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S_A_V [24]
3 years ago
10

A mercury manometer (rho = 13,600 kg/m3) is connected to an air duct to measure the pressure inside. The difference in the manom

eter levels is 30 mm, and the atmospheric pressure is 100 kPa. (a) Judging from Fig. P1–65, determine if the pres- sure in the duct is above or below the atmospheric pressure. (b) Determine the absolute pressure in the duct
Physics
1 answer:
beks73 [17]3 years ago
3 0

Answer:

P = 104 kPa

Explanation:

a) To know if the pressure is higher or lower than the atmospheric, we only have to see the tubes if the right side is higher the pressure is higher than the atmospheric, otherwise it is lower.

b) The pressure in a system is

          P = P_{atm} + ρ g (Y2-y1)

Where P_{atm} is the atmospheric pressure in the open part of the tuno,ρ is the liquid density and (y2-y1) is the difference height between the two sides of the tube.

Let's reduce to SI units

         (y2-Y1) = 30 mm (1m / 1000 mm) = 30 10-3 m

        P_{atm} = 100 KPa = 100 10 3 Pa

Let's calculate the pressure

       P =  P_{atm}  + 13.6 9.8 30 10⁻³

       P = 100 10³ + 3,998

       P = 103,998 Pa

      P = 104 kPa

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Alkaline earth metals have a low density<br><br><br> true<br> false
Marta_Voda [28]

true

Explanation:

this is because melting point and boiling point decreases down the group because they are held together by attractions between positive nuclei and delocalised electrons

6 0
3 years ago
To win the game, a place kicker must kick a
Dafna11 [192]

Answer:

1.86 m

Explanation:

First, find the time it takes to travel the horizontal distance.  Given:

Δx = 52 m

v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s

a = 0 m/s²

Find: t

Δx = v₀ t + ½ at²

52 m = (22.2 m/s) t + ½ (0 m/s²) t²

t = 2.35 s

Next, find the vertical displacement.  Given:

v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s

a = -9.8 m/s²

t = 2.35 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²

Δy = 4.91 m

The distance between the ball and the crossbar is:

4.91 m − 3.05 m = 1.86 m

5 0
4 years ago
How does friction affect the distance of an object?
stiv31 [10]
It slows the object down so it cannot move well and evetually the object cannot be pushed and farther
4 0
3 years ago
A camera takes a properly exposed photo with a 4.0 mm diameter aperture and a shutter speed of 1/1000 s. If the photographer foc
hoa [83]

Answer:

option E

Explanation:

given,

diameter = 4 mm

shutter speed = 1/1000 s

diameter of aperture = ?

shutter speed = 1/250 s

exposure time to the shutter time

E V = log_2(\dfrac{N^2}{t})

N is the diameter of the aperture and t is the time of exposure

now,

log_2(\dfrac{N^2}{t_1})=log_2(\dfrac{N^2}{t_2})

\dfrac{N_1^2}{t_1}=\dfrac{N_2^2}{t_2}

inserting all the values

\dfrac{4^2}{1000}=\dfrac{N_2^2}{250}

      N₂² = 4

      N₂ = 2 mm

hence , the correct answer is option E

4 0
3 years ago
A 4.10 g bullet moving at 837 m/s strikes a 820 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
atroni [7]

Answer:

(a) 1.85 m/s

(b) 4.1 m/s

Explanation:

Data

  • bullet mass, Mb = 4.10 g
  • initial bullet velocity, Vbi = 837 m/s
  • wooden block mass, Mw = 820 g
  • initial wooden block  velocity, Vwi = 0 m/s
  • final bullet velocity, Vbf = 467 m/s

(a) From the conservation of momentum:

Mb*Vbi + Mw*Vwi = Mb*Vbf + Mw*Vwf

Mb*(Vbi - Vbf)/Mw = Vwf

4.1*(837 - 467)/820 = Vwf

Vwf = 1.85 m/s

(b) The speed of the center of mass speed is calculated as follows:

V = Mb/(Mb + Mw) * Vbi

V = 4.1/(4.1 + 820) * 837

V = 4.1 m/s

6 0
3 years ago
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