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3 years ago

AP PHYSICS please answer

Physics

1 answer:

mamaluj [8]3 years ago

4 0

I think it’s d not sure

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On planet X, the absolute pressure at a depth of 2.00 m below the surface of a liquid nitrogen lake is 5.00 × 105 N/m2 . At a de

eimsori [14]

Answer:

300000.01008 Pa

123.76237 m/s²

Explanation:

$\rho$ = Density of liquid nitrogen = 808 kg/m³

h = Depth

g = Acceleration due to gravity

P = Atmospheric pressure

Absolute Pressure is given by

Below 2 m from surface

$P_a=P+\rho gh\\\Rightarrow 5\times 10^5=P+808g\times 2$

Below 5 m from surface

$P_a=P+\rho gh\\\Rightarrow 8\times 10^5=P+808g\times 5$

Subtracting the above equations we get

$-3\times 10^5=-808g3\\\Rightarrow g=\frac{3\times 10^5}{808\times 3}\\\Rightarrow g=123.76237\ m/s^2$

The acceleration due to gravity on the planet is 123.76237 m/s²

Equating the value of g in the first equation

$5\times 10^5=P+808\times 123.76237\times 2\\\Rightarrow P=5\times 10^5-808\times 123.76237\times 2\\\Rightarrow P=300000.01008\ Pa$

The atmospheric pressure on the planet is 300000.01008 Pa

7 0

4 years ago

A weight of 1400 pounds is suspended from two cables as shown in the figure. What is the tension in the left cable? _________ po

juin [17]

Answer:

Following are the solution to this question:

Explanation:

Law:

$\to \theta= 180^{\circ}- 50^{\circ}- 25^{\circ}$

$= 180^{\circ}- 75^{\circ}\\\\= 105^{\circ}$

$\to \frac{T_{L}}{\sin (90+50)}= \frac{T_{R}}{\sin (25+90)}=\frac{1400}{\sin (105)}$

$\to T_L=931.65 \ pounds \\\\ \to T_R=1313.59 \ pounds \\\\$

4 0

3 years ago

Why should you do you should

uranmaximum [27]

One does not simply should you do you should

8 0

3 years ago

PLSSSS Jenny runs every morning for exercise she takes 25 minutes to run 1,000 m and walk 1,600 m or 20 minutes to run 2,000 m a

ELEN [110]

The running speed of Jenny willl be 2.5 times the velocity she walks.

<h3 /><h3>What is velocity?</h3>

The change of distance concerning to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

Condition 1;

She takes 25 minutes to run 1,000 m and walk 1,600 m The total time to run as well as the walk is 25 minutes.

$\rm t = t_r +t_w \\\\ 25 = t_r +t_w$

As we know that;

Distance = speed × time

$\rm x_r = v_r \times t_r \\\\ t_r = \frac{x_r}{v_r}$

For walking;

$\rm t_w = \frac{x_w}{v_w}$

$\rm t = t_r +t_w \\\\ 25 = \frac{x_r}{v_r} +\frac{x_w}{v_w} \\\\ 25 = \frac{1000 \ m}{v_r} +\frac{1600 \ }{v_w} \\\\ 40 v_w +64 v_r = v_rv_w$

Same as for condition 2;

$\rm 100 v_w + 40 v_r = v_rv_w$

$\rm 40 v_w +64 v_r =\rm 100 v_w + 40 v_r \\\\ v_r = 2.5 v_w$

Condition 3;

$\rm d_w = 800 \ m$

$\rm d_w = v_w \times t_w$

$\rm t_w = \frac{d_w}{v_w } \\\\ \rm t_w = \frac{800}{v_w } \\\\$

Hence their running speed of Jenny willl be 2.5 times the velocity at she walks.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

5 0

2 years ago

Sphere A is attached to the ceiling of an elevator by a string. A second sphere is attached to the first one by a second string.

MrMuchimi

Answer:

Part a)

$T_1 = 53.13 N$

$T_2 = 26.6 N$

Part b)

$T_1 = 70.1 N$

$T_2 = 35 N$

Part c)

$a = 4.62 m/s^2$

Explanation:

Part a)

As we know that the elevator is accelerating downwards

so we have force equation for sphere A given as

$m_ag + T_2 - T_1 = m_a a$

also for second sphere we have

$m_bg - T_2 = m_b a$

from above equations we have

$(m_a + m_b)g - T_1 = (m_a + m_b) a$

$(2m)g - T_1 = (2m)a$

so we have

$T_1 = (2m)(g - a)$

$T_1 = (2\times 3.14)(9.81 - 1.35)$

$T_1 = 53.13 N$

Now from other equation we have

$T_2 = m_2(g - a)$

$T_2 = 3.14(9.81 - 1.35)$

$T_2 = 26.6 N$

Part b)

Now the elevator is accelerating upwards

so we have force equation for sphere A given as

$T_1 - (m_ag + T_2) = m_a a$

also for second sphere we have

$T_2 - m_b g = m_b a$

from above equations we have

$T_1 - (m_a + m_b)g = (m_a + m_b) a$

$T_1 - 2mg = (2m)a$

so we have

$T_1 = (2m)(g + a)$

$T_1 = (2\times 3.14)(9.81 + 1.35)$

$T_1 = 70.1 N$

Now from other equation we have

$T_2 = m_2(g + a)$

$T_2 = 3.14(9.81 + 1.35)$

$T_2 = 35 N$

Part c)

Now we know that maximum possible tension in the string is

T = 92.6 N

so we have

$T_1 = (2m)(g + a)$

$92.6 = 2(3.14)(9.81 + a)$

$a = 4.62 m/s^2$

6 0

3 years ago