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2 years ago

What is the speed of a wave in (m/s) with a 5 meter wavelength and a period of 20 seconds?

Physics

1 answer:

arlik [135]2 years ago

6 0

Answer: 0.25 m/s

Explanation: Speed = wavelengt · frequency

v = λf and frequency is 1/period f = 1/T

Then v = λ/T = 5 m / 20 s = 0.25 m/s

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Light travels fastest when moving through

vitfil [10]

A vaccum unlike sound,light can travel through any matter including a great vacuum of nothing (space)

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3 years ago

A voltmeter was used to check the coolant and a reading of 0.2 volt with the engine off was measured. A reading of 0.8 volt was

Julli [10]

Answer:

C. Technician B

Explanation:

Excessive Galvanic activity:

To check for excessive galvanic activity, voltmeter is used to check the coolant. If the voltmeter is giving a reading greater than 0.5 V, there is excessive galvanic activity. Excessive galvanic activity is solved by flushing the coolant fluid from engine and refiling it.

Electrolysis problem:

When the system is not properly ground, the cooling system accepts stray current and the coolant becomes an electrolyte which might eat up the radiator. To test for excessive electrolysis, start the engine and turn on all electrical accessories, if the reading is more than 0.5 V, there is electrolysis problem. Ground wires and connections should be checked at this point to stop stray current.

In our case, the first reading is 0.2 V(engine turned off) which is normal and there is no excessive galvanic activity. This means that Technician A is not correct. The second reading is 0.8 V when the engine and all electrical accessories are turned on. This reading is greater than 0.5 V which means there is an electrolysis problem. This means that Technician B is correct and ground wires and connections should be inspected and repaired.

7 0

3 years ago

block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8

dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: $\sqrt{800} \,\,\,N$

Explanation:

First try to write all forces in vector component form:

The force F1 acting at 45 degrees would have multiplication factors of $\frac{\sqrt{2} }{2}$ on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is $F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

y-component of F1 is $F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

y-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is: $v_x=4\,*\,2=8\,m/s$

y-component of the velocity is: $v_y=4\,*\,2=8\,m/s$

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3 years ago

The coach has not seen improvement as expected in his team and plans to change practice session. To help the players he changes

Allushta [10]

Answer:

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Explanation:

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3 years ago

A 1.2 g pebble is stuck in a tread of a 0.76 m diameter automobile tire, held in place by static friction that can be at most 3.

Maksim231197 [3]

Answer:

$v=33.764m/s$

Explanation:

Given data

Mass m=1.2 g=0.0012 kg

diameter d=0.76 m

Friction Force F=3.6 N

To find

Velocity v

Solution

From the Centripetal force we know that

$F_{c}=\frac{mv^{2} }{r}$

Where m is mass

v is velocity

r is radius

Substitute the given values to find velocity v

$F_{c}=\frac{mv^{2} }{r}\\v^{2}=\frac{F_{c}(r)}{m}\\ v=\sqrt{\frac{F_{c}(r)}{m}}\\ v=\sqrt{\frac{F_{c}(diameter/2)}{m}}\\v=\sqrt{\frac{(3.6N)(0.76/2)m}{(0.0012kg)}}\\v=33.764m/s$

4 0

3 years ago