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raketka [301]
2 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]2 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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An individual white LED (light-emitting diode) has an efficiency of 20% and uses 1.0 W of electric power. a. How many LEDs must
Neporo4naja [7]

Answer:

8, 8 W

Explanation:

The useful power of 1 Light Emitting Diode is

0.2\times 1=0.2\ W

Total power required is 1.6 W

Number of Light Emitting Diodes would be

n=\dfrac{1.6}{0.2}\\\Rightarrow n=8

The number of Light Emitting Diodes is 8.

Power would be

P=8\times 1=8\ W

The power that is required to run the Light Emitting Diodes is 8 W

7 0
2 years ago
A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/
Elena-2011 [213]

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

KE_i + PE_i = KE_f + PE_f

\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0

\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2

So by solving above equation we will have

v_f = 84.3 m/s

now in order to find the momentum we can use

P = mv

P = 1.25 \times 84.3

P = 105.4 kg m/s

6 0
2 years ago
 If the gauge pressure of a gas is 114 kPa, what is the absolute pressure?
Anastasy [175]

Answer:

D. 214 kPa

Explanation:

The absolute pressure is given by:

p = p_a + p_g

where

p is the absolute pressure

p_a \sim 100 kPa is the atmospheric pressure

p_g is the gauge pressure

In this problem, we have

p_g = 114 kPa

So, the atmospheric pressure is

p = 100 kPa + 114 kPa = 214 kPa

4 0
2 years ago
Read 2 more answers
A cubical box, 5.00 cm on each side, is immersed in a fluid. The gauge pressure at the top surface of the box is 594 Pa and the
Zolol [24]

Answer:

The density of the fluid is 1100 kg/m³.

Explanation:

Given that,

Height = 5.00 cm

Pressure at top =594 Pa

Pressure at bottom = 1133 Pa

We need to calculate the change in pressure

Using formula of change in pressure

\Delta P=P_{b}-P_{t}

Where, P_{b} = Pressure at bottom

P_{t} = Pressure at top

put the value into the formula

\Delta P=1133-594

\Delta P=539\ Pa

Using formula of pressure for density

\Delta P = \rho g h

\rho =\dfrac{\Delta P}{gh}

Where, \rho = density

P = pressure

h = height

Put the value in to the formula

\rho =\dfrac{539}{5.00\times10^{-2}\times9.8}

\rho =1100\ kg/m^3

Hence, The density of the fluid is 1100 kg/m³.

4 0
2 years ago
A 14000N car traveling at 25m/s rounds a curve of radius 200m. Find the following: a. The centripetal acceleration of the car.
tamaranim1 [39]

Answer:

Explanation:

Given

Weight of car W=14,000\ N

mass of car m=\frac{14,000}{9.8}=1428.57\ N

velocity of car v=25\ m/s

radius r=200\ m

(a)Centripetal acceleration is given by

a_c=\frac{v^2}{r}

a_c=\frac{25^2}{200}

a_c=3.125\ m

(b)Force that provide centripetal acceleration

F=F_c=\frac{mv^2}{r}

F=\frac{1428.57\times 25^2}{200}

F=4464.285\ N

(c)Friction force between car and tires is given by

=\mu N

where \mu=coefficient of static friction

N=normal reaction

Centripetal force will balance the friction force

F_c=F_r

4464.285=\mu \times 1428.57\times 9.8

\mu =0.318

6 0
2 years ago
Read 2 more answers
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