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Zolol [24]
3 years ago
7

Which term describes the number of crests that pass a point in a given amount of time

Physics
2 answers:
joja [24]3 years ago
7 0
The term is frequency.

The frequency is the number of vibrations per unit of time or the number of waves that passes a point per unit of time.

Every crest (and every trough) represents a pass of the wave so you can count the number of crests in an intervavl of time to find the frequency as the number of crests divided by the time elapsed. 
anzhelika [568]3 years ago
5 0
The correct answer is Frequency.
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Which statement best describes what energy transfer diagrams show?
monitta

Answer:

Every energy transformation results in a reduction of energy

Explanation:

6 0
4 years ago
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A train moving at a constant speed of 50.0 km/h moves east for 45.0 min, then in a direction 45.0° east of due north for 10.0 mi
Yuki888 [10]

Answer:

3.47km/h with a direction of 67.8 degrees north of west.

Explanation:

First we need to calculate the displacement on the X axis, so:

d_{x1}=50.0km/h*45min(\frac{1h}{60min})=37.5km\\d_{x2}=50.0km/h*10min(\frac{1h}{60min})*cos(45^o)=5.90km\\d_{x1}=-50.0km/h*55.0min(\frac{1h}{60min})=45.8km\\D_x=37.5+5.90-45.8=-2.4km

then on the Y axis:

D_y=50.0km/h*10min(\frac{1h}{60min})*sin(45^o)=5.90km

The magnitud of the displacement is given by:

D=\sqrt{D_x^2+D_y^2} \\D=6.37km

and the angle:

\alpha =arctg(\frac{5.90}{2.41})=67.8^o

that is 67.8 degrees north of west.

v=\frac{D}{t}\\v=\frac{6.37km}{110min*\frac{1h}{60min}}\\v=3.47km/h

8 0
4 years ago
If an object accelerating at −1.5m/s^2 takes 1.2s to reach 5.0m/s, what was its initial speed?
USPshnik [31]

-1.5 m/s^2 x 1.2 seconds = -1.8 m/s

It is a negative value which means the object slowed down. The object would have originally been going that amount more.


5.0 + 1.8 = 6.8 m/s


answer: 6.8 m/s

3 0
2 years ago
A piece of aluminum has a volume of 4.89 x 10-3 m3. The coefficient of volume expansion for aluminum is = 69 x 10-6(C°)-1. The t
Leno4ka [110]

Answer:

11.515 Joule

Explanation:

Volume of aluminium = V = 4.89×10⁻³ m³

Coefficient of volume expansion for aluminum = α = 69×10⁻⁶ /°C

Initial temperature = 19.1°C

Final temperature = 357°C

Pressure of air = 1.01×10⁵ Pa

Change in temperature = ΔT= 357-19.1 = 337.9 °C

Change in volume

ΔV = αVΔT

⇒ΔV = 69×10⁻⁶×4.89×10⁻³×337.9

⇒ΔV = 114010.839×10⁻⁹ m³

Work done

W = PΔV

⇒W = 1.01×10⁵×114010.839×10⁻⁹

⇒W = 11.515 J

∴ Work is done by the expanding aluminum is 11.515 Joule

7 0
3 years ago
An amusement park ride called the Rotor debuted in 1955 in Germany. Passengers stand in the cylindrical drum of the Rotor as it
steposvetlana [31]

Answer:

μs = 0.36

Explanation:

  • While the drum is rotating, the riders, in order to keep in a circular movement, are accelerated towards the center of the drum.
  • This acceleration is produced by the centripetal force.
  • Now, this force is not a different type of force, is the net force acting on the riders in this direction.
  • Since the riders have their backs against the wall, and the normal force between the riders and the wall is perpendicular to the wall and aiming out of it, it is easily seen that this normal force is the same centripetal force.
  • In the vertical direction, we have two forces acting on the riders: the force of gravity (which we call weight) downward, and the friction force, that will oppose to the relative movement between the riders and the wall, going upward.
  • When this force be equal to the weight, it will have the maximum possible value, which can be written as follows:

       F_{frmax} = \mu_{s}* F_{n}  = m * g  (1)

  • where μs= coefficient of static friction (our unknown)
  • As  we have already said Fn = Fc.
  • The value of the centripetal force, is related with the angular velocity ω and the radius of the drum r, as follows:

      F_{n} = m* \omega^{2} * r  (2)

  • Replacing (2) in (1), simplifying and rearranging terms, we can solve for μs, as follows:

       \mu_{s} = \frac{g}{\omega^{2} r}  (3)

  • Prior to replace ω for its value, is convenient to convert it from rev/min to rad/sec, as follows:

       \omega = 26.0 \frac{rev}{min} * \frac{1min}{60 sec} *\frac{2*\pi rad }{1 rev} = 2.72 rad/sec (4)

  • Replacing g, ω and r in (3):
  • \mu_{s} = \frac{g}{\omega^{2} r} = \frac{9.8m/s2}{(2.72rad/sec)^{2} *3.7 m} = 0.36 (5)

3 0
3 years ago
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