c) the moment of inertia is an additive scalar quantity therefore

I_{total} = I₁ + I₂

I_{total} = 0.2941 + 0.2963

I_{total} = 0.5904 kg m²

3) ask the value of the angular acceleration, that is, the second derivative of the angle with respect to time squared

indicate the angular velocity of the system w = 400 rev / s

Let's reduce the SI system

w = 400 rev / s (2π rad / rev) = 2513.27 rad / s

as the system is rotating we can calculate the centripetal acceleration

a = w² R

a = 2513.27² 0.633

a = 3.998 10⁶ m / s²

the linear and angular variable are related

a = α r

α = a / r

α = 3.998 10⁶ / 0.633

α = 6.31 10⁶ rad / s²

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Is it true or false that A proton traveling to the right moves in between the two large plates. A vertical electric field, point

Lelu [443]

Answer:

False

Explanation:

In the question it is given that a proton travels in the right moves in between the two plates. An electric field that is pointing in the vertically upward direction has magnitude of 3 N/C. The electric field is produced by the plates.

Now from here we can conclude that the direction of the force acting on the proton when the proton is in between the plates acts in the downward direction. While the electric field lines points in the upward direction. Thus the electric point will not be in the direction of the force that acts on the proton.

3 0

3 years ago

A tank of water is in the shape of a cone (assume the ""point"" of the cone is pointing downwards) and is leaking water at a rat

Inessa05 [86]

Answer:

a) dh/dt = -44.56*10⁻⁴ cm/s

b) dr/dt = -17.82*10⁻⁴ cm/s

Explanation:

Given:

Q = dV/dt = -35 cm³/s

R = 1.00 m

H = 2.50 m

if h = 125 cm

a) dh/dt = ?

b) dr/dt = ?

We know that

V = π*r²*h/3

and

tan ∅ = H/R = 2.5m / 1m = 2.5 ⇒ h/r = 2.5

⇒ h = (5/2)*r

⇒ r = (2/5)*h

If we apply

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒ d(r²*h)/dt = 3*35/π = 105/π ⇒ d(r²*h)/dt = -105/π

a) if r = (2/5)*h

⇒ d(r²*h)/dt = d(((2/5)*h)²*h)/dt = (4/25)*d(h³)/dt = -105/π

⇒ (4/25)(3*h²)(dh/dt) = -105/π

⇒ dh/dt = -875/(4π*h²)

b) if h = (5/2)*r

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒ d(r²*h)/dt = d(r²*(5/2)*r)/dt = (5/2)*d(r³)/dt = -105/π

⇒ (5/2)*(3*r²)(dr/dt) = -105/π

⇒ dr/dt = -14/(π*r²)

Now, using h = 125 cm

dh/dt = -875/(4π*h²) = -875/(4π*(125)²)

⇒ dh/dt = -44.56*10⁻⁴ cm/s

then

h = 125 cm ⇒ r = (2/5)*h = (2/5)*(125 cm)

⇒ r = 50 cm

⇒ dr/dt = -14/(π*r²) = - 14/(π*(50)²)

⇒ dr/dt = -17.82*10⁻⁴ cm/s

4 0

3 years ago

Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at

dexar [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

The amplitude of the lateral force is $F = 25 \ N$

The frequency is $f = 1 \ Hz$

The mass of the bridge per unit length is $\mu = 2000 \ kg /m$

The length of the central span is $d = 144 m$

The oscillation amplitude of the section considered at the time considered is $A = 75 \ mm = 0.075 \ m$

The time taken for the undriven oscillation to decay to $\frac{1}{e}$ of its original value is t = 6T

Generally the mass of the section considered is mathematically represented as

$m = \mu * d$

=> $m = 2000 * 144$

=> $m = 288000 \ kg$

Generally the oscillation amplitude of the section after a time period t is mathematically represented as

$A(t) = A_o e^{-\frac{bt}{2m} }$

Here b is the damping constant and the $A_o$ is the amplitude of the section when it was undriven

So from the question

$\frac{A_o}{e} = A_o e^{-\frac{b6T}{2m} }$

=> $\frac{1}{e} =e^{-\frac{b6T}{2m} }$

=> $e^{-1} =e^{-\frac{b6T}{2m} }$

=> $-\frac{3T b}{m} = -1$

=> $b = \frac{m}{3T}$

Generally the amplitude of the section considered is mathematically represented as

$A = \frac{n * F }{ b * 2 \pi }$

=> $A = \frac{n * F }{ \frac{m}{3T} * 2 \pi }$

=> $n = A * \frac{m}{3} * \frac{2\pi}{25}$

=> $n = 0.075 * \frac{288000}{3} * \frac{2* 3.142 }{25}$

=> $n = 1810 \ people$

3 0

3 years ago

A small object with a 5.0-mC charge is accelerating horizontally on a friction-free surface at 0.0050 m/s2 due only to an electr

kolbaska11 [484]

Answer:

$0.002 N/C$

Explanation:

Parameters given:

Charge of object, q = 5 mC = $5 * 10^{-3} C$

Acceleration of object, a = $0.005 m/s^2$

Mass of object, m = 2.0 g

The Electric field exerts a particular force on the object, causing it to accelerate (Electrostatic force).

We know that Electrostatic force, F, is given in terms of Electric field, E, as:

F = qE

This means that the object exerts a force of -qE on the Electric force (Action with equal and opposite reaction).

The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:

F = ma

Equating the two forces of the object, we get:

-qE = ma

=> $E = \frac{-ma}{q}$

Solving for E, we have:

$E = \frac{-2 * 10^{-3} * 0.005}{5 * 10^{-3}} \\\\\\E = -0.002 N/C$

The magnitude will be:

$|E| = |-0.002| N/C = 0.002 N/C$

The electric field has a magnitude of 0.002 N/C.

4 0

4 years ago

What is the gauge pressure at point A when spigot C is closed?

Nesterboy [21]

Given,

$\begin{gathered} l_1=0.2m \\ l_2=0.07m \\ h=0.47m \\ \rho=\frac{1g}{cm^3} \end{gathered}$

Atm

3 0

2 years ago