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lozanna [386]
3 years ago
6

Step 4: Solve the equation. KE = 27.5 kg x 4 m²/s2 =

Chemistry
1 answer:
harkovskaia [24]3 years ago
4 0
Kinetic energy is 1/2m*v^2
They already halved the mass for you and they already squared the velocity for you. I’m unsure why you’re even asking this question. It’s simple multiplication at this point.
(27.5kg)*(4m^2/s^2) = 110 Jules
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Consider the two electron arrangements for neutral atoms A and B. Are atoms A and B the same element? A - 1s2, 2s2, 2p6, 3s1 B -
iogann1982 [59]

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3 0
3 years ago
A 75 g piece of gold (Au) at 1000 K is dropped into 200 g of H2O at 300K in an insulatedcontainer at 1 bar. Calculate the temper
Arturiano [62]

Answer:

the final temperature is T final = 308 K

Explanation:

since all heat released by gold is absorbed by water

Q gold + Q water = Q surroundings =0 (insulated)

Assuming first that no evaporation of water occurs , and denoting g as gold and w as water , then

Q gold = m g*cp g* ( T final - T initial g)

Q gold = m w*cp w* ( T final - T initial w)

where

m= mass

cp = specific heat capacity

T final = final temperature

T initial g and T initial w =  initial temperature of gold and water respectively

thus

Q gold + Q water = 0

m g*cp g* ( T final - T initial g) + m w*cp w* ( T final - T initial w) =0

m g*cp g* T final + m w*cp w* T final =  m g*cp g* T initial g+ m w*cp w* T initial w

T final = (m g*cp g* T initial g+ m w*cp w* T initial w)/(m g*cp g+ m w*cp w)

replacing values and assuming cp w = 1 cal/gK = 4.816 J/gK and cp g = 0.129 J/gK (from tables), then

T final =  (75 g*0.129 J/gK* 1000 K + 200 g * 4.816 J/gK * 300 K )/(75 g*0.129 J/gK*+ 200 g * 4.816 J/gK ) = 308 K

T final = 308 K

since T boiling water = 373 K and T final = 308 K , we confirm that water does not evaporate

therefore the final temperature is T final = 308 K

3 0
3 years ago
How might the biodiversity of a mowed lawn compare to that of huge weedy field?
Dennis_Churaev [7]

Answer: The mowed lawn is the one from where the grasses are removed by using the machines or tools.

Explanation:

The mowed lawn is expected to have low number of species as the grasses may be few or scanty thus can support the population of few species like insects, mice, birds, and small number of grazing animals. On the other hand the weedy field can be hub of insects, reptiles like snakes, small mammals, and large mammals. Large weed field can provide food, and habitat to the large number of species. This will support the increase in biodiversity as compared to the mowed lawn.

5 0
3 years ago
A stock solution of 1.00 M Fe2(SO4)3 is available. How many milliliters are
Andrews [41]

Answer:

375 mL

Explanation:

M1*V1 = M2*V2

M1 = 1.00 M

V1 = ?

M2 = 0.750 M

V2 = 0.500 L

1.00 M * V1 = 0.750 M * 0.500 L

V1 = 0.750*0.500/1.00 = 0.375 L = 375 mL

5 0
3 years ago
One tank of gold fish is fed the normal amount of food once a day. A second tank is fed twice a day. A third tank is fed four ti
Arturiano [62]

The question is incomplete, the complete question is;

One tank of goldfish is feed the normal amount which is once a day, a second tank is fed twice a day, and a third tank is fed four times a day during a 6 week study. The fishes' body fat is recorded daily.

Independent Variable-

Dependent Variable-

Constants

Control Group-

Answer:

A) The amount of food the gold fish receives

B) Body fat of the gold fish

C) -Type of fish used in the study (gold fish)

Time period within which the fishes were fed (Six week period)

Shape and size of tank

D) group of gold fish fed the normal amount

Explanation:

The purpose of the study is to determined the impact of amount of feed on the body fat of gold fish. Hence, the amount of feed is the independent variable while the body fat of the feed is the dependent variable.

The control group receives the normal amount of feed (once a day). The fishes are all gold fish, fed within a six week period. All the tanks were of the same shape and size. These are the constants in the study.

4 0
3 years ago
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