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Anna11 [10]
3 years ago
12

The amount of force needed to keep a 0.1 kg basketball moving at a constant speed of 6 m/s on frictionless ice is how much N?

Physics
1 answer:
FromTheMoon [43]3 years ago
3 0
  • Mass=m=0.1kg
  • Acceleration=0(Constant speed)

\\ \sf\longmapsto F=ma

\\ \sf\longmapsto F=0.1(0)

\\ \sf\longmapsto F=0N

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Distinguishing sound and electromagnetic waves
Angelina_Jolie [31]

Electromagnetic waves are not mechanical and transverse in nature, while

Sound waves are mechanical and longitudinal

Explanation:

In physics, waves are classified into two types:

  • Mechanical waves are those where the vibration is produced by the vibrations of the particles in a medium, therefore they need a medium to propagate
  • Electromagnetic waves are those produced by the oscillations of electric and magnetic fields; they do not a medium to propagate, since they can travel through a vacuum

Moreover, depending on the direction of the vibration, waves are further classified into:

  • Transverse waves: those where the direction of the vibration is perpendicular to the direction of propagation of the wave
  • Longitudinal waves: those where the direction of the vibration is parallel to the direction of propagation of the wave

Electromagnetic waves are not mechanical and transverse in nature (the oscillations of the electric and magnetic fields occur in a plane perpendicular to the direction of the wave), while

Sound waves are mechanical and longitudinal (the vibrations of the particles of the medium is back-and-forth along the direction of propagation of the wave)

Learn more about waves:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

8 0
3 years ago
if a 1-m diameter sewer pipe is flowing at a depth of 0.4 m and has a flow rate of 0.15 m^3/s, what will be the flow rate when t
Anna35 [415]

Answer:

0.35 m³/s

Explanation:

When the pipe's depth is 0.4 m, the area of the circular segment is:

A = ½ R² (θ − sin θ)

The depth of the water is:

h = R (1 − cos(θ/2))

Solving for θ:

0.4 = 0.5 (1 − cos(θ/2))

0.8 = 1 − cos(θ/2)

cos(θ/2) = 0.2

θ/2 = acos(0.2)

θ = 2 acos(0.2)

θ ≈ 2.74 rad

The area is therefore:

A = ½ (0.5 m)² (2.74 − sin 2.74)

A = 0.338 m²

The cross-sectional area when the pipe is full is:

A = π (0.5 m)²

A = 0.785 m²

The flow velocity is constant:

v = v

Q / A = Q / A

(0.15 m³/s) / (0.338 m²) = Q / (0.785 m²)

Q = 0.35 m³/s

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Answer:

D

Explanation:

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