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puteri [66]
3 years ago
8

un serbatoio d'acqua di forma cilindrica è riempito con acqua dolce fino a un livello di 6,40 m calcola la pressione dell'acqua

alla profondità di 4,50 m e di 5,50 m dalla superficie dell'acqua
Physics
1 answer:
ohaa [14]3 years ago
6 0
Per calcolare la pressione a una certa profondità, devi considerare la legge di Stevino:
<span>p = ρ · g · h

Tenendo conto che:
g = 9,81 m/s²
ρ = 1000 kg/m³

Troviamo:
p(h</span>₁) = ρ · g · h₁ = 1000 · 9,81 · 4,50 = 44145 Pa
p(h₂) = ρ · g · h₂ = 1000 · 9,81 · 5,50 = 53955 Pa


You might be interested in
Machmer Hall is 400 m North and 180 m West of Witless.
yan [13]

Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

Machmer Hall is 400 m North and 180 m West of Witless.

We need to calculate the distance

Using Pythagorean theorem

D = \sqrt{(d_{m})^2+(d_{w})^2}

Where, d_{m} =distance of Machmer Hall

d_{w} =distance of Witless

Put the value into the formula

D = \sqrt{(400)^2+(180)^2}

D=438.63\ m

Hence, The distance from Witless to Machmer is 438.63 m.

5 0
3 years ago
Students want to investigate the inverse relationship between the pressure and temperature of an ideal gas as predicted by the i
grin007 [14]

Answer:

Option B, Fix the piston in place so the volume of the pas remains constant

Explanation:

As we know

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

The effect on variable due to another variable can be studied by keeping the third variable constant.

Hence, in order the study the variation of temperature with pressure or vice versa, the volume needs to fixed at a certain value.

Hence, option B is correct

7 0
3 years ago
Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as
viva [34]

Answer:

a) L=0. b) L = 262 k ^   Kg m²/s and c)  L = 1020.7 k^   kg m²/s

Explanation:

It is angular momentum given by

      L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]

Let's apply this relationship to our case

Let's start by breaking down the speed

      v₀ₓ = v₀ cosn 45

      voy =v₀ sin 45

      v₀ₓ = 9 cos 45

      voy = 9 without 45

      v₀ₓ = 6.36 m / s

      voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

   vfy² = voy²- 2 g y

   y = voy² / 2g

   y = (6.36)²/2 9.8

   y = 2.06 m

Let's calculate the angular momentum

L= \left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^   Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

     R = vo² sin 2θ / g

     R = 9² sin (2 45) /9.8

     R = 8.26 m

L = \left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^   kg m² /s

5 0
3 years ago
I need help
musickatia [10]

Answer:

C

Explanation:

4 0
2 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
GarryVolchara [31]

I'm in physics, but I think the answer is 22. The angle of reflection is reflecting whatever the angle was so, in this case, it must be 22

3 0
2 years ago
Read 2 more answers
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