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iris [78.8K]
3 years ago
6

Two forces are applied to a 17kg box, as shown. The box is on a smooth surface

Chemistry
1 answer:
HACTEHA [7]3 years ago
5 0

Answer:

A. the box accelerated to at 1.0 m/s² to the right because the net force is 17N to the right.

Explanation:

To solve this problem, we need to find the net force acting on the body;

   Net force  = Forward force  - Backward force

  Forward force  = 35N

  Backward force  = 18N

 Net force  = 35N  - 18N = 17N

 So;

   Net force  = mass x acceleration

        17  = 17 x acceleration

        Acceleration  = 1m/s²

Therefore, the body moves in the forward direction with an acceleration of  1m/s²

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Avoiding an accident when driving can depend on reaction time. That time, measured in seconds from the moment the driver see dan
crimeas [40]

Answer:

The answer is below

Explanation:

A normal model is represented as (μ, σ). Therefore for (1.5, 0.18), the mean (μ) = 1.5 and the standard deviation (σ) = 0.18

The z score shows by how many standard deviations the raw score is above or below the mean. It is given as:

z=\frac{x-\mu}{\sigma}

a) For x < 1.35 s

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.35-1.5}{0.18}=-0.83

From the normal distribution table, the percent of drivers have a reaction time less than 1.35 seconds = P(x < 1.35) = P(z < -0.83) = 0.2033 = 20.33%

b) For x > 1.9 s

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.9-1.5}{0.18}=2.22

From the normal distribution table, the percent of drivers have a reaction time greater than 1.9 seconds = P(x > 1.9) = P(z > 2.22) = 1 - P(z<2.22) = 1 - 0.9868 = 0.0132 = 1.32%

c) For x = 1.45

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.45-1.5}{0.18}=-0.28

For x = 1.75

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.75-1.5}{0.18}=1.39

From the normal distribution table, P(1.45 < x < 1.75) = P(-0.28 < z < 1.39) = P(z < 1.39) - P(z< - 0.28) = 0.9177 - 0.3897 = 0.528 = 52.8%

d) A percentage of 10% corresponds to a z score of -1.28

z=\frac{x-\mu}{\sigma}\\\\-1.28=\frac{x-1.5}{0.18}\\\\x-1.5=-0.2034\\\\x=1.27

e) P(z < z1) - P(z< -z1) = 60%

P(z < z1) - P(z< -z1) = 0.6

P(z < -z1) = 1 - P(z < z1)

P(z<z1) - (1 - P(z < z1)) = 0.6

2P(z<z1) - 1= 0.6

2P(z<z1) = 1.6

P(z<z1) = 0.8

From the z table, z1 = 0.85

0.85=\frac{x-1.5}{0.18}and-0.85=\frac{x -1.5}{0.18}  \\\\x=1.65 \ and\ x=1.35

The reaction time between 1.35 and 1.65 seconds

8 0
3 years ago
Why do they call Cleveland "the mistake by the lake"?
algol13
Mostly because the lake smell really bad and the waters really dirty

3 0
3 years ago
At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ / mol. When 1.697 g of compound
melisa1 [442]

Answer:

13.85 kJ/°C

-14.89 kJ/g

Explanation:

<em>At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter? </em>

<em />

The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:

1.697g.\frac{1mol}{101.67g} .\frac{(-3039.0kJ)}{mol} =-50.72kJ

According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.

Qcomb + Qcal = 0

Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ

The heat capacity (C) of the calorimeter can be calculated using the following expression.

Qcal = C . ΔT

where,

ΔT is the change in the temperature

Qcal = C . ΔT

50.72 kJ = C . 3.661 °C

C = 13.85 kJ/°C

<em>Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?</em>

Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ

The heat of combustion per gram of B is:

\frac{-56.09 kJ}{3.767g} =-14.89 kJ/g

4 0
3 years ago
The molarity of a 2 liter aqueous solution that contains 222.2 grams of dissolved calcium chloride ( CaCl2), expressed with two
sveta [45]

Answer:

The answer to your question is 1 M

Explanation:

Data

Molarity = ?

mass of CaCl₂ = 222.2 g

Volume = 2 l

Process

1.- Calculate the molar mass of CaCl₂

CaCl₂ = 40 + (35.5 x 2) = 40 + 71 = 111 g

2.- Calculate the moles of CaCl₂

                    111g of CaCl₂ ---------------- 1 mol

              222.2 f of CaCl₂  ----------------  x

                      x = (222.2 x 1) / 111

                      x = 222.2 / 111

                      x = 2 moles

3.- Calculate the Molarity

Molarity = moles / Volume

-Substitution

Molarity = 2/2

-Result

Molarity = 1

3 0
3 years ago
What is the coefficient in this chemical formula? 3C6H12O6
erma4kov [3.2K]

Answer:

C-3

Explanation:

7 0
2 years ago
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