Ask question

Ask question

All categories

3 years ago

8

Air pressure may be represented as a function of height above the surface of the Earth as shown below.. P(h) = P_0 e^-.00012h. .

. . In this function, P0 is air pressure at sea level, and h is measured in meters. Which of the following equations will find the height at which air pressure is 85% of the air pressure at sea level?. . P_o =.85P_o e^-.00012h. .85p_o=p_o e^-.00012h. h= .85 e^-.00012h. .85 =h x e ^-.00012h

2 answers:

Rufina [12.5K]3 years ago

5 0

Answer. Second Option: .85p_o=p_o e^-.00012h

Solution:

P(h)=Po e^(-0.00012h)

Air pressure: P(h)

Height above the surface of the Earth (in meters): h

Air pressure at the sea level: Po

Height at which air pressure is 85% of the air pressure at sea level:

h=?, P(h)=85% Po

P(h)=(85/100) Po

P(h)=0.85 Po

Replacing P(h) by 0.85 Po in the formula above:

P(h)=Po e^(-0.00012h)

0.85 Po = Po e^(-0.00012h)

Lisa [10]3 years ago

5 0

The equation that will find the height if the pressure is 85% of the air pressure at sea is: P(h)= .85*P0e^-0.00012h

You might be interested in

The___is the time it takes for a wave to complete one cycle.

nignag [31]

Answer:

Period

Explanation:

we know that

The period of a wave is the time required for one complete cycle of the wave to pass by a point.

7 0

3 years ago

Read 2 more answers

Which value is represented by the slope of the line?

netineya [11]

The slope of the line is

(change in ' y ' between the ends) / (change in ' x ' between the ends)

Slope = (630g - 0) / (70 cm^3 - 0)

Slope = (630 / 70) g/cm^3

<em>Slope = 9.0 g/cm^3</em>

5 0

3 years ago

As shown in the figure below, Justin walks from the house to his truck on a windy day. He walks 20 m toward

juin [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity is $v =0.333 \ m/s$ in positive x -direction

The speed is $s = 0.733 \ m/s$

Explanation:

From the question we are told that

The distance from the house to truck is D = 20 m

The distance traveled back to retrieve wind-blown hat is d = 15

The distance from the wind-blown hat position too the truck is k = 20 m

The total time taken is t = 75 s

Generally when calculating the displacement the Justin's backward movement to collect his wind - blown hat is taken as negative

Generally Justin's displacement is mathematically represented as

$L = 20 - 15 + 20$

=> $L = 25 \ m$

Generally the average velocity is mathematically represented as

$v = \frac{L}{t}$

=> $v = \frac{25}{75}$

=> $v =0.333 \ m/s$

Generally the distance covered by Justin is mathematically represented as

$R = D+ d + k$

=> $R = 20 + 15 +20$

=> $R = 55 \ m$

Generally Justin's average speed over a 75 s period is mathematically represented as

$s = \frac{R}{ t}$

=> $s = \frac{55}{ 75}$

=> $s = 0.733 \ m/s$

8 0

3 years ago

A motorcycle that is slowing down uniformly. The motorcycle covers 1 ????m=1000 m in 80 sec⁡. The motorcycle then covers the nex

Lynna [10]

Answer:

Part a)

acceleration = -0.042 m/s/s

Part b)

initial speed = 14.17 m/s

final speed = 5.77 m/s

Explanation:

Part a)

Let the initial velocity of the motorcycle is

$v_i = v_o$

now at the end of 80 s let the speed is

$v_f = v_1$

after another 120 s let the speed will be

$v_f' = v_2$

now we know that

$d = \frac{v_i + v_f}{2} (t)$

$d = \frac{v_o + v_1}{2}(80)$

$1000 = 40(v_o + v_1)$

also we know that

$v_1 - v_o = a(80)$

also we have

$1000 = \frac{v_1 + v_2}{2}(120)$

$1000 = 60(v_1 + v_2)$

now we can say

$(v_2 + v_1) - (v_o + v_1) = \frac{50}{3} - \frac{50}{2}$

also we know

$v_2 - v_o = a(120 + 80)$

$-8.33 = 200 a$

$a = -0.042 m/s^2$

Part b)

now we have

$v_1 + v_o = 25$

$v_1 - v_o = (-0.042)(80)$

$v_1 = 10.83 m/s$

so the starting velocity of the trip is

$v_o = 25 - 10.83 = 14.17 m/s$

now speed after t = 200 s is given as

$v_2 = v_o + at$

$v_2 = 14.17 - (0.042)(200)$

$v_2 = 5.77 m/s$

5 0

3 years ago

At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex

telo118 [61]

Answer:

346 * 10⁶ m

Explanation:

The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let $F_{e}$ be the force of gravity exerted by the earth

and let $F_{m}$ be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d, is given by:

$F = \frac{Gm_{1} m_{2} }{d^{2} }$

Mass of the earth, $m_{e} = 5.97 * 10^{24} kg$

Mass of the moon, $m_{m} = 7.348 * 10^{22} kg$

Mass of the satellite, $m_{s} = ?$

$F_{e} = \frac{G*5.97 * 10^{24} M }{d^{2} }$ ...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

$F_{m} = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }$ ............................(2)

Equating equations (1) and (2)

$\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }$

$(5.97 * 10^{24})(14.78 * 10^{16} -7.688*10^{8}d + d^{2}) = 7.348 * 10^{24} d^{2} \\88.24*10^{40} - 45.9 * 10^{32}d + 5.97 * 10^{24}d^{2} = 7.348 * 10^{24} d^{2}\\ 1.378 * 10^{24}d^{2} + 45.9 * 10^{32}d + 88.24*10^{40} = 0\\$

Factorising out $10^{24}$

$1.378d^{2} + 45.9 * 10^{8}d + 88.24*10^{16} = 0$

Solving for d in the quadratic equation above:

d = 346 * 10⁶ m

4 0

3 years ago