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Nimfa-mama [501]
3 years ago
14

[W]?

Physics
1 answer:
natali 33 [55]3 years ago
5 0

87 km I think .........

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What is the net force acting upon this object? *
AnnZ [28]

2 N Right

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4 0
3 years ago
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An open pipe, 0.29 m long, vibrates in the second overtone with a frequency of 1,227 Hz. In this situation, the fundamental freq
Andru [333]

Answer:

f = 409 Hz

Explanation:

We have,

Length of the open organ pipe, l = 0.29 m

Frequency of vibration of second overtone, f_2 = 1227 Hz

It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :

f_2=\dfrac{3v}{2l}

v is speed of sound

Let f is the fundamental frequency. It is given by :

f=\dfrac{v}{2l}

The relation between f and f₂ can be written as :

f_2=3f\\\\f=\dfrac{f_2}{3}\\\\f=\dfrac{1227}{3}\\\\f=409\ Hz

So, the fundamental frequency of the pipe is 409 Hz.              

6 0
3 years ago
When a 5.0kg cart undergoes a 2.2m/s increase in speed, what is the impulse of the cart
Karolina [17]

Answer:

11.0 kg m/s

Explanation:

The impulse exerted on the cart is equal to its change in momentum:

I=\Delta p=m\Delta v

where

m = 5.0 kg is the mass of the cart

\Delta v=2.2 m/s is its change in speed

Substituting numbers into the equation, we find

I=(5.0kg)(2.2 m/s)=11 kg m/s

7 0
3 years ago
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The bus is heading southwest at 56 miles per hour is it veloctiy or speed
Anvisha [2.4K]

Answer:

Its velocity

Explanation:

Velocity is speed with a direction.

Speed is the rate at which someone or something is able to move or operate.

In the question its telling us that the bus is going 56 miles per hour but it aslo gives us a direction, in this case is southwest.

4 0
3 years ago
A flywheel with a diameter of 1.42 m is rotating at an angular speed of 207 rev/min. (a) What is the angular speed of the flywhe
Archy [21]

Answer:

a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions

Explanation:

a. Its angular speed in radians per second  ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s

b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m

So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s

c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min

α = (ω₁ - ω)/t

  = (1410 - 207)/(80.5/60)

  = 60(1410 - 207)/80.5

  = 60(1203)80.5

  = 896.65 rev/min² ≅ 897 rev/min²

d. Using θ = ωt + 1/2αt²

where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and  t = 80.5/60 min = 1.342 min

θ = ωt + 1/2αt²

  = 207 × 1.342 + 1/2 × 896.65 × 1.342²

  = 277.725 + 807.417

  = 1085.14 revolutions ≅ 1085 revolutions

5 0
3 years ago
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