‼️HELP HELP HELP‼️

Productivity increases when A. inputs increase while outputs remain the same. B. inputs decrease while outputs remain the same.

Blizzard [7]

Answer: Productivity increases when inputs and outputs increase proportionately.

Explanation:

Productivity increases when inputs and outputs increase proportionately. Input has to be directly proportional to output to be productive. This means increase in input to a system must leads to drastic increase in the output. When the output is not balanced with the amount of input, it leads to unproductivity.

Being productive can be business wise or in terms if personal growth and development.

3 0

3 years ago

Why is a protective apron or lab coat important to use when working with acids?

Kipish [7]

Answer:

Acids break down fabrics and can cause burns if the acids are strong.

Explanation:

A protective apron or lab coat is important when working with acids because acids break down fabrics and can cause burns if the acids are strong.

An acid is a substance that interacts with water to produce excess hydroxonium ions in an aqueous solution.
A strong acid ionizes completely in solution.
When they come in contact with a fabric, they break them down violently.
So, if they come in contact with the skin, it causes a violent break down of body tissues.
The apron acts a protective layer.

4 0

3 years ago

Read 2 more answers

Calculate the approximate volume of a uranium nucleus, 23592u. (you can ignore the mass defect in this calculation and simply ta

tester [92]

Radius of nuclei is given by formula

$R = R_oA^{1/3}$

now we can say volume of the nuclei is given as

$V = \frac{4}{3}\pi R_o^3* A$

now the density is given as

density = mass / volume

mass of nuclei = mass of neutron + mass of protons

$m = z*m_p + (A- z)*m_n$

$m_p = m_n = 1.008u$

$m = A*1.008u$

Now density is given as

$\rho = \frac{A*1.008u}{\frac{4}{3}\pi R_0^3* A}$

here we know that

$R_0$ = 1.2 fm

$\rho = \frac{1.008u}{\frac{4}{3}\pi*(1.2*10^{-15})^3}$

$\rho = 2.31 * 10^{17} kg/m^3$

So from above we can say that density of all nuclei is almost same.

5 0

3 years ago

A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o

masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, $F_{f2}$ = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = $F_{f2}$ × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

$T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27$

$Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92$

The frictional force on the block W₂, $F_{f2}$ ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ $F_{w2}$ = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + $\mathbf{F_{w2}}$

The frictional force from the ground, $\mathbf{F_{f1}}$ = N×μ + $\mathbf{F_{f2}}$ = P

Where;

P = The horizontal force applied to the block

P = (W₁ + $\mathbf{F_{w2}}$ ) × μ + $\mathbf{F_{f2}}$

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

brainly.com/question/18038995

3 0

3 years ago

Why does the lens need to be thicker for viewing nearby objects?

Maurinko [17]

Answer: To focus on a near object – the lens becomes thicker, this allows the light rays to refract (bend) more strongly. To focus on a distant object – the lens is pulled thin, this allows the light rays to refract slightly.

Explanation:

5 0

3 years ago