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kiruha [24]
2 years ago
10

A 5 inch tall balloon shoot doubles in height every 3 days. if the equation y=ab^x, where is x is the number of doubling periods

,represents the height of the bamboo shoot,what are the values of a and b?
Physics
1 answer:
VashaNatasha [74]2 years ago
3 0
In this item if we let x be the number of doubling period, that is number of days divided by 3, and y be the height of the bamboo shoot, in the equation
                                     y  = ab^x
The value of a is 5 because that is the original height of the shoot and b is 2 because we are talking about the doubling rate. The equation is therefore,
                                    y = 5(2^x)
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In an adiabatic process oxygen gas in a container is compressed along a path that can be described by the following pressure p,
viktelen [127]

Answer:

Explanation:

In case of gas , work done

W = ∫ p dV , p is pressure and dV is small change in volume

the limit of integration is from Vi to Vf .

= ∫ p dV

=  ∫ p₀V^{-\frac{6}{5}  dV

= p₀ V^{-\frac{6}{5} +1} / ( \frac{-6}{5} +1 )

=  - 5p₀ V^{-\frac{1}{5}

Taking limit from Vi  to Vf

W = - 5 p₀ ( V_f^\frac{-1}{5} - V_i^{\frac{-1}{5}  ) ltr- atm.

7 0
2 years ago
In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad
Crank

Answer:

a)   Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c)      Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

        P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

         A₁ v₁ = A₂ v₂

         v₂ = v₁ A₁ / A₂

We replace

         ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

         ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

         v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

           Q = A v

           Q = A₁ v₁

           Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]

The areas are

            A₁ = π r₁

            A₂ = π r₂

We replace

        Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]

Let's calculate for the different pressures

      r₁ = d₁ / 2 = 1.00 / 2

      r₁ = 0.500 10⁻² m

      r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

      Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))

       Q = 1.57 10⁻² √(12 10³/2550)

        Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

        Q = 1.57 10⁻² √(2 12 10³ / (850 3)

         Q = 4.8 10⁻² m³ / s

5 0
2 years ago
A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force resisting the carts moti
-BARSIC- [3]

The force applied by the man is 60 N

Explanation:

We can solve this problem by applying Newton's second law, which states that:

\sum F = ma (1)

where

\sum F is the net force acting on the child+cart

m is the mass of the child+cart system

a is their acceleration

In this problem, we have:

m = 30.0 kg is the mass

a=1.50 m/s^2

And there are two forces acting on the child+cart system:

  • The forward force of pushing, F
  • The force resisting the cart motion, R = 15.0 N

Therefore we can write the net force as

\sum F = F -R

where R is negative since its direction is opposite to the motion

So eq.(1) can be rewritten as

F-R=ma

And solving for F,

F=ma+R=(30.0)(1.50)+15.0=60 N

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

4 0
2 years ago
A spring has a spring constant of 48 N/m. The end of the spring hangs 8 m above the ground. How much weight can be placed on the
Setler79 [48]
The answer is 96 N .....................................
7 0
2 years ago
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An 80-g particle moving with an initial speed of 50 m/s in the positive x direction strikes and sticks to a 60-g particle moving
liubo4ka [24]

The collision is a form of inelastic collision because the it forms a single mass after is collides. So it can be solve by momentum balance

( 0.08 kg * 50 m/s ) + ( 0.06 kg * 50 m/s) = ( 0.08 + 0.06 kg ) v

V = 50 m/s

So the kinetic energy lost is

KE = 0.5 (50 m/s)^2) *( 0.14 – 0.08kg )

KE = 75 J

8 0
2 years ago
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