Which example involves the transformation of chemical energy directly into light energy?

An upright spring with a 96g mass on it is compressed 2 cm. When

Alexeev081 [22]

Answer:

I only know answer A and it's 2825.28 N/m, with rounding it's 2825.5

Explanation:

Use the m*g*h=1/2*k*x^2 equation

96*9.81*60=1/2*k*2^2

5650.56=2k

5650.56/2=2825.28N/m

8 0

3 years ago

A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp

amm1812

Answer:

the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Explanation:

Given that,

mass, m = 1kg

spring constant k = 21N/M

damping force = $-\beta\frac{dx}{dt} = \frac{-10dx}{dt}$

$\beta = 10$

By Newtons second law ,

The diffrential equation of motion with damping is given by

$m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}$

substitute the value of m =1kg, k = 21N/M, and $\beta = 10$

$1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}$

$\frac{d^2x}{dt^2} + 10\frac{dx}{dt} + 21x = 0$

suppose the equation of the form $x =e^m^t$ ,

and the auxilliary equation is given by

$m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3$

The general solution for the above differential equation is

$x(t) =C_1e^{-3t}+C_2e^{-7t}$

Derivate with respect to t

$x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}$

(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

$x'(0) = 0$

substitute the initial condition

$C_1 +C_2 = 1$

$-3C_1-7C_2=0$

Solve the above equation to get C₁ and C₂

$C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}$

substitute for C₁ and C₂ in general solution

$x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Thus the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

3 0

3 years ago

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A man stands still on a moving walkway that is going at a speed of 0.3 m/s to the south. What is the velocity of the man accordi

RSB [31]

The velocity is 0.3 m/s South.

3 0

3 years ago

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Which equation is correct according to Ohm’s law? Which equation is correct according to Ohm’s law? A.) V = IR B.) I = R/V C.) R

vodomira [7]

Answer:

$V = IR$

Explanation:

Required

Which equation represents ohm's law?

Literally, ohm's law implies that current (I) is directly proportional to voltage (V) and inversely proportional to resistance (R).

Mathematically, this can be represented as:

$I\ \alpha\ \frac{V}{R}$

Convert the expression to an equation

$I\ =\ \frac{V}{R}$

Multiply both sides by R to make V the subject

$I\ * R\ =\ \frac{V}{R} * R$

$I\ * R\ =V$

Reorder

$V = I\ * R$

$V = IR$

<em>Option (a) is correct; Others are not</em>

6 0

3 years ago

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Define the reflection with diagram

LekaFEV [45]

Answer:

The incident light ray which lands upon the surface is said to be reflected off the surface. The ray that bounces back is called the reflected ray. If a perpendicular were to be drawn on reflecting surface, it would be called normal. The figure below shows the reflection of an incident beam on a plane mirror.

Explanation:

6 0

3 years ago