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Mazyrski [523]
3 years ago
8

According to Bernoulli's equation, the pressure in a fluid will tend to decrease if its velocity increases. Assuming that a wind

speed of 1 m/s causes a pressure drop of 0.645 Pa, what pressure drop is predicted by Bernoulli's equation for a wind speed of 5 m/s

Physics
1 answer:
Pie3 years ago
7 0

Answer:

The pressure drop predicted by Bernoulli's equation for a wind speed of 5 m/s

= 16.125 Pa

Explanation:

The Bernoulli's equation is essentially a law of conservation of energy.

It describes the change in pressure in relation to the changes in kinetic (velocity changes) and potential (elevation changes) energies.

For this question, we assume that the elevation changes are negligible; so, the Bernoulli's equation is reduced to a pressure change term and a change in kinetic energy term.

We also assume that the initial velocity of wind is 0 m/s.

This calculation is presented in the attached images to this solution.

Using the initial conditions of 0.645 Pa pressure drop and a wind speed of 1 m/s, we first calculate the density of our fluid; air.

The density is obtained to be 1.29 kg/m³.

Then, the second part of the question requires us to calculate the pressure drop for a wind speed of 5 m/s.

We then use the same formula, plugging in all the parameters, to calculate the pressure drop to be 16.125 Pa.

Hope this Helps!!!

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A 3-kg rock is thrown upward with a force of 200 N at a location where the local gravitational acceleration is 9.79 m/s2 . Deter
expeople1 [14]

Answer: 56.87m/s^{2}

Explanation:

If we make an analysis of the net force F_{net} of the rock that was thrown upwards, we will have the following:

F_{net}=F_{up}-W  (1)

Where:

F_{up}=200N is the force with which the rock was thrown

W is the weight of the rock

Being the weight the relation between the mass m=3kg of the rock and the acceleration due gravity g=9.79m/s^{2} :

W=m.g=(3kg)(9.79m/s^{2}) (2)

W=29.37 N (3)

Substituting (3) in (1):

F_{net}=200N-29.37 N  (4)

F_{net}=170.63 N  (5) This is the net Force on the rock

On the other hand, we know this force is equal to the multiplication of the mass with the acceleration, according to Newton's 2nd Law:

F_{net}=m.a  (6)

Finding the acceleration a:

a=\frac{F_{net}}{m}  (7)

a=\frac{170.63 N}{3kg} (8)

Finally:

a=56.87m/s^{2}

3 0
2 years ago
What happens when a resultant electric field exist in conductor​
klio [65]

Answer:

When an electric field exists in a conductor a current will flow.

This implies a voltage difference between two points on the conductor.

Electrostatics pertains to static charge distributions.

That means that an object such as a charged spherical conductor will be at the same potential (voltage) on both its outer and inner surfaces.

7 0
3 years ago
Tom has two pendulums with him. Pendulum 1 has a ball of mass 0.2 kg attached to it and has a length of 5 m. Pendulum 2 has a ba
mars1129 [50]

Given Information:

Pendulum 1 mass = m₁ = 0.2 kg

Pendulum 2 mass = m₂ = 0.6 kg

Pendulum 1 length = L₁ = 5 m

Pendulum 2 length = L₂ = 1 m

Required Information:

Affect of mass on the frequency of the pendulum = ?

Answer:

The mass of the ball will not affect the frequency of the pendulum.

Explanation:

The relation between period and frequency of pendulum is given by

f = 1/T

The period of pendulum is given by

T = 2π√(L/g)

Where g is the acceleration due to gravity and L is the length of the string

As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.

Bonus:

Pendulum 1:

T₁ = 2π√(L₁/g)

T₁ = 2π√(5/9.8)

T₁ = 4.49 s

f₁ = 1/T₁

f₁ = 1/4.49

f₁ = 0.22 Hz

Pendulum 2:

T₂ = 2π√(L₂/g)

T₂ = 2π√(1/9.8)

T₂ = 2.0 s

f₂ = 1/T₂

f₂ = 1/2.0

f₂ = 0.5 Hz

So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.

3 0
2 years ago
Read 2 more answers
7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh
pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

F= \mu R

where \mu is the coefficient of friction = 0.02

R = Normal reaction of the load = mgcos\theta = 25 \times 9.81 \times cos 10 = 241.52N

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

F=0.02 \times 241.52N

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0
3 years ago
Find equivalent resistance. <br><br>Answer asap and please, please don't spam.​
Alinara [238K]

Answer:

R = 4.77 ohms

Explanation:

Four resistors are given such that,

R₁ = 2 ohms

R₂ = 3 ohms

R₃ = 5 ohms

R₄ = 10 ohms

Here, R₁ and R₂ in series. The equivalent is given by :

R₁₂ = R₁ + R₂

= 2 + 5

R₁₂ = 7 ohms

Similarly, R₃ and R₄ are in series. so,

R₃₄ = R₃ + R₄

= 10+5

R₃₄ = 15 ohms

Now, R₁₂ and R₃₄ are in parallel. So,

\dfrac{1}{R}=\dfrac{1}{R_{12}}+\dfrac{1}{R_{34}}\\\\\dfrac{1}{R}=\dfrac{1}{7}+\dfrac{1}{15}\\\\R=4.77\ \Omega

So, the equivalent resistance s 4.77 ohms.

4 0
3 years ago
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