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3 years ago

A term used to describe water that is safe to drink?

1 answer:

8 0

The Safe Drinking Water Act<span> (SDWA) is the federal law that protects public </span>drinking water<span> supplies throughout the nation. Under the SDWA, EPA sets standards for </span>drinking water<span> quality and with its partners implements various technical and financial programs to ensure </span>drinking water<span> safety.</span>

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If A = 5 and C = 13, what is sin x?

lisabon 2012 [21]

<span>By pythagorean theorem then, the vertical side of the right triangle must be 12.
Then if x is the angle between the horizontal side and the hypotenuse, sin(x) = 12/13 but also the anser should be in this sentences.
</span>

6 0

3 years ago

A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

Answer:

The value is $T_R = 11.8 \ days$

Explanation:

From the question we are told that

The semi - major axis of the rocky debris $a_R = 45.0\ AU$

The semi - major axis of Planet D is $a_D = 60 \ AU$

The orbital period of planet D is $T_D = 18.164 \ days$

Generally from Kepler third law

$T \ \ \alpha \ \ a^{\frac{3}{2} }$

Here T is the orbital period while a is the semi major axis

$\frac{T_D}{T_R} = \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}$

=> $T_R = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }$

=> $T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }$

=> $T_R = 11.8 \ days$

7 0

3 years ago

I need to make a project that represents united together any ideas?

Orlov [11]

Make a puzzle and each puzzle piece represent a state

6 0

3 years ago

The following is regarding Momentum Change and Force. help is needed please!

mixer [17]

The distance travelled is 10 m and the velocity gained at the end of this time is 2 m/s.

<h3>Velocity of the object at the end of the time</h3>

F = mv/t

where;

m is mass of the object
v is velocity of the object
t is time

Ft = mv

v = Ft/m

v = (50 x 10)/250

v = 2 m/s

<h3>Distance traveled by the object</h3>

v² = u² + 2as

where;

u is initial velocity = 0

a is acceleration

a = F/m

a = 50 N/ 250 kg

a = 0.2 m/s²

v² = 0 + 2as

s = v²/2a

s = (2²)/(2 x 0.2)

s = 10 m

Thus, the distance travelled is 10 m and the velocity gained at the end of this time is 2 m/s.

Learn more about distance here: brainly.com/question/2854969

#SPJ1

4 0

2 years ago

Consider a railroad bridge over a highway. A train passing over the bridge dislodges a loose bolt from the bridge, which proceed

schepotkina [342]

Answer:

12.35m

Explanation:

Hello! To solve this problem we must consider the following:

1. The car moves with constant speed, which means that the distance traveled is equal to the multiplication of time by speed.

X = VT

we solve the equation for time

$T=\frac{x}{V} =\frac{27}{17} =1.588s$

2. The bolt moves with constant acceleration, with acceleration of 9.81m / s ^ 2, so we could apply the following equation.

note=remember that "a uniformly accelerated motion", means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

$Y= VoT-\frac{1}{2}gt^{2}$

where

Vo = Initial speed

T = time

=1.588s

g =gravity=9.8m/s^2

Y = bridge height

solving

$Y= \frac{1}{2}gt^{2}\\Y=(0.5)(9.8)(1.588^2)=12.35m$

4 0

3 years ago