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3 years ago

When you sit in a chair, your body exerts a downward force on the chair. What

2 answers:

7 0

I feel like the answer would be B. The chair pushes down on the floor becuase the question says when you sit in a chair, your body exerts a downward force on the chair so it would be pushing downward meaning that the chair would also go down making it push onto the floor.

Rufina [12.5K]3 years ago

6 0

Answer D

Explanation:just did it on apex

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Two people are talking at a distance of 3.0 m from where you are, and

tigry1 [53]

Answer:

The sound intensity that the other student measure, I₂ is expected to be;

A) 6.2 × 10⁻⁸ W/m²

Explanation:

The measure of sound intensity is given by the following formula;

$I = \dfrac{P}{4 \cdot \pi \cdot R^2}$

$\dfrac{I_2}{I_1} = \dfrac{R_1^2}{R_2^2}$

Where;

I = The intensity

R = The radius

P = The power of the sound

Whereby we have;

The distance of the two people talking, R₁ = 3.0 m

The measure of the sound intensity, I₁ = 1.1 × 10⁻⁷ W/m² (from an online source)

The distance of the other student from the two people talking, R₂ = 4.0 m

Therefore, the estimate of the sound intensity, I₂, is given as follows;

${I_2} = \dfrac{R_1^2}{R_2^2} \times {I_1}$

${I_2} = \dfrac{(3.0 \, m)^2}{(4.0 \, m)^2} \times 1.1 \times 10^{-7} \ W/m^2 = 6.1875 \times 10^{-8} \ W/m^2$

I₂ = 6.1875 × 10⁻⁸ W/m²

∴ The sound intensity that the other student measure, I₂ ≈ 6.2 × 10⁻⁸ W/m²

5 0

3 years ago

Ball is dropped from the height H the total distance covered in last second of its motion is equal to distance covered in first

Alik [6]

$\rm{\gray{\underline{\underline{\blue{GIVEN:-}}}}}$

A ball is dropped from a height = H

The total distance covered in last second of its motion is equal to the distance covered in first 3 second .

$\rm{\gray{\underline{\underline{\blue{TO\:FIND:-}}}}}$

The height of the journey .

$\rm{\gray{\underline{\underline{\blue{SOLUTION:-}}}}}$

We have know that,

$\green\bigstar\:\rm{\gray{\overbrace{\underbrace{\red{S\:=\:ut\:+\:\dfrac{1}{2}\:at^2\:}}}}}$

Where,

S = Distance .

u = initial velocity = 0m/s

t = time = 3s

a = acceleration

[Note :- Here, acceleration is ‛acceleration due to gravity’ .]

a = 10m/s^2

=> S = 1/2 × 10 × (3)^2

=> S = 5 × 9

=> S = 45m -----(1)

✒ If the ball takes ‛n’ second to fall the ground, then distance covered in nth second is,

$\green\bigstar\:\rm{\gray{\overbrace{\underbrace{\red{S_n\:=\:u\:+\:\dfrac{g}{2}\:(2n\:-\:1)\:}}}}}$

=> Sn = 0 + 10/2 (2 × n - 1)

=> Sn = 5 (2n - 1)

=> Sn = 10n - 5 -----(2)

Therefore,

10n - 5 = 45

=> 10n = 45 + 5

=> n = 50/10

=> n = 5

Now put the value of ‛n = 5’ in equation(2), we get

=> Sn = 10 × 5 - 5

=> Sn = 50 - 5

=> Sn = 45m

$\rm{\pink{\therefore}}$ The height of the journey is “ 45m ” .

7 0

3 years ago

Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinal and transverse modul

gregori [183]

Answer:

It is not possible to produce continued and oriented fiber.

Explanation:

To solve the problem it is necessary to take into account the concepts related to Fiber volume ratio. The amount of fiber in a fiber reinforced compound corresponds directly to the mechanical properties of the compound. Given the fiber volume fraction, the theoretical elastic properties of a compound can be determined. The elastic modulus of a compound in the fiber direction of a unidirectional compound can be calculated using the following equation:

$E = (1-V_f)E_m+V_fE_f$

Where,

E is the longitudinal modulus of Elasticity

$V_f$ is the fiber volume ratio

$E_m$ is the elastic modulus of the matrix

$E_f$ is the elastic modulus of the fibers

We need to consult the table of characteristics of Fibers and Reinforcements of Materials, in which they specify that the modulus of elasticity of the aramid fiber-epoxy is

$E_f = 131Gpa$

Moreover from the statement,

$E = 35Gpa$

$E_m = 3.4Gpa$

Replacing in the previous equation,

$35 = 3.4 (1-V_f)+131V_f$

$V_f = 0.25 \rightarrow longitudinal$

To make the comparison we now calculate the Fiber volume ratio through the transverse elastic modulus,

$E = \frac{E_mE_f}{(1-V_f)E_f+V_fE_f}$

Our values are given in this case as:

$E = 5.17Gpa\\E_m = 3.4Gpa \\E_f = 131Gpa$

Replacing,

$5.17 = \frac{3.4*131}{(1-V_f)(131)+V_f*3.4}$

$V_f = 0.351 \rightarrow transversal$

From both cases it is possible to conclude that it is not possible to produce a fiber of the specified material in a continuous and oriented manner, as long as the volume fraction is different in the different cases.

5 0

3 years ago

What are three basic conditions for a hurricane to form

tester [92]

High ocean temperatures of above 26.5 degrees Celsius and coriolis effect( deflection of winds create spiraling motion)

3 0

3 years ago

How much work is done when a very large force is applied to an object but does not move the object?

Ierofanga [76]

Work = (force) x (distance)

If the distance is zero (the object doesn't move), then it
doesn't matter how great the force is. The work is still zero.

4 0

2 years ago

Read 2 more answers