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Hitman42 [59]
2 years ago
11

The period of a wave is 20 ms (milliseconds) and its wavelength is 4 cm. Calculate:

Physics
1 answer:
IgorLugansk [536]2 years ago
8 0

Answer:

A. 50 Hz

B. 2 m/s

Explanation:

We'll begin by converting 20 ms to s. This can be obtained as follow:

1000 ms = 1 s

Therefore,

20 ms = 20 ms × 1 s / 1000 ms

20 ms = 0.02 s

Next, we shall convert the value of the wavelength (i.e 4cm) to m. This can be obtained as follow:

100 cm = 1 m

Therefore,

4 cm = 4 cm × 1 m / 100 cm

4 cm = 0.04 m

A. Determination of the frequency.

Period (T) = 0.02 s

Frequency (f) =?

f = 1 / T

f = 1 / 0.02

f = 50 Hz

Therefore, the frequency of the wave is 50 Hz

B. Determination of the velocity.

Wavelength (λ) = 0.04 m

Frequency (f) = 50 Hz

Velocity (v) =?

v = λf

V = 0.04 × 50

v = 2 m/s

Therefore, the velocity of the wave is 2 m/s

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a. Acetyl-CoA combines with a pyruvic acid to make glucose in the Krebs cycle.


Explanation:

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b. A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kg  m2. If the string is pulled with a force
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Answer:

f = 8 N

Explanation:

Data provided in the question

Radius of the pulley  = r = 0.05 m

Moment of inertia = (I) = 0.2 kg.m^{2}

Angular acceleration = ∝ = 2 rad/sec

Based on the above information

As we know that

Torque is

= force \times  radius

= f \times r

And,

Torque is also

= moment\ of\ inertia \times angular\ acceleration

= I \times \alpha

So,

We can say that

f \times r = I \times \alpha

f \times 0.05 = 0.2 \times 2

0.05f = 0.4

f = 8 N

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3 years ago
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
puteri [66]

Answer:

t=5.3687\ s  is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is 0\ mph\ to\ 70\ mph.

  • Initial velocity of the car, v_i=0\ mph
  • final velocity of the car during the test, v_f=32\ mph=14.3052\ m.s^{-1}
  • Time taken to accelerate form zero to 32 mph at full power, t=1.2\ s
  • initial velocity of the car, u_i=0\ mph
  • final desired velocity of the car, u_f=64\ mph=28.6105\ m.s^{-1}

Now the acceleration of the car:

a=\frac{v_f-v_i}{t}

a=\frac{14.3052-0}{1.2}

a=11.921\ m.s^{-1}

Now using the equation of motion:

u_f=u_i+a.t

64=0+11.921\times t

t=5.3687\ s is the time taken by the car to accelerate the desired range of the speed from zero at full power.

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