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3 years ago

An automobile accelerates from rest at 1+ 3* sqrt (t) mph/sec for 9 seconds.

1 answer:

3 0

Writing the acceleration as a function of time:
a(t) = 1 + 3√t

Integrating acceleration, we obtain velocity:
v(t) = t + 2(t)^(3/2) + c;
object at rest so velocity at t = 0 is 0 so c = 0.
v(t) = t + 2(t)^(3/2)

Integrating velocity to obtain an equation for displacement:
d(t) = t²/2 + 4/5 t^(5/2) + c
Applying limits from t = 0 to t = 9
d = 9²/2 + 4/5 9^(5/2)
d = 234.9 m

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3 years ago

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3 years ago

We measure the loudness of sound in decibels.<br> a. True<br> b. False

Tatiana [17]

Answer: The correct answer is True.

Explanation:

Loudness of sound is referred to how soft or loud a sound is for the listener.

This term is measured in a unit known as decibels referred to as dB.

This unit is used to measure the relative intensity of sounds on a scale from zero to 100 dB.

More the value of decibels, it will be uncomfortable for a person to hear that sound.

So Yes, the loudness of sound is measured in decibels.

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4 years ago

(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.

faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let call "x₁" distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁ ( force between earth and the object )

F₁ = K * Mt * m₀/ ( x₁)² K is a gravitational constant

F₂ (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂ K*Mt*m₀ / x₁² = K*Ml*m₀ /x₂²

Or simplifying the expression

Mt/ x₁² = Ml/ x₂²

We know that x₁ + x₂ = 384000 Km then

x₁ = 384000 - x₂

Mt/( 384000 - x₂)² = Ml / x₂²

Mt * x₂² = Ml *( 384000 - x₂)²

We need to solve for x₂

Mt * x₂² = Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂² = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂² = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂² + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂ = -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 254880*10∧10 / 1160

x₂ ₁ = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁ = -56.43*10∧5 + 508* 10∧5 / 1160

x₂ ₁ = 451.27*10∧5/1160

x₂ ₁ = 4512.7*10∧4 /1160

x₂ ₁ = 3.89*10∧4 km (distance between the moon and the object)

x₂ ₁ = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁ = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

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3 years ago

The depth of the Pacific Ocean in the Mariana Trench is 36,198 ft. What is the gauge pressure at this depth

FinnZ [79.3K]

Answer:

the pressure at the depth is 1.08 × $10^{8}$ Pa

Explanation:

The pressure at the depth is given by,

P = h $\rho$ g

Where, P = pressure at the depth

h = depth of the Pacific Ocean in the Mariana Trench = 36,198 ft = 11033.15 meter

$\rho$ = density of water = 1000 $\frac{kg}{m^{3} }$

g = acceleration due to gravity ≈ 9.8 $\frac{m}{s^{2} }$

P = 11033.15 × 9.8 × 1000

P = 1.08 × $10^{8}$ Pa

Thus, the pressure at the depth is 1.08 × $10^{8}$ Pa

4 0

3 years ago