1. What function does the skull serve for the skeleton?

Aleks [24]

Answer:

first order date and most recent order date

Explanation:

it was switched. column 5 should be most recent order date because it's 2020 while column 6 should be first order date because it was in 2019

8 0

2 years ago

Read 2 more answers

"On a good day, it takes Mr. Hess 23 minutes (1380 seconds) to drive the 12 miles (19,300 meters). What is his average velocity

il63 [147K]

Answer:

his average velocity is

$v = \frac{19300}{1380} = 13.9855(m \div s)$

5 0

2 years ago

Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24

Oksi-84 [34.3K]

Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
$E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}$
Now At xp we have:
$\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0$
$\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0$
Which is a second order equation, using the quadratic formula to solve for xp would give us:
$xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
or
$xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.

6 0

3 years ago

So far, you’ve been working with an "ideal" pulley system. How do you think real pulley systems are different, and how would tha

almond37 [142]

Answer:

In an ideal pulley system is assumed as a perfect system, and the efficiency of the pulley system is taken as 100% such that there are no losses of the energy input to the system through the system's component

However, in a real pulley system, there are several means through which energy is lost from the system through friction, which is converted into heat, sound, as well as other forms of energy

Given that the mechanical advantage = Force output/(Force input), and that the input force is known, the energy loss comes from the output force which is then reduced, and therefore, the Actual Mechanical Advantage (AMA) is less than the Ideal Mechanical Advantage of an "ideal" pulley system

The relationship between the actual and ideal mechanical advantage is given by the efficiency of the pulley system as follows;

$Efficiency \, \% = \dfrac{AMA}{IMA} \times 100$

Explanation:

8 0

3 years ago

PLEASEEE HELP ME WITH THIS ALSO. I DONT WANT TO FAIL. You push a merry-go-round on which Kim and Katie are riding. Kim weighs 45

Serjik [45]

Answer:

The body weight

Explanation:

5 0

3 years ago