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3 years ago

Define Bernoulli's principle

2 answers:

8 0

The principle in hydrodynamics that an increase in the velocity of a stream of fluid results in a decrease in pressure. Also called Bernoulli effect or Bernoulli theorem.

Serhud [2]3 years ago

6 0

It's when an increase in a liquid's speed creates an el pressure decrease in a liquid'ss speed and creates a pressure increase. <3

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Compare and contrast the energy transfer of a roller coaster to that of a pendulum

Softa [21]

When the pendulum and roller coaster move to the top, its has more potential energy whereas when comes to the bottom has more kinetic energy.

<h3>Compare and contrast the energy transfer of a roller coaster to that of a pendulum:</h3><h3>What is the transfer of energy in a roller coaster?</h3>

The transfer of potential energy to kinetic energy occur when the roller coaster move along the track. As the motor pulls the cars to the top, the body has more potential energy whereas when the body comes to the bottom , it has kinetic energy in the object.

<h3>What is the energy transfer in a pendulum?</h3>

As a pendulum swings, its potential energy changes to kinetic energy and kinetic energy changes into potential energy. At the top more potential energy is present.

So we can conclude that When the pendulum and roller coaster move to the top, its has more potential energy whereas when comes to the bottom has more kinetic energy.

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8 0

2 years ago

A mysterious rocket-propelled object of mass 49.5 kg is initially at rest in the middle of the horizontal, frictionless surface

Romashka-Z-Leto [24]

The distance covered by the rocket in the first 5 seconds is 7.13 m

Explanation:

In order to solve the problem, we have to find the acceleration of the rocket. We can do it by using Newton's second law, which states that the net force on the rocket is equal to the product between its mass (m) and its acceleration (a):

$F(t)=ma(t)$

Where here we have:

$F(t)=16.8t$ is the time-dependent force

m = 49.5 kg is the mass

Solving for a(t),

$a(t)=\frac{F(t)}{m}=\frac{16.8t}{49.5}=0.34t[m/s^2]$

This is the time-dependent acceleration of the rocket.

By integrating this expression, we find the velocity of the rocket at time t:

$v(t)=\int a(t)dt=\int (0.34t)dt=\frac{0.34t^2}{2}=0.17t^2$

where we didn't add the constant term since the velocity at t = 0 is zero (the rocket starts from rest).

By integrating this expression, we then find an expression for the position of the rocket at time t:

$x(t)=\int v(t) dt = \int (0.17t^2)dt=\frac{0.17t^3}{3}=0.057t^3$

Therefore, the distance covered by the rocket in the first t = 5.00 s is:

$x(5)=0.057(5.0)^3=7.13 m$

Learn more aboiut force and acceleration:

brainly.com/question/11411375

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8 0

3 years ago

A rectangular tank that is 4 feet long, 2 feet wide and 15 feet deep is filled with a heavy liquid that weighs 60 pounds per cub

V125BC [204]

Answer:

a) W₁ = 54000 Lb-ft

b) W₂ = 77760 Lb-ft

c) W₃ = 24000 Lb-ft

W₄ = 40560 Lb-ft

Step by step

W= ∫₁² ydF 1 and 2 are the levels of liquid

Where dF is the differential of weight of a thin layer

y is the height of the differential layer and

ρ*V = F

Then

dF = ρ* A*dy*g

ρ*g = 60 lb/ft³

A= Area of the base then

Area of the base is:

A(b) = 4*2 = 8 ft²

Now we have the liquid weighs 60 lb/ft³

Then the work is:

W₁ = ∫₀¹⁵ 8*60*y*dy ⇒ W₁ =480* ∫₀¹⁵ y*dy

W₁ =480* y² /2 |₀¹⁵ ⇒ 480/2 [ (15)² - 0 ]

W₁ = 240*225

W₁ = 54000 Lb-ft

b) The same expression, but in this case we have to pump 3 feet higher, then:

W₂ = ∫₀¹⁸ 480*y*dy ⇒ 480*∫₀¹⁸ydy ⇒ 480* y²/2 |₀¹⁸

W₂ = 480/2 * (18)²

W₂ = 240*324

W₂ = 77760 Lb-ft

c) To pump two-thirds f the liquid we have

2/3* 15 = 10

W₃ = 480*∫₀¹⁰ y*dy ⇒ W₃ = 480* y²/2 |₀¹⁰

W₃ = 240*(10)²

W₃ = 24000 Lb-ft

W₄ =480*∫₀¹³ y*dy

W₄ =480* y²/2 |₀¹³

W₄ = 240*(13)²

W₄ = 240*169

W₄ = 40560 Lb-ft

3 0

3 years ago

If two automobiles have the same velocity do they have the same acceleration?

Mademuasel [1]

Nope not really one could have a faster acceleration than the other.

8 0

3 years ago

Two flat conductors are placed with their inner faces separated by 9mm. If the surface charge density on inner face A is 69 pC/m

ahrayia [7]

Answer:

0.07 V

Explanation:

The electric field between two parallel plates is uniform, and its magnitude is given by

$E=\frac{\sigma}{\epsilon_0}$

where

$\sigma$ is the magnitude of the charge density on each face

$\epsilon_0 = 8.85419 \cdot 10^-12 C^2/Nm^2$ is the vacuum permittivity

In this problem, we have

$\sigma=69 pC/m^2 = 69\cdot 10^{-12} C/m^2$

So, the electric field is

$E=\frac{69\cdot 10^{-12} C/m^2}{8.85419\cdot 10^-12 C^2/Nm^2}=7.79 N/C$

So now we can calculate the potential difference between the two plates, given by:

$\Delta V=E d$

where $d=9 mm=0.009 m$ is the distance between the two plates. Substituting, we find

$\Delta V=(7.79 N/C)(0.009 m)=0.07 V$

5 0

4 years ago