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eduard
3 years ago
12

A block of mass m attached to the end of a spring of spring constant ???? undergoes simple harmonic motion with amplitude ???? a

nd angular frequency ???? . The position of the block is described by a cosine function with an initial phase angle phi=0. Suppose that each of the variables m,????,????, and phi is increased in value.
1. Classify which increases produce an increase in the frequency of oscillation and which do not.Increases frequency Does not increase frequencyOptions:a. increase Ab. increase kc. increase md. increase phi
Physics
1 answer:
nikklg [1K]3 years ago
4 0

Answer:

90

Explanation:

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A toxin that inhibits the production of gtp would interfere with the function of a signal transduction pathway that is initiated
Vaselesa [24]

Answer: G protein linked

Explanation:

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3 years ago
What should a model of the rock cycle show as the process that changes metamorphic rock to igneous rock?
Ratling [72]

Answer:

Rock cycle describes the concept in geology that shows the changes in the rock type in the geologic time.

Many processes undergoes in the rock cycle that leads to transition of rocks, the process that changes metamorphic rock to igneous rock is melting and crystallization. The metamorphic rock melts under high pressure and temperature and which crystallizes into a completely different kind of rock, called igneous rock.

7 0
3 years ago
A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. what is the angular acceleration?
SCORPION-xisa [38]

Answer:

α = -π/3 rad/s²

θ = 1.5π rad ≈ 4.71 rad

θ = 0.75 rev

Explanation:

30 rev/min (2π rad/rev) / (60 s/min) = π rad/s

α = (ωf - ωi) / t = (0 - π) / 3 = -π/3 rad/s²

θ = ½αt² = ½(π/3)3² = 1.5π rad ≈ 4.71 rad

θ = 1.5π rad / 2π rad/rev = 0.75 rev

5 0
2 years ago
The function x = (1.2 m) cos[(3πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 9.7 s, what are the (a) di
nlexa [21]

Answer and Explanation:

Let:

x(t)=Acos(\omega t+ \phi)

The equation representing a simple harmonic motion, where:

x=Displacement\hspace{3}from\hspace{3}the\hspace{3}equilibrium\hspace{3}point\\A=Amplitude \hspace{3}of\hspace{3} motion\\\omega= Angular \hspace{3}frequency\\\phi=Initial\hspace{3} phase\\t=time

As you may know the derivative of the position is the velocity and the derivative of the velocity is the acceleration. So we can get the velocity and the acceleration by deriving the position:

v(t)=\frac{dx(t)}{dt} =- \omega A sin(\omega t + \phi)\\\\a(t)=\frac{dv(t)}{dt} =- \omega^2 A cos(\omega t + \phi)

Also, you may know these fundamental formulas:

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

Now, using the previous information and the data provided by the problem, let's solve the questions:

(a)

x(9.7)=1.2 cos((3 \pi *(9.7))+\frac{\pi}{5} ) \approx -0.70534m

(b)

v(9.7)=-(3\pi) (1.2) sin((3\pi *(9.7))+\frac{\pi}{5} ) \approx 9.1498 m/s

(c)

a(9.7)=-(3 \pi)^2(1.2)cos((3\pi*(9.7))+\frac{\pi}{5} )\approx -62.653m/s^2

(d)

We can extract the phase of the motion, the angular frequency and the amplitude from the equation provided by the problem:

\phi = \frac{\pi}{5}

(e)

f=\frac{\omega}{2 \pi} =\frac{3\pi}{2 \pi} =\frac{3}{2} =1.5 Hz

(f)

T=\frac{2 \pi}{\omega} =\frac{2 \pi}{3 \pi} =\frac{2}{3} \approx 0.667s

8 0
2 years ago
Read 2 more answers
A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the
kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂  : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F  : Force ( N)

d  : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the  bar

mm = 1.5 kg :  mass of the  monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey  (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances  from forces the point 2

d₁₂ = (3-0.6) m = 2.4m  : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium  of moments at the point  2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

5 0
2 years ago
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