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raketka [301]
3 years ago
12

At a certain depth in the ocean. the absolute pressure is p. If you 90 to twice that depth (treating the water as incompressible

) A) the absolute pressure will be 2p.B) the absolute pressure will be less than 2p. C) the absolute pressure will be greater than 2p. D) the gauge pressure will not change. E) Use gauge pressure will increase but will not double.
Physics
1 answer:
KatRina [158]3 years ago
5 0

Answer:

correct option is B

the absolute pressure will be less than 2p

Explanation:

given data

absolute pressure = p

depth = twice

to find out

what is correct option

solution

we know here that p absolute pressure so we consider atmospheric pressure p1

so we can say at some depth h pressure will be

pressure p = p1 + ρ g h

so if depth is double

pressure will be

2p = 2p1 +2 (ρ g h )

so

2p - p1 = p1 + 2 (ρ g h )

so correct option is B

the absolute pressure will be less than 2p

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During a very quick stop, a car decelerates at 6.8 m/s^2. Assume the forward motion of the car corresponds to a positive directi
geniusboy [140]

Answer:

-24.28571 rad/s²

29.57239 revolutions

3.91176 seconds

52.026478 m

Explanation:

a_t = Tangential acceleration = -6.8 m/s²

r = Radius of wheel = 0.28

\omega_i = Initial angular velocity = 95 rad/s

\theta = Angle of rotation

\omega_f = Final angular velocity

t = Time taken

Angular acceleration is given by

\alpha=\frac{a_t}{t}\\\Rightarrow \alpha=\frac{-6.8}{0.28}\\\Rightarrow \alpha=-24.28571\ rad/s^2

The angular acceleration is -24.28571 rad/s²

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-95^2}{2\times -24.28571}\\\Rightarrow \theta=185.80885\ rad=185.80885\times \frac{1}{2\pi}\\\Rightarrow \theta=29.57239\ rev

The number of revolutions is 29.57239

\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-95}{-24.28571}\\\Rightarrow t=3.91176\ s

The time it takes for the car to stop is 3.91176 seconds

Linear distance

s=r\theta\\\Rightarrow s=0.28\times 185.80885\\\Rightarrow s=52.026478\ m

The distance the car travels is 52.026478 m

8 0
3 years ago
Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit
Ede4ka [16]

Answer:

386.2^{\circ}F

Explanation:

We are given that

P_1=200lbf/in^2

P_2=60lbf/in^2

v_1=200ft/s

v_2=1700ft/s

T_1=500^{\circ}F

Q=0

C_p=1BTU/lb^{\circ}F

We have to find the exit temperature.

By steady energy flow equation

h_1+v^2_1+Q=h_2+v^2_2

C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}

1BTU/lb=25037ft^2/s^2

Substitute the values

1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}

500+1.598=T_2+115.4

T_2=500+1.598-115.4

T_2=386.2^{\circ}F

7 0
3 years ago
.You have always been impressed by the speed of the elevators in your apartment building. You wonder about the maximum accelerat
AlexFokin [52]

Answer:

5.51 m/s^2

Explanation:

Initial scale reading = 50 kg  

assume the greatest scale reading = 78.09 kg

<u>Determine the maximum acceleration for these elevators</u>

At rest the weight is = 50 kg

Weight ( F ) = mg = 50 * 9.81 = 490.5 N<u> </u>

<u> </u>At the 10th floor weight = 78.09 kg

Weight at 10th floor ( F ) = 78.09 * 9.81 = 766.11 N

F = change in weight

Change in weight( F ) = ma = 766.11 - 490.5 (we will take the mass as the starting mass as that mass is calculated when the body is at rest)

50 * a = 275.61

Hence the maximum acceleration ( a ) = 275.61 / 50 = 5.51 m/s^2

3 0
3 years ago
A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with
gulaghasi [49]

Answer:

8.8 cm

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Explanation:

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k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm

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At x = 0.7 A

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\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s

The block's speed is 31.422 cm/s

4 0
3 years ago
why can scientists ignore the forces of attraction between particles in a gas under ordinary conditions?
Anna35 [415]
<span>The particles in a gas are apart and moving fast, so the forces of attraction are too weak to have a noticeable effect.</span>
5 0
3 years ago
Read 2 more answers
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