Ask question

Ask question

All categories

3 years ago

15

You have an aquarium which holds 2.65 gallons of water and you would like to place it on an antique stand. Before placing the aq

uarium on the stand, as a precaution, you decide to determine the weight of the water in the aquarium. Determine the weight of water you will be placing in the aquarium.

2 answers:

snow_tiger [21]3 years ago

8 0

Answer:

10.0 kg

Explanation:

1 gallon of water is equal to 3.79 liters of water. Then, 2.65 gallons of water are:

2.65 gal water × (3.79 L water / 1 gal water) = 10.0 L water

The density of water is about 1.00 kg/L. The mass corresponding to 10.0 L of water is:

10.0 L × (1.00 kg/L) = 10.0 kg

The mass of 2.65 gallons of water is 10.0 kg.

qwelly [4]3 years ago

6 0

Answer:

The weight of the water is 10.0 kg

Explanation:

Hi there!!

Since the density of water is 1 g/ml, knowing the volume of water, we can determine its weight. First, let´s convert the unit gallon into liters:

1 gallon = 3.785 l

Then:

2.65 gallons · (3.785 l / 1 gallon) = 10.0 l

Since 1 l = 1000 ml, 10.0 l will be 10000 ml

The density of water is 1 g/ml, then the weight of 10.0 l is 10000 g or 10.0 kg.

You might be interested in

The image below The image below The image below The image below The image below The image below The image below The image below

larisa [96]

Answer:

wow i have know idea

Explanation:

5 0

2 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

3 years ago

An object weighing 49 N is dropped from a height of 30 m. It is found to be moving with a velocity of 24m/s just before it hits

faltersainse [42]

The frictional force will be 0.22N.

<h3>What is Frictional force?</h3>

Frictional force is the force generated between two surfaces that are in contact and slide against each other.

Given,

Weight=4N

mass =4.9/9.8=0.5kg

Hieght =30m

velocity=24m/s

Acceration , v²-u²=2as

24²/2×30 =a , u is zero

a= 1.5m/s²

By Using conservation of energy ,

30F+1/2mv²=mgh

30F=1150-144

F= 6/30

F=0.2N

The force will be 0.2N

to learn more about Friction clickhttps://brainly.com/question/17608236

#SPJ9

6 0

2 years ago

Hey! I found this question quite interesting. Check it out - https://www.meritnation.com/ask-answer/question/to-raise-a-200kg-st

Yanka [14]

Answer:

yes yes so do I

6 0

2 years ago

A 1.5 kg ball is dropped from a height of 2.Gm. Assuming energy is

Vikki [24]

Answer:

Plug in the given values and solve for the final velocity. Remember, when the ball is on the ground it has a height of zero.

Explanation:

6 0

2 years ago