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Sergio039 [100]
3 years ago
15

Please I really need help on these

Physics
1 answer:
blagie [28]3 years ago
4 0

Answer:

  3, 4, 2

Explanation:

I can't see the final ones.

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The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0
katen-ka-za [31]

Answer:

a)  F = 2.66 10⁴ N, b)   h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

          P₂ = P₀ = 1.01 105 Pa

inside

         P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

         P₁ = 0.85   1.01 10⁵ + 1000   9.8  5

         P₁ = 85850 + 49000

         P₁ = 1.3485 10⁵ Pa

the net force is

         ΔP = P₁- P₂

         Δp = 1.3485 10⁵ - 1.01 10⁵

         ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

         P = Fe / A

         F = P A

the area of ​​a circle is

         A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

         d = 100 cm (1 m / 100 cm) = 1 m

         F = 3.385 104 pi / 4 (1) ²

         F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

        P₁ = P₂

        0.85 P₀ + ρ g h = P₀

        h = \frac{P_o ( 1-0.850)}{\rho \ g}

        h = \frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}

        h = 1.55 m

4 0
2 years ago
Determine the amount of potential energy of a 4.2 kg book that is placed on a shelf with a height of 0.9 meters. Round your answ
swat32

Answer:

that is a correct answer maybe .

5 0
3 years ago
Jacob is riding his bike down a tall hill. At the top of the hill, he has 2800J of potential energy. At the bottom of the hill,
elena55 [62]

Answer:

Just find the formula then you can do it

4 0
2 years ago
Read 2 more answers
If the momentum of an object is doubled then kinetic energy is ...?​
stiks02 [169]

Answer:

increased with the same rate as momentum

7 0
3 years ago
An electron of mass 9.11×10−31 kgkg leaves one end of a TV picture tube with zero initial speed and travels in a straight line t
ki77a [65]

Explanation:

We will use the equations of constant acceleration to find out a_{x} and time t.

As we know that the initial speed is zero. So

(a)  

v_{0x} = 0

x - x_{o} = 1.25×10^{-2}m

v_{x} = 3.3×10^{6}m/s

v^{2} _{x} =  v^{2} _{x_{o} } + 2a_{x} (x - x_{o} )

a_{x} = \frac{v^{2} _{x} - v^{2} _{ox} }{2(x - x_{o}) }

   = \frac{(3.3 * 10^{6})^{2}  - 0 }{2(1.25 * 10^{-2}) }

   = 4.356×10^{14} m/s²

(b)

v_{x} = v_{ox} + a_{x}t

t = v_{x} - vo_{x}/a_{x}

t = \frac{3.00 * 10^{6} }{4.356*10^{14} } = 6.8870×10^{-9}s

(c)

ΣF_{x} = ma_{x}

       = (9.11×10^{-31})(4.356×10^{14}m/s²)

       = 3.968×10^{-16} N

6 0
2 years ago
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