All categories

2 years ago

Imagine a glass box that is completely sealed and in that box is a cactus in a small pot which statement is most accurate?

2 answers:

6 0

I THINK C BECAUSE IF IT IS A GLASS BOX HOW DID A CACTUS GET IN AND NOTHING CAN GET IN OR OUT OF THE BOX SO THERE IS NO CACTUS IN THE BOX

Whitepunk [10]2 years ago

3 0

Answer:

The answer is: C. This is a closed system because of no substance and entering or leaving the box.

Explanation:

The most accurate statement if you start from the basis that the cactus is inside a glass box that is completely closed and sealed, you can hardly enter anything, so the best answer is C.

The answer is: C. This is a closed system because of no substance and entering or leaving the box.

You might be interested in

. a boat can travel in still water. (a) if the boat points directly across a stream whose current is what is the velocity (magni

mixer [17]

a) The velocity of the boat relative to the shore is 3.40 m/s and b) The position of the boat relative to its point of origin after 3s is 10.20m.

Here it is given that the speed of the boat (x) = 2.20m/s

The speed of the stream current (y) = 1.20m/s

a) We have to find the velocity of the boat relative to the shore.

The speed of the boat = x + y

= 2.20 + 1.20

= 3.40m/s

b) Now we have to find the position of the boat after 3s

The formula for speed:

Speed = Distance/ Time

distance = speed × time

speed = 3.40m/s

Time = 3s

distance = 3.40 × 3

= 10.20 m

Therefore we get a) speed as 3.40m/s and b) distance as 10.20m.

To know more about the boat and stream refer to the link given below:

brainly.com/question/382952

#SPJ4

8 0

1 year ago

Maya and Kenzie are discussing oil spills and how they impact the environment. How can humans help reduce the impact of oil spil

madam [21]

The correct answer is A.

6 0

2 years ago

Read 2 more answers

A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro

tekilochka [14]

Answer:

distance between object and image = 18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm ......................3

so from equation 2 and 3

distance between object and image = 5.4 + 13.5

distance between object and image = 18.9 cm

6 0

3 years ago

A body falls from the top of the tower and during the last second of its fall it fall through 23mvfind height of tower.

DanielleElmas [232]

Answer:

39.7 m

Explanation:

First, we conside only the last second of fall of the body. We can apply the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where, taking downward as positive direction:

s = 23 m is the displacement of the body

t = 1 s is the time interval considered

$a=g=9.8 m/s^2$ is the acceleration

u is the velocity of the body at the beginning of that second

Solving for u, we find:

$ut=s-\frac{1}{2}at^2\\u=\frac{s}{t}-\frac{1}{2}at=\frac{23}{1}-\frac{1}{2}(9.8)(1)=18.1 m/s$

Now we can call this velocity that we found v,

v = 18 m/s

And we can now consider the first part of the fall, where we can apply the following suvat equation:

$v^2-u^2 = 2as'$

where

v = 18 m/s

u = 0 (the body falls from rest)

s' is the displacement of the body before the last second

Solving for s',

$s'=\frac{v^2-u^2}{2a}=\frac{18.1^2-0}{2(9.8)}=16.7 m$

Therefore, the total heigth of the building is the sum of s and s':

h = s + s' = 23 m + 16.7 m = 39.7 m

7 0

3 years ago

A point charge with a charge q1 = 4.00 μC is held stationary at the origin. A second point charge with a charge q2 = -4.10 μC mo

GrogVix [38]

Answer:

W = -0.480 J

Explanation:

given,

q₁ = 4 μC

q₂ = -4.10 μC

$W = kq_1q_2(\dfrac{1}{a}+\dfrac{1}{b})$

$b = \sqrt{(0.27-0)^2+(0.27-0)^2}$

b = 0.381

k = 8.99 × 10⁹ Nm²/C²

$W = 8.99\times 10^9\times 4\times 10^{-6}\times (-4.1 \times 10^{-6})(\dfrac{1}{0.17}+\dfrac{1}{0.381})$

$W = [-147.436\times (5.88-2.62)\times 10^{-3}]J$

W = -0.480 J

Work done by the electric force W = -0.480 J

4 0

2 years ago