All categories

3 years ago

The gravitational strength on Earth is less than the gravitational strength on Neptune. Which statement is correct?

1 answer:

5 0

The mass of an object will never change no matter what, however, the weight relies on gravity. So therefore, the answer would be D.

You might be interested in

A plane flying over the ocean is on course to land on an aircraft carrier. The diagonal distance between the plane and the aircr

fomenos

Answer:

the plane's current height above the ocean is 3.42 miles

Step-by-step explanation:

Given that the diagonal distance between the plane and the aircraft carrier is 10 miles.

What we need to find is the vertical distance between the plane and the aircraft carrier which is the plane's current height above the ocean. This can be gotten using sine rule.

For sine rule, you need a side and its opposite angle. For this question the triangle formed is a right angle triangle with diagonal distance, vertical distance and horizontal distance.

let a = diagonal distance = 10 miles

let the opposite angle to the diagonal distance be A = 90° (right angle)

let the vertical distance = b

let the opposite angle to the vertical distance be B = 20°

Sine rule states that $\frac{a}{sin(A)} = \frac{b}{sin(B)}$

∴ $\frac{10}{sin(90)} = \frac{b}{sin(20)}$

$b = sin(20)\frac{10}{sin(90)}$

$b = 0.342*\frac{10}{1}$

⇒ b = 3.42 miles

Therefore the plane's current height above the ocean is 3.42 miles

4 0

3 years ago

Determine the x- and y-intercepts of the graph of y=1/4x−2 .

Serggg [28]

8 is the x intercept and -2 is the y

4 0

3 years ago

Read 2 more answers

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

3 0

3 years ago

50 pts please help me due today links or incorrect answers will get reported

Bumek [7]

Using the normal distribution, the area underneath the shaded region <u>between the two z-scores</u> is given by:

C. 0.6766.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

Hence, for this problem, the area is the <u>p-value of z = 0.75 subtracted by the p-value of z = -1.3</u>.

Looking at the z-table, the p-values are given as follows:

z = 0.75: 0.7734.

z = -1.3: 0.0968.

Then:

0.7734 - 0.0968 = 0.6766.

Which means that option C is correct.

More can be learned about the normal distribution at brainly.com/question/15181104

#SPJ1

3 0

1 year ago

A building was planned to be built in an area north of 8th Street and east of Main Street, as shown below.

dalvyx [7]

Answer:

Step-by-step explanation:

4 0

2 years ago