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Artist 52 [7]
3 years ago
13

What are the magnitude (size) and direction of the cumulative force acting on the car shown in the picture A. 30 N to the right

B. 2500 N down C. 30 N to the left D. 3500 N up

Chemistry
2 answers:
Komok [63]3 years ago
7 0

Answer:30n to the right

Explanation:

UNO [17]3 years ago
4 0

Answer: The correct answer is option(A).

Explanation:

Total forces exerting on the car = F

F_r=50 N= Force on car exerting in right direction

F_l=20 N= Force on car exerting in left direction

F_u=2500 N= Force on car exerting in upward direction

F_d=2500 N= Force on car exerting in downward direction

F=F_r+F_l+F_u+F_d

(-F_u)=F_d (negative sign shows the direction)

Since, upward force are equal in magnitude but opposite in direction by which they will balance out each other.so, the net force car will be due to two forces F_r and F_l

F=F_r+F_l=50 N+(-20 N)=30 N(negative shows the direction)

The magnitude (size) and direction of the cumulative force acting on the car will 30 N towards right direction.Hence, correct answer is option(A).

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How many formula units of silver fluoride, AgF, are equal to 42.15 g of this substance?
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2.05 × 10²³ formula units

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3 0
2 years ago
mass of 6.584 g. After it is heated, it has a mass of 4.194 g. What is the percentage by mass of water in the hydrate?
Bad White [126]
M₁=6.584 g
m₂=4,194 g

m(H₂O)=m₁-m₂

w(H₂O)=m(H₂O)/m₁

w(H₂O)=(m₁-m₂)/m₁

w(H₂O)=(6.584-4.194)/6.584=0.3630  (36.30%)

the percentage by mass of water in the hydrate 36.30

8 0
3 years ago
Fe2O3(s) + 3CO(g) ---> 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast
baherus [9]

Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Fe_2O_3=\frac{450}{160}=2.8moles

\text{Moles of} CO=\frac{260}{28}=9.3moles

Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)

According to stoichiometry :

1 mole of Fe_2O_3 require 3 moles of CO

Thus 2.8 moles of Fe_2O_3 will require=\frac{3}{1}\times 2.8=8.4moles  of CO

Thus Fe_2O_3 is the limiting reagent as it limits the formation of product and CO is the excess reagent.

As 1 mole of Fe_2O_3 give = 2 moles of Fe

Thus 2.8 moles of Fe_2O_3 give =\frac{2}{1}\times 2.8=5.6moles of Fe

Mass of Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g

Theoretical yield of liquid iron = 313.6 g

Experimental yield = 288 g

Now we have to calculate the percent yield

\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{288g}{313.6g}\times 100=91.8\%

Therefore, the percent yield is, 91.8%

6 0
3 years ago
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