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sdas [7]
3 years ago
12

A boy pushes on a wagon so that it accelerates at a rate of 0.50 m/s 2 . The wagon has a mass of 24 kg. What is the magnitude of

the boy's pushing force? (Ignore frictional effects.)
Chemistry
2 answers:
deff fn [24]3 years ago
6 0
The answer is 12 N. I hope this helps. 
Effectus [21]3 years ago
5 0

Answer:

12n

Explanation:

it was on my test

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Please Help! Hurry!
Rasek [7]

Answer:

A. the jar covers

Explanation:

An independent variable is what you, as the scientist, changes.

6 0
3 years ago
Read 2 more answers
How many moles are in 442.8 g of strontium carbonate (SrCO3)?
GuDViN [60]

Answer:

3 moles

Explanation:

SrCO3

Mass = 442.8g

Molar mass = (87.6 + 12 * [16*3]) = 147.6g/mol

Number of moles = mass / molar mass

Number of moles = 442.8 / 147.6

Number of moles = 3

7 0
3 years ago
Match the salt with the acid and base used to form it in a neutralizing reaction.
nignag [31]
<h3><u>Answer;</u></h3>

K2SO4 - KOH and H2SO4

NaBr - NaOH and HBr

<h3><u>Explanation;</u></h3>
  • Salts are compounds that contain positively charged ions and negatively charged ions. That is; they consist of the positive ion of a base and the negative ion of an acid.
  • Salts may be produced when acids and bases are combined together in equal proportions, a process called neutralization.
  • During neutralization reactions an acid reacts with a base to form a salt and water as the only products.
4 0
3 years ago
Read 2 more answers
A 45-g aluminium spoon(specific heat 0.80 / J/gdegree Celsius) at 24 degree celsius is placed in 180 ml(180 grams) of coffee at
Vlad [161]

Explanation:

a) The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, heat_{absorbed}=heat_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 45 g

m_2 = mass of coffee = 180 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 85^oC

c_1 = specific heat of aluminium = 0.80J/g^oC

c_2 = specific heat of coffee= 4.186 J/g^oC

Putting all the values in equation 1, we get:

45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]

T_{final}=80.30^oC

80.30 °C is the final temperature.

b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.

So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.

8 0
3 years ago
Which of the following is true about two neutral atoms of the element gold? (1 point) The nucleus is missing in both. Each has a
klio [65]

i have the same question idk the anwser tho

5 0
3 years ago
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