Please if anyone know this question, tell me its answer. I will be very thankful to you<br>

If you use a horizontal force of 32.0 N to slide a 12.5 kg wooden crate across a floor at a constant velocity, what is the coeff

sergejj [24]

Answer:

Value of coefficient of kinetic friction is 0.26 .

Explanation:

Given:

Mass of wooden crate, m = 12.5 kg.

Horizontal force to keep the block moving with constant velocity, F = 32.0 N.

Since, the block is moving with constant velocity.

So, net force experience by it is zero.

Therefore, fore of friction is equal to applied force.

Now, force of friction , $F=\mu_kN$ ( here $\mu_k$ is coefficient of kinetic friction and N is normal force)

Therefore, $\mu_kN=\mu_k\times mg=\mu_k\times 12.5 \times 9.8=122.5\times \mu_k$

Now, both forces are equal.

$122.5\times \mu_k=32\\\mu_k=\dfrac{32}{122.5}=0.26$

The value of coefficient of kinetic friction is 0.26 .

Hence, it is the required solution.

6 0

3 years ago

What is the concentration of NaCI in an aqueous solution that contains 0.032 grams of NaCI in 600. Grams of the solution

jolli1 [7]

Answer:

0.00032 Grams of NaCl per 1 gram of the solution

Explanation:

8 0

3 years ago

Phosgene, a poisonous gas, when heated will decompose into carbon monoxide and chlorine in a reversible reaction: COCl2 (g) <

Advocard [28]

Answer:

8.08 × 10⁻⁴

Explanation:

Let's consider the following reaction.

COCl₂(g) ⇄ CO (g) + Cl₂(g)

The initial concentration of phosgene is:

M = 2.00 mol / 1.00 L = 2.00 M

We can find the final concentrations using an ICE chart.

COCl₂(g) ⇄ CO (g) + Cl₂(g)

I 2.00 0 0

C -x +x +x

E 2.00 -x x x

The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.

The concentrations at equilibrium are:

[COCl₂] = 2.00 -x = 1.96 M

[CO] = [Cl₂] = 0.0398 M

The equilibrium constant (Keq) is:

Keq = [CO].[Cl₂]/[COCl₂]

Keq = (0.0398)²/1.96

Keq = 8.08 × 10⁻⁴

4 0

3 years ago

Cyclopropene decomposes to propene when heated to 500 C, calculate rate constant for the first order reaction 0 min 1.48 mmol/L

Dmitry_Shevchenko [17]

Answer:

k = 0.0306 min-1

Explanation:

The table is given as;

Time, Concentration

0 1.48

5 1.27

10 0.98

15 0.84

The integrated rate law for a first order reaction is given as;

ln [A] = -kt + ln [Ao]

where;

[A] = Final Concentration

[Ao] = Initial Concentration

k = rate constant

t = time

In the table, taking the first two sets of values;

t = 5

k = ?

[Ao] = 1.48

[A] = 1.27

Inserting into the equation;

ln(1.27) = - k (5) + ln(1.48)

ln(1.27) - ln(1.48) = -5k

-0.1530 = -5k

k = -0.1530 / -5

k = 0.0306 min-1

6 0

3 years ago

You work for a cutlery manufacturer who wants to electrolytically precipitate 0.500 g of silver onto each piece of a batch of 25

ICE Princess25 [194]

Answer:

373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.

Explanation:

Mass of silver to be precipitated on ecah spoon = 0.500 g

Number of silver spoons = 250

Total mass of silver = 250 × 0.500 g = 125 g

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of AgCN = n = $\frac{125 g}{134 g/mol}=0.9328 mol$

Volume of AgCN solution =V

Molarity of the AgCN = 2.50 M

$V=\frac{0.9328 mol}{2.50 M}=0.3731 L=373.1 mL$

(1 L = 1000 mL)

373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.

6 0

3 years ago

Please if anyone know this question, tell me its answer. I will be very thankful to you​

Please if anyone know this question, tell me its answer. I will be very thankful to you