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kirill [66]
3 years ago
7

g The bottom end of a long vertical tube filled with liquid is opened in a basin exposed to air having pressure 100.8 kilo-Pasca

ls. The top end of the tube remains closed. The maximum height of liquid in the tube that this air pressure can support is 10.5 meters. Determine the density of the liquid in units of bevelled fraction numerator k g over denominator m cubed end fraction.
Physics
1 answer:
Masja [62]3 years ago
3 0

Answer:

979.6 kg/m³

Explanation:

We know pressure P = hρg where h = height of liquid = 10.5 m, ρ = density of liquid and g = acceleration due to gravity = 9.8 m/s²

So, density ρ = P/hg

Since P = 100.8 kPa = 100.8 × 10³ Pa

substituting the values of the variables into the equation for ρ, we have

ρ = P/hg

= 100.8 × 10³ Pa ÷ (10.5 m × 9.8 m/s²)

= 100.8 × 10³ Pa ÷ 102.9 m²/s²

= 0.9796 × 10³ kg/m³

= 979.6 kg/m³

So, the density of the liquid is 979.6 kg/m³

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The light from Star 1 reaches Star 3 in 138 years and Star 2 in 63 years. Which of these conclusions about the stars is correct?
krek1111 [17]

The correct answer is A


6 0
3 years ago
Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is
Llana [10]

Answer:

K_e_q=22.75878093\frac{N}{m}

f=1.363684118Hz

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}

Replacing the data provided:

\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}

K_e_q=22.75878093\frac{N}{m}

Finally, to calculate the frequency of oscillation we use this:

f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }

Replacing m and k:

f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz

4 0
3 years ago
A cart is moving to the right with a constant speed of 20 m s . A box of mass 80 kg moves with the cart without slipping. The co
nignag [31]

Answer:

Explanation:

Given

mass of box m=80 kg

coefficient of kinetic friction \mu _k=0.15

coefficient of Static friction \mu _s=0.30

cart is moving with constant velocity therefore Net Force is zero

Since there is no net acceleration therefore friction force will be zero

mathematically

f_r=ma

where f_r=frictional\ Force

a=acceleration

a=0

f_r=0

4 0
3 years ago
A rifle has a mass of 7-kg and the bullet has a mass of 0.7-kg. If the velocity of the bullet is 350-m/s after the rifle is fire
nevsk [136]

Answer:

-35 m/s

Explanation:

Momentum is conserved.

Momentum before firing = momentum after firing

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Before the bullet is fired, the bullet and rifle have no velocity, so u₁ and u₂ are 0.

0 = m₁v₁ + m₂v₂

Given m₁ = 0.7 kg, v₁ = 350 m/s, and m₂ = 7 kg:

0 = (0.7 kg) (350 m/s) + (7 kg) v

v = -35 m/s

The rifle recoils at 35 m/s in the opposite direction.

8 0
3 years ago
Gardeners understand that plants have specific pH requirements. Most plants thrive in soil with a pH that is slightly acidic, ne
guapka [62]

Answer:

the answer is d if u r using usatesprep

6 0
3 years ago
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