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2 years ago

Section 2 match the graphs below with one of the graph descriptions

Physics

1 answer:

Juli2301 [7.4K]2 years ago

3 0

The distance-time graphs are labelled accordingly:

Graph one: Option C

Graph Two: Option A

Graph Three: Option D

Graph Four: Option B

<h3>What is a distance time graph?</h3>

A distance-time graph depicts how far a person travels at any given time.

A distance-time graph can be used to illustrate the distance traveled by an item moving in a straight line.

The gradient of the line in a distance-time graph equals the speed of the item.

The quicker the item moves, the higher the gradient (and the steeper the line).

Learn more about distance-time graph:
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A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves o

brilliants [131]

Answer:

$\mathbf{x = 3te ^{-2 \sqrt{2t}}}$

Explanation:

From the given information concerning the spring-mass system:

Let us apply Hooke law.

Then, we have:

mg = ks

8 = k4

k = 8/4

k = 2

Provided that the mass weighing 8 lbs is attached to a spring.

Then, we can divide it by gravity 32 ft/s².

∴

m = 8/32

m = 1/4 slugs

The medium that offers the damping force $\beta = \sqrt{2}$

Now, let us set up a differential equation that explains the motion of the spring-mass system.

The general equation is:

$mx '' + \beta x' + kx = 0$

where;

$m = \dfrac{1}{4}$

k = 2, and

$\beta = \sqrt{2}$

Then;

$\dfrac{1}{4}x'' + \sqrt{2} x' + 2x = 0$

By solving the above equation, the auxiliary equation is:

$m^2 + 4 \sqrt{2} m + 8n = 0$

Using quadratic formula:

$\dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}$

$m = \dfrac{-4 \sqrt{2} \pm \sqrt{(4 \sqrt{2})^2 -4(1)(8) }}{2}$

$m = \dfrac{-4 \sqrt{2} \pm \sqrt{32 -32 }}{2}$

$m = -2 \sqrt{2}$

Since this is a repeated root, the solution to their differential equation took the form.

$x_c = c_1 e^{mt} + c_2 t e^{mt}$

$x_c = c_1 e^{-2\sqrt{2} t} + c_2 t e^{-2\sqrt{2} t}$

From the initial condition.

At equilibrium position where the mass is being from:

x(0) = 0

Also, at the downward velocity of 3 ft/s

x'(0) = 3

Then, at the first initial condition:

$x_c (0) = c_1 e^{-2\sqrt{2} *0} + c_2 (0) e^{-2\sqrt{2} *0}$

$0= c_1 e^{0} + 0$

$0= c_1$

At the second initial condition;

$x' = -2 \sqrt{2} c_1 e^{-2 \sqrt{2} t } -2 \sqrt{2} c_2 t e^{-2 \sqrt{2} t } + c_2 e^{-2 \sqrt{2} t}$

where;

x'(0) = 3

$x' (0) = -2 \sqrt{2} c_1 e^{-2 \sqrt{2}* 0 } -2 \sqrt{2} c_2 (0) e^{-2 \sqrt{2} *0 } + c_2 e^{-2 \sqrt{2} *0}$

$3 = -2 \sqrt{2} * 0 *e^0 - 0 + c_2 e^0$

$3 = 0 + c_2$

$3 = c_2$

Replacing in the constraints, the equation of the motion is:

$\mathbf{x = 3te ^{-2 \sqrt{2t}}}$

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stepan [7]

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A system absorbs 186 kj of heat and the surroundings do 110 kj of work on the system. what is the change in internal energy of t

hichkok12 [17]

296kJ

add the two, there you go

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