Hellllppppppp please

What variables show a direct relationship? Check all that apply. the speed of a car and the distance traveled the elevation abov

Llana [10]

The variables that show a direct relationship are :
- The speed of a car and the distance traveled
- Number of students in a cafeteria and the amount of food consumed
- The distance a planet is from the sun and that planet's temperature
- The mass of a space shuttle and its acceleration through space

In direct relationship, when one factor is increased/decreased , it will directly cause the other factor to be increased/decreased

3 0

3 years ago

Read 2 more answers

488 J of work is done to a box which is moved across the floor for a distance of 8.9 m. What net force is required to act on the

kvv77 [185]

According to the formula
$a = f \times d$
Where a is work, f is force and d is the distance that box was moved over. And from that formula, you can get that f = a/d and that is 54.83N of force

8 0

3 years ago

How much energy is used when a 110kw appliance is used for 3 hours

Bogdan [553]

We could take the easy way out and just say

(110 kW) x (3 hours) = 330 kilowatt hours .

But that's cheap, and hardly worth even 5 points.
If we want to talk energy, let's use the actual scientific unit of energy.
________________________________________________

" 110 kw " means 110,000 watts = 110,000 joules/second .

(3 hours) x (3600 sec/hour) = 10,800 seconds.

(110,000 joules/second) x (10,800 seconds) = 1.188 x 10⁹ Joules

That's

==> 1,188,000,000 joules

==> 1,188,000 kilojoules

==> 1,188 megajoules

==> 1.188 gigajoules

Atsa nawfulotta energy !
It goes back to that "110 kw appliance" that we started with.
That's no common ordinary household appliance. 110 kw is something like
147 horsepower. In order to bring 110 kw into your house, you'd need to
take 458 Amperes through the 240-volt line from the pole. Most houses
are limited to 100 or 200 Amperes, tops. And the TRANSFORMER on
the pole, that supplies the whole neighborhood, is probably a 50 kw unit.

6 0

3 years ago

Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0

3 years ago

Statement that combines energy and mass into one law

inysia [295]

Answer:

The law of conservation of mass or principle of mass conservation

Explanation:

It states that for any system closed to all transfers of matter and energy, the mass of the system must remain constant over time, as the system's mass cannot change, so quantity can neither be added nor be removed.

5 0

3 years ago