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ValentinkaMS [17]

4 years ago

5

If the average distance between bumps on a road is about 10 m and the natural frequency of the suspension system in the car is a

bout 0.90 hz, at what speed will you feel the bumps the most?

1 answer:

Mnenie [13.5K]4 years ago

3 0

When you hit a bump every 0.9 seconds.

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What type of circuit is illustrated PLEASE HELPPP

kumpel [21]

Well the shown circuit is definitely open because the lines are not connected. And it is not a parallel circuit because there would be more lines. Like a string of Christmas lights. So it must be an "open series circuit".

7 0

3 years ago

6) If nuclei A has a probability P of decaying in time t and nucleus B has a probability 2p of decaying in time t, which stateme

vredina [299]

Answer:

sorry dont know so so sorry

6 0

4 years ago

A student measures the diameter of a small cylindrical object and gets the following readings: 4.32, 4.35, 4.31, 4.36, 4.37, 4.3

Zinaida [17]

Answer:

a. $\bar{d}=4.34 cm$

b. $\sigma=0.023 cm$

c. $\rho=(0.0089\pm 0.00058) kg/cm^{3}$

Explanation:

a) The average of this values is the sum each number divided by the total number of values.

$\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}$

$x_{i}$ is values of each diameter
N is the total number of values. N=6

$\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}$

$\bar{d}=4.34 cm$

b) The standard deviation equations is:

$\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}$

If we put all this values in that equation we will get:

$\sigma=0.023 cm$

Then the mean diameter will be:

$\bar{d}=(4.34\pm 0.023)cm$

c) We know that the density is the mass divided by the volume (ρ = m/V)

and we know that the volume of a cylinder is: $V=\pi R^{2}h$

Then:

$\rho=\frac{m}{\pi R^{2}h}$

Using the values that we have, we can calculate the value of density:

$\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}$

We need to use propagation of error to find the error of the density.

$\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}$

δm is the error of the mass value.
δd is the error of the diameter value.
δh is the error of the length value.

Let's find each partial derivative:

1. $\frac{\partial\rho}{\partial m}=\frac{4m}{\pi d^{2}h}=\frac{4*1.66}{\pi 4.34^{2}*12.6}=0.0089$

2. $\frac{\partial\rho}{\partial d}=-\frac{8m}{\pi d^{3}h}=-\frac{8*1.66}{\pi 4.34^{3}*12.6}=-0.004$

3. $\frac{\partial\rho}{\partial h}=-\frac{4m}{\pi d^{2}h^{2}}=-\frac{4*1.66}{\pi 4.34^{2}*12.6^{2}}=-0.00071$

Therefore:

$\delta\rho=\sqrt{\left(0.0089)^{2}*0.05^{2}+\left(-0.004)^{2}*0.023^{2}+\left(-0.00071)^{2}*0.5^{2}}$

$\delta\rho=0.00058$

So the density is:

$\rho=(0.0089\pm 0.00058) kg/cm^{3}$

I hope it helps you!

3 0

3 years ago

When a ball player throws a ball straight up, by how much does the speed of the ball decrease each second while ascending?

Alexxandr [17]

-9.8 m/s^s because thats the earth gravity so it will lose 9.8 m/s^2 until its stop and thats because its the opposite of the force towards the earth!
Hope it helps

6 0

3 years ago

Which action can humans take to reduce wave erosion?

mezya [45]

The action that humans can take to reduce wave erosion is to build breakwaters in the ocean.

<h3>What is erosion?</h3>

The term erosion has to do with the washing away of top soil either by water or by wind. The control of erosion is a very important soil conservation effort.

Hence, the action that humans can take to reduce wave erosion is to build breakwaters in the ocean.

Learn more about erosion:brainly.com/question/3852201?

#SPJ1

3 0

2 years ago