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Reptile [31]
3 years ago
7

Need Help.

Physics
1 answer:
zaharov [31]3 years ago
7 0
Newton’s fifth law says so i’m sorry it’s just logic
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A helium tank holds a volume of .02 m^3 at a pressure of 15.5*10^6 Pa and a temperature of 293 K. How many spherical balloons wi
lukranit [14]

Answer:

Explanation:

We shall find the volume of stored gas at atmospheric pressure .

P₁ V₁ = P₂ V₂

15.5 x 10⁶ x .02 = 10⁵ x V₂  ( atmospheric pressure is 10⁵ Pa )

V₂ = 3.1 m³

volume of one balloon = 4/3 x π r³ , r is radius of balloon

= 4/3  π x .11³

= .050658 m³

no of balloon = total volume to be filled / volume of one balloon

= 3.1 / .050658

= 61 .2

= 61

7 0
3 years ago
Neglecting air resistance, which of the following statements is true regarding an object in freefall?Lector inmersivo A. An obje
Dafna1 [17]

Answer:

E. An object’s velocity changes at a constant rate, and its acceleration remains constant.

Explanation:

When an object is in freefall, it implies that the object is falling freely under gravity. If it falls towards the earth surface, the fall is in the direction of the Earth's gravitational force.

At the point of release of the object, its initial velocity is zero because it is at rest. But when released, its velocity increases at a constant rate until it is acted upon by an external force. But its acceleration remains constant, acceleration due to gravity.

3 0
4 years ago
Consider two soap bubbles with radius r1 and r2 (r1 <r2) connected via a valve. What happens if we open the valve​
Sedaia [141]

Complete Question

The  complete question is shown on the first uploaded image

Answer:

The pressure difference of the first bubble is   \Delta  P _1 =10  J/m^3

The pressure difference of the second bubble is  \Delta  P _2 =20  J/m^3

The pressure difference on the second bubble is higher than that of the first bubble so when the valve is opened pressure from second bubble will cause air to flow toward the first bubble making is bigger

Explanation:

From the question we are told that

    The  radius of the first bubble is  r_1 =  10 \ mm=0.01 \ m

      The radius of the second bubble is  r_2  =  5 \ mm  =  0.005 \ m

      The surface tension of the soap solution is  s =  25 \ mJ/m^2 = 25*10^{-3} J/m^2

Generally according to the Laplace's Law for a spherical membrane the pressure difference is mathematically represented as

         \Delta  P  =  \frac{4 s}{R}

Now the pressure difference for the first bubble is  mathematically evaluated as

        \Delta  P _1 =  \frac{4 s}{r_1}

substituting values  

       \Delta  P _1 =  \frac{4 *25 *10^{-3}}{0.01}

       \Delta  P _1 =10  J/m^3

Now the pressure difference for the second bubble is  mathematically evaluated as

        \Delta  P _2 =  \frac{4 s}{r_1}

       \Delta  P _2 =  \frac{4 *25 *10^{-3}}{0.005}

       \Delta  P _2 =20  J/m^3

3 0
3 years ago
A ufo was detected on the radar flying 7400 miles in 3 min, what is the estimated speed per hour?
Ghella [55]
S=7400mi
t=3 min= 0.05h
v=7400mi/0.05h
v=148000mph
8 0
4 years ago
A boy jumps from rest, straight down from the top of a cliff. He falls halfway down to the water below in 0.883 s. How much time
Salsk061 [2.6K]

Answer:

1.25 s

Explanation:

We are given that

Initial velocity of boy=u=0

Time=0.883 s

Let x be the total distance covered by boy in entire trip.

s=\frac{x}{2}

We know that

s=ut+\frac{1}{2}gt^2

Where g=10m/s^2

Substitute the values

\frac{x}{2}=0\times 0.883+\frac{1}{2}(10)(0.883)^2

\frac{x}{2}=3.898

x=3.898\times 2=7.796m

Again using the above formula

7.796=0(t)+\frac{1}{2}(10)t^2

7.796=5t^2

t^2=\frac{7.796}{5}=1.5592

t=\sqrt{1.5592}=1.25 s

Hence, 1.25 s passes during his entire tripe from the top down to the water.

8 0
3 years ago
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