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3 years ago

Where is date that can be controlled placed on a line graph ?

1 answer:

5 0

Question:Where is date that can be controlled placed on a line graph ? Answer: independent variable and can include things like temperature, time, depth, etc. This data is placed on the horizontal X axis.

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If a wave is traveling at a certain speed and it's frequency is doubled,what happens to the wavelength of that water

Elena L [17]

Answer:

If the frequency is doubled the wavelength is only half as long.If you put your fingertip in a pool of water and repeatedly move it up and down,you will create circular water waves.

Hope this helps!!!

4 0

2 years ago

"determine the resultant internal loadings acting at the cross sections at points f and g of the frame. set θ = 27º and t = 178

Leviafan [203]

Hi you didn't provide any images to solve the question, hence I am going to solve a different question of same concept so you can have an idea how to tackle such types of questions.(please refer to the attachment for question)

Answer:

<u> Please refer to the attachment for answers and explanation</u>

Explanation:

<u> Please refer to the attachment for answers and explanation</u>

5 0

3 years ago

At a distance of 8 m, the sound intensity of one speaker is 66 dB. If we were to place 3 speakers in a circle of radius 8 m, wha

ludmilkaskok [199]

Answer:

dβ = 70. 77 dβ

Explanation:

The intensity of sound in decibels is

dβ = 10 log I/I₀

let's look for the intensity of this signal

I / I₀ = 10 dβ/10

I / I₀ = 3.981 10⁶

the threshold intensity of sound for humans is I₀ = 1 10⁻¹² W / m²

I = 3.981 10 ⁶ 1 10⁻¹²

I = 3,981 10⁻⁶ W / m²

It is indicated that 3 cornets are placed in the circle, for which total intensity is

I_total - 3 I

I_total = 3 3,981 10⁻⁶

I_total = 11,943 10⁻⁶ W / m²

let's reduce to decibels

dβ = 10 log (11,943 10⁻⁶/1 10⁻¹²)

dβ = 10 7.077

dβ = 70. 77 dβ

3 0

2 years ago

A ballplayer standing at home plate hits a baseball that is caught by another player at the same height above the ground from wh

kvv77 [185]

Answer:

Maximum height reached = 39.39 m

Explanation:

At maximum height

We have equation of motion v² = u² + 2as

Consider the vertical motion of baseball,

Initial velocity, u = 32.1sin60=27.8 m/s

Final velocity, v = 0 m/s

Acceleration, a = -9.81 m/s²

Substituting

v² = u² + 2as

0 = 27.8² + 2 x (-9.81) x s

s = 39.39 m

Maximum height reached = 39.39 m

4 0

2 years ago

Newton’s Law of Cooling. Newton’s law of cooling states that the rate of change in the temperature T(t) of a body is proportiona

nexus9112 [7]

Answer:

After 30 minutes, the temperature of the body is: T₁₀ = 311.60 K

After 60 minutes, the temperature of the body is: T₂₀ = 298.18 K

Explanation:

Given that:

$\dfrac{dT}{dt}= K \bigg [ M(t) -T(t)\bigg]$

where;

K = 0.04

M(t) = 293

Then;

$\dfrac{dT}{dt}= 0.04 \bigg [ 293 -T\bigg]$

$\dfrac{dT}{dt}= 11.72 -0.04 \ T$

Using Euler's Formula;

$T_{n+1} = T_n + hf( t_n, T_n)$

where;

$f(t_n,T_n) = 11.72 - 0.04 T_n$

Then;

$T_{n+1} = T_n + 3.0 (11.72-0.04 \ T_n)$

$T_{n+1} = 0.88T_n + 35.16 \ ---(1)$

At initial state $t_0$ (0); $T_0$ = 360

At t₁ = 3.0 when T₀ = 360

$T_1= 0.88 T_o + 35.16$

$T_1= 0.88 (360) + 35.16$

$T_1 = 351.96 \ K$

At t₂ = 6.0 when T₂ = 0.88T₁ + 35.16

T₂ = 0.88(351.96) + 35.16

T₂ = 344.89 K

At t₃ = 9.0 when T₃ = 0.88T₂ + 35.16

T₃ = 0.88(344.89) + 35.16

T₃ =338.66 K

At t₄ = 12.0 when T₄ = 0.88T₃ + 35.16

T₄ = 0.88(338.66) + 35.16

T₄ = 333.18 K

At t₅ = 15.0 when T₅ = 0.88T₄ + 35.16

T₅ = 0.88(333.18) + 35.16

T₅ = 328.36 K

At t₆ = 18.0 when T₆ = 0.88T₅ + 35.16

T₆ = 0.88(328.36) + 35.16

T₆ = 324.12 K

At t₇ = 21.0 when T₇ = 0.88T₆ + 35.16

T₇ = 0.88(324.12) + 35.16

T₇ = 320.39 K

At t₈ = 24.0 when T₈ = 0.88T₇ + 35.16

T₈ = 0.88(320.29) + 35.16

T₈ = 317.02 K

At t₉ = 27.0 when T₉ = 0.88T₈ + 35.16

T₉ = 0.88(317.02) + 35.16

T₉ = 314.14 K

At t₁₀ = 30 when T₁₀ = 0.88T₉ + 35.16

T₁₀ = 0.88(314.14) + 35.16

T₁₀ = 311.60 K

At t₁₁ = 33.0 when T₁₁ = 0.88T₁₀ + 35.16

T₁₁ = 0.88(311.60) + 35.16

T₁₁ = 309.37 K

At t₁₂ = 36.0 when T₁₂ = 0.88T₁₁ + 35.16

T₁₂ = 0.88(309.37)+ 35.16

T₁₂ = 307.41 K

At t₁₃ = 39.0 when T₁₃ = 0.88T₁₂ + 35.16

T₁₃ = 0.88( 307.41) + 35.16

T₁₃ = 305.68 K

At t₁₄ = 42.0 when T₁₄ = 0.88T₁₃ + 35.16

T₁₄ = 0.88(305.68) + 35.16

T₁₄ = 304.16 K

At t₁₅ = 45.0 when T₁₅ = 0.88T₁₄ + 35.16

T₁₅ = 0.88(304.16) + 35.16

T₁₅ = 302.82 K

At t₁₆ = 48.0 when T₁₆ = 0.88T₁₅ + 35.16

T₁₆ = 0.88(302.82) + 35.16

T₁₆ = 301.64 K

At t₁₇ = 51.0 when T₁₇ = 0.88T₁₆ + 35.16

T₁₇ = 0.88(301.64) + 35.16

T₁₇ = 300.60 K

At t₁₈ = 54.0 when T₁₈ = 0.88T₁₇ + 35.16

T₁₈ = 0.88(300.60) + 35.16

T₁₈ = 299.69 K

At t₁₉ = 57.0 when T₁₉ = 0.88T₁₈ + 35.16

T₁₉ = 0.88(299.69) + 35.16

T₁₉ = 298.89 K

At t₂₀ = 60 when T₂₀ = 0.88T₁₉ + 35.16

T₂₀ = 0.88(298.89) + 35.16

T₂₀ = 298.18 K

4 0

2 years ago