Help please ASAP ONLY 1 question

A person is diving in a lake in the depth of h = 5.5 m. The density of the water is rho = 1.0 x10^3 kg/m^3. The pressure of the

Cloud [144]

Answer:a) P = Po + rho×h×g

b) P = 5.4 × 10^9 pa

c) F = P/A = (Po + rho×h×g)/A

d) 1.174×10^11N

Explanation: Using the formula

P = Po + rho×h×g

P = 1.0 x 10^5 + 1000 × 5.5 × 9.81

P = 5.4 × 10^9pa

The magnitude of the force exerted by water on the top of the person's head F at the depth h in terms of P

F = P/A = (Po + rho×h×g)/A

Using the above formula

Where A = 0.046m^2

F = P/ A = 5.4×10^9/0.046

F = 1.174×10^11N

3 0

2 years ago

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju

denpristay [2]

Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

upward acceleration of the helicopter, $a=5\ m.s^{-2}$

time after the takeoff after which the engine is shut off, $t_a=10\ s$

a)

Maximum height reached by the helicopter:

using the equation of motion,

$h=u.t+\frac{1}{2} a.t^2$

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

$h=0+0.5\times 5\times 10^2$

$h=250\ m$

b)

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

acceleration after deploying the parachute, $a_p=2\ m.s^{-2}$

height fallen freely by Austin:

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

$u=$ initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

$t_f=$ time of free fall

$h_f=0+0.5\times 9.8\times 7^2$

$h_f=240.1\ m$

Velocity just before opening the parachute:

$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

Time taken by the helicopter to fall:

$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

$h=$ height from where it falls = 250 m

now,

$250=0+0.5\times 9.8\times t_h^2$

$t_h=7.1429\ s$

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

remaining time,

$t'=t_h-t_f$

$t'=7.1428-7$

$t'=0.1428\ s$

Now the height fallen in the remaining time using parachute:

$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

$h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2$

$h'=9.8165\ m$

Now the height of Austin above the ground when the helicopter crashed on the ground:

$\Delta h=h-(h_f+h')$

$\Delta h=250-(240.1+9.8165)$

$\Delta h=0.0835\ m$

5 0

2 years ago

A passenger train left station A at 6:00 p.m. Moving with the average speed 45 mph, it arrived at station B at 10:00 p.m. A tran

marishachu [46]

<h2>Average speed of transit train is 60 mph</h2>

Explanation:

Average speed of passenger train = 45 mph

Time taken from station A to station B for passenger train = 10:00 - 6:00 = 4 hours

Distance between station A to station B = 45 x 4 = 180 miles.

Time taken from station A to station B for transit train = 4 - 1 = 3 hours

Distance between station A to station B = Average speed of transit train x Time taken from station A to station B for transit train

180 = Average speed of transit train x 3

Average speed of transit train = 60 mph

Average speed of transit train is 60 mph

8 0

3 years ago

In Trial III, a different, looser, spring is used; its force constant is 23.1 N/m. The suspended mass is the same as the one in

cricket20 [7]

Answer:

T=0.827s

Explanation:

The period of a spring can be calculated with the equation

$T=2\pi w$

But we know as well that w is given by,

$w=\sqrt{\frac{k}{m}}$

Replacing,

$w=\frac{2\pi}{T}= \sqrt{\frac{k}{m}}\\T=2\pi\sqrt{\frac{k}{m}}\\T=2\pi\sqrt{\frac{0.4}{23.1}}$

So we have that

$T=0.827s$

7 0

2 years ago

While moving a box 10.0 m across the floor, Sam pushes with a force of 150 N to the right. The force friction acting on the box

svet-max [94.6K]

Answer:

Explanation:required formula is

W 1=F*S

W1=work done by Sam =?

F=force applied by sam=150N

S=displacement =10m

again

W2=F*S

W2=work done by friction =?

S=displacement =10m

F=friction =25N

W=W1-W2=net work done

please feel free to ask if you have any questions

4 0

3 years ago