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mylen [45]
3 years ago
7

A horizontal board of negligible thickness and area 2.0 m2 hangs from a spring scale that reads 80 N when a 4.0 m/s wind moves b

eneath the board. The air above the board is stationary. What does the scale read when the wind stops? The density of air is 1.25 kg/m3 .
Physics
1 answer:
ki77a [65]3 years ago
6 0

Answer:

Scale reading for no wind W'=60N

Explanation:

From the question we are told that

Area A= 2.0 m^2

Weight of board W=80

Velocity V=4.0m/s

Density of air \delta= 1.25 kg/m3 .

Generally the equation for pressure difference by Bernoulli equation is mathematically given by

  dP=\frac{1}{2}pv^2

  dP=10Pa

Generally force acting on the board by air is mathematically given by

F=\triangle PA

F=(10)2=>20N

Therefore

Scale reading for no wind W'

W'=W-F\\W'=80-20

W'=60N

Scale reading for no wind W'=60N

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A toy cannon uses a spring to project a 5.38-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force
yawa3891 [41]

(a) 1.43 m/s

We can solve this problem by using the law of conservation of energy.

The initial total energy stored in the spring-mass system is

E=U=\frac{1}{2}kx^2

where

k = 7.91 N/m is the spring constant

x=5.08 cm = 0.0508 m

Substituting,

E=\frac{1}{2}(7.91)(0.0508)^2=0.0102 J

The final kinetic energy of the ball is equal to the energy released by the spring + the work done by friction:

E+W_f=K

where

K_f=\frac{1}{2}mv^2 is the kinetic energy of the ball, with

m=5.38 g = 5.38\cdot 10^{-3} kg being the mass of the ball

v being the final speed

W_f = -F_f d is the work done by friction (which is negative since the force of friction is opposite to the motion), where

F_f = 0.0323 N is force of friction

d = 14.5 cm = 0.145 m is the displacement

Substituting,

W_f = -(0.0323)(0.145)=-4.68\cdot 10^{-3} J

So, the kinetic energy of the ball as it leaves the cannon is

K_f = E+W_f = 0.0102 - 4.68\cdot 10^{-3}=0.00552 J

And so the final speed is

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.00552)}{0.00538}}=1.43 m/s

(b) +5.08 cm

The speed of the ball is maximum at the instant when all the elastic potential energy stored in the spring has been released: in fact, after that moment, the spring does no longer release any more energy, so the kinetic energy of the ball from that moment will start to decrease, due to the effect of the work done by friction.

The elastic potential energy of the spring is

U=\frac{1}{2}kx^2

And this has all been released when it becomes zero, so when x = 0 (equilibrium position of the spring). However, the spring was initially compressed by 5.08 cm, so the ball has maximum speed when

x = +5.08 cm

with respect to the initial point.

(c) 1.78 m/s

The maximum speed is the speed of the ball at the moment when the kinetic energy is maximum, i.e. when all the elastic potential energy has been released.

As we calculated in part (a), the total energy released by the spring is

E = 0.0102 J

The work done by friction here is just the work done to cover the distance of

d = 5.08 cm = 0.0508 m

Therefore

W_f = -(0.0323)(0.0508)=-1.64\cdot 10^{-3} J

So, the kinetic energy of the ball at the point of maximum speed is

K_f = E+W_f = 0.0102 - 1.64\cdot 10^{-3}=0.00856 J

And so the final speed is

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.00856)}{0.00538}}=1.78 m/s

7 0
3 years ago
Calculate the frequency if the number of revolutions is 300 and the paired poles are 50.
Andrei [34K]

Answer: A

Explanation: We know that f=p*n

f=50*300=15000 Hz = 15kHz.

Have a great day! <3

3 0
2 years ago
PLEASE HELP ASAP!!
Gwar [14]
The third one.

The nucleus of an atom is made up of protons and neutrons. Protons and neutrons have the same relative mass (rounded to 1). Electrons have an extremely small mass.

Therefore, the particles in the nucleus are more massive than those in the electron cloud.
4 0
2 years ago
Astronomers discover an exoplanet, a planet obriting a star other than the Sun, that has an orbital period of 3.27 Earth years i
Naddik [55]

Answer:

  r = 3.787 10¹¹ m

Explanation:

We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration

    F = ma

    G m M / r² = m a

The centripetal acceleration is given by

    a = v² / r

For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship

    v = d / t

The distance traveled Esla orbits, in a circle the distance is

    d = 2 π r

Time in time to complete the orbit, called period

     v = 2π r / T

Let's replace

    G m M / r² = m a

    G M / r² = (2π r / T)² / r

    G M / r² = 4π² r / T²

    G M T² = 4π² r3

     r = ∛ (G M T² / 4π²)

Let's reduce the magnitudes to the SI system

     T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)

     T = 1.03 10⁸ s

Let's calculate

      r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]

      r = ∛ (21.44 10³⁵ / 39.478)

      r = ∛(0.0543087 10 36)

      r = 0.3787 10¹² m

      r = 3.787 10¹¹ m

7 0
3 years ago
What are alkaline earth metals used for?
Tomtit [17]

Answer:

C firework

Explanation:

from Quizlet

6 0
3 years ago
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