Ask question

Ask question

All categories

3 years ago

8

A scientist is testing the seismometer in his lab and has created an apparatus that mimics the motion of the earthquake felt in

part (a) by attaching the test mass to a spring. If the test mass weighs 13 N, what should be the spring constant of the spring the scientist use to simulate the relative motion of the test mass and the ground from part (a)?

1 answer:

otez555 [7]3 years ago

7 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$a_{max} = 0.00246 \ m/s^2$

b

$k =722.2 \ N/m$

Explanation:

From the question we are told that

The amplitude is $A = 1.8 \ cm = 0.018 \ m$

The period is $T = 17 \ s$

The test weight is $W = 13 \ N$

Generally the radial acceleration is mathematically represented as

$a = w^2 r$

at maximum angular acceleration

$r = A$

So

$a_{max} = w^2 A$

Now $w$ is the angular velocity which is mathematically represented as

$w = \frac{2 * \pi }{T}$

Therefore

$a_{max} = [\frac{2 * \pi}{T} ]^2 * A$

substituting values

$a_{max} = [\frac{2 * 3.142}{17} ]^2 * 0.018$

$a_{max} = 0.00246 \ m/s^2$

Generally this test weight is mathematically represented as

$W = k * A$

Where k is the spring constant

Therefore

$k = \frac{W}{A}$

substituting values

$k = \frac{13}{0.018}$

$k =722.2 \ N/m$

You might be interested in

A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a

Svet_ta [14]

A) The car overtakes the truck after 7.56 s

B) Initial distance between car and truck: 37.1 m

C) Speed of the truck: 15.9 m/s, speed of the car: 25.7 m/s

D) See graph in attachment

Explanation:

A)

The truck starts from rest and has a constant acceleration, so its position at time t can be written as

$x_t(t)=d+\frac{1}{2}a_tt^2$

where

d is the initial distance between the truck and the car (the truck starts some distance ahead of the car)

$a_t=2.10 m/s^2$ is the acceleration of the truck

The car position instead it is given by the equation

$x_c(t)=\frac{1}{2}a_ct^2$

where

$a_c=3.40 m/s^2$ is the acceleration of the car

The car overtakes the truck when the truck has moved 60.0 m, so when

$x_t(t') = d + 60$

Therefore, solving the equation, we find the time t when this occurs:

$d+\frac{1}{2}a_t t'^2 = d+60\\\frac{1}{2}a_tt'^2=60\\t'=\sqrt{\frac{2\cdot 60}{a_t}}=\sqrt{\frac{120}{2.1}}=7.56 s$

B)

In order to find the initial distance between the car and the truck (d), we have to calculate first the distance covered by the car during these 7.56 s. It is given by:

$x_c(t')=\frac{1}{2}a_c t'^2=\frac{1}{2}(3.40)(7.56)^2=97.2 m$

This means that after 7.56 s, when the car reaches the truck, the car has covered 97.2 m while the truck has covered 60 m. However, their positions are now equal, so we can write:

$x_c(t')=x_t(t')$

And by solving the equation, we find the value of d, the initial distance between car and truck:

$\frac{1}{2}a_c t'^2 = d + \frac{1}{2}a_t t'^2\\d=\frac{1}{2}(a_c-a_t)t'^2 = \frac{1}{2}(3.40-2.10)(7.56)^2=37.1 m$

C)

In order to find the speed of each vehicle, we use the following suvat equation:

$v=u+at$

where

u is the initial velocity

a is the acceleration

t is the time

For the truck, we have:

u = 0

$a_t = 2.10 m/s^2$

So its speed after t = 7.56 s is

$v_t = 0+(2.10)(7.56)=15.9 m/s$

For the car, we have

u = 0

$a_c=3.40 m/s^2$

So its speed after t = 7.56 s is

$v_c=0+(3.40)(7.56)=25.7 m/s$

D)

Find the graph required in attachment.

On the x-axis, it is represented the time in seconds. On the y-axis, it is represented the position in meters.

Both curves are in the shape of a parabola since the motion of both vehicles is an accelerated motion.

The curve that starts at -37.1 m is the curve representing the car: in fact, the car starts behind the truck by 37.1 m. The curve that starts from x = 0, t= 0 is that of the truck.

The two curves meets when t = 7.56 s: at that time, the two vehicles have reached the same position, and we see that occurs when x = 60 m, which means that this happens when the truck has covered 60 meters.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0

3 years ago

Express in words AND mathematically the relationship between period and frequency

inessss [21]

The wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength. That number, also known as the frequency, will be larger for a short-wavelength wave than for a long-wavelength wave.

8 0

3 years ago

Read 2 more answers

Aliens come blasting into our solar system and wipe out everything but the Sun, the Earth, and Jupiter. Discuss (conceptually) w

KATRIN_1 [288]

Answer:

There is no short answer.

Explanation:

In theory, the gravitational pull between the planets depends on their total mass and the distance between them. The sun has a mass of $1.989x10^{30}$ kg, Jupiter has a mass of $1.898x10^{27}$ kg and Earth has a mass of $5.972x10^{24}$ kg.

Earth is closer to Jupiter than it is to the sun. The other planets that are behind jupiter are gone so that means the sun's pulling force's effect on jupiter is increased drastically since there is no force the balance that and that means that jupiter is going to get closer to earth.

The new center of mass of the new solar system is roughly located between earth and the sun, closer to earth.

I hope this answer helps.

4 0

3 years ago

A toy car accelerates at a constant rate from rest to a speed of 4 m/s in a time of 0.55 s. What was the magnitude of the accele

Korvikt [17]

initial velocity=u=0m/s
Final velocity=v=4m/s
Time=t=0.5s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{4-0}{0.5}$

$\\ \sf\longmapsto Acceleration=\dfrac{4}{0.5}$

$\\ \sf\longmapsto Acceleration=8m/s^2$

3 0

3 years ago

A 0.600 m long pendulum is used to determine the acceleration due to gravity on a distant plane. If 20 oscillations are complete

katrin2010 [14]

Answer:

7.50 m/s^2

Explanation:

The period of a pendulum is given by:

$T=2\pi \sqrt{\frac{L}{g}}$ (1)

where

L = 0.600 m is the length of the pendulum

g = ? is the acceleration due to gravity

In this problem, we can find the period T. In fact, the frequency is equal to the number of oscillations per second, so:

$f=\frac{N}{t}=\frac{20}{35.5 s}=0.563 Hz$

And the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{0.563 Hz}=1.776 s$

And by using this into eq.(1), we can find the value of g:

$g=\frac{4 \pi^2 L}{T^2}=\frac{4 \pi^2 (0.600 m)}{(1.776 s)^2}=7.50 m/s^2$

6 0

3 years ago