All categories

4 years ago

ASAP!! please What is the answer? MCQ. Thank you

Physics

2 answers:

sdas [7]4 years ago

8 0

Answer:

Id say C

Explanation:

The hypotenuse for c is up and to the right so the unit vectors show that

siniylev [52]4 years ago

3 0

Answer:

Option c

Explanation:

The diagram which follow" head to tail rule" show equilibrium

You might be interested in

Three magnets are placed on a plastic stick as shown in the image. Explain how the magnets need to be rearranged so that they st

dimaraw [331]

Answer:

The magnets need to be placed with red closest to blue.

Opposite poles attract.

The magnets will be attracted to each other with enough force to stick together.

Explanation:

6 0

3 years ago

Read 2 more answers

Which data set has the largest standard deviation

AfilCa [17]

Answer:

<em>The data set marked as B has the largest standard deviation</em>

Explanation:

<u>Standard Deviation</u>

It's a number used to show how a set of measurements is spread out from the average value. A low standard deviation means that most of the values are close to the average. A high standard deviation means that the numbers are more spread out.

The formula for the standard deviation is

$\displaystyle \sigma=\sqrt{\frac{\sum (x_i-\mu)^2}{n}}$

Where $x_i$ is the value of each measurement, n is the number of elements in the set, and $\mu$ is the average or media of the values, defined as

$\displaystyle \mu=\frac{\sum x_i}{n}$

Let's analyze each set of data:

A.3,4,3,4,3,4,3

The average is

$\displaystyle \mu=\frac{3+4+3+4+3+4+3}{7}=3.43$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2}{7}}$

$\sigma=0.5$

B.1,6,3,15,4,12,8

The average is

$\displaystyle \mu=\frac{1+6+3+15+4+12+8}{7}=7$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(1-7)^2+(6-7)^2+(3-7)^2+(15-7)^2+(4-7)^2+(12-7)^2+(8-7)^2}{7}}$

$\sigma=4.7$

C. 20, 21,23,19,19,20,20

The average is

$\displaystyle \mu=\frac{20+21+23+19+19+20+20}{7}=20.29$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(20-20.29)^2+(21-20.29)^2+(23-20.29)^2+(19-20.29)^2+(19-20.29)^2+(20-20.29)^2+(20-20.29)^2}{7}}$

$\sigma=1.3$

D.12,14,13,14,12,13,12

The average is

$\displaystyle \mu=\frac{12+14+13+14+12+13+12}{7}=12.86$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(12-12.86)^2+(14-12.86)^2+(13-12.86)^2+(14-12.86)^2+(12-12.86)^2+(13-12.86)^2+(12-12.86)^2}{7}}$

$\sigma=0.8$

We can see the data set marked as B has the largest standard deviation

5 0

4 years ago

Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

Hitman42 [59]

The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

$h = \frac{v^{2}}{2g}$

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

Since some of the Kinetic energy is converted to potential energy of the body;

$\frac{mv^{2}}{2} = mgh$

$h = \frac{v^{2}}{2g}$

Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.

Learn more about elastic collisions at: brainly.com/question/7694106

7 0

2 years ago

Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot

frozen [14]

Answer:

0.285

Explanation:

Given two forces of different magnitude, it is important to note that the product of normal force and coefficient of kinetic friction should be equal to the sum of these two forces at equilibrium. Therefore, this can be Mathematically expressed as:

$N/\mu_k=F_1+F_2\\\\$

where N is normal force, $\mu$ is coefficient of static friction, F is force and subscripts 1 and 2 represent larger and smaller magnitude forces respectively. Making $\mu$ the subject of the formula then

$\mu_k=\frac{F_1+F_2}{N}$