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pychu [463]
2 years ago
15

How do we calculate the "range" of the measurements?

Physics
1 answer:
Alika [10]2 years ago
8 0

Answer: I think its A

Explanation:

If not, what is the answer, beause I'm on the same quiz.

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Analyze how friction can be both a positive and negative aspect on our lives
denpristay [2]
Positive - Friction allows us to create heat in a desperate situation, like being lost in the woods. If we didn't have friction, we would probably freeze to death.
Negative - Friction can also cause unwanted fires, such as forest fires. If friction didn't exist, we wouldn't have these.
8 0
3 years ago
A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at
storchak [24]

Answer

given,

wavelength = λ = 18.7 cm

                    = 0.187 m

amplitude , A = 2.34 cm

v = 0.38 m/s

A)  angular frequency = ?

     f = \dfrac{v}{\lambda}

     f = \dfrac{0.38}{0.187}

     f =2.03\ Hz

angular frequency ,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) the wave number ,

      K = \dfrac{2\pi}{\lambda}

     K= \dfrac{2\pi}{0.187}

    K =33.59\ m^{-1}

C)

as the wave is propagating in -x direction, the sign is positive between x and t

y ( x ,t) = A sin(k  x - ω t)

y ( x ,t) = 2.34  x  sin(33.59 x - 12.75 t)

4 0
3 years ago
A cello string vibrates in its fundamental mode with a frequency of 335 1/s. The vibrating segment is 28.5 cm long and has a mas
Inga [223]

Answer:

The tension in string is found to be 188.06 N

Explanation:

For the vibrating string the fundamental frequency is given as:

f1 = v/2L

where,

f1 = fundamental frequency = 335 Hz

v = speed of wave

L = length of string = 28.5 cm = 0.285 m

Therefore,

v = f1 2L

v = (335 Hz)(2)(0.285)

v = 190.95 m/s

Now, for the tension:

v = √T/μ

v² = T/μ

T = v² μ

where,

T = Tension

v = speed = 190.95 m/s

μ = linear mass density of string = mass/L = 0.00147 kg/0.285 m = 5.15 x 10^-3 kg/m

Therefore,

T = (190.95 m/s)²(5.15 x 10^-3 kg/m)

<u>T = 188.06 N</u>

4 0
3 years ago
Read 2 more answers
How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck
Katyanochek1 [597]

Answer:

Explanation:

From the given information:

the car's momentum = momentum of the truck

∴

(a) 816 kg × v = 2650 kg × 16.0 km/h

v = (2650 kg × 16.0 km/h) /  816 kg

v = 51.96 km/hr

(b) 816 kg × v = 9080 kg × 16.0 km/h

v = (9080 kg × 16.0 km/h) /  816 kg

v = 178.04 km/hr

8 0
3 years ago
A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.
Tcecarenko [31]

Answer:

a) 1.89 m/s  b) 172.1 N

Explanation:

a)

  • Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
  • This work, is just 4,475 J.
  • So we can write the following equation:

        \Delta K = \frac{1}{2} * m*v^{2} = 4,475 J

  • where m= mass of the truck = 2.5*10³ kg.
  • So, we can find the speed v, as follows:

        v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} }  = 1.89 m/s

b)

  • The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

        W = F*d

  • We can solve for F, as follows:

        F = \frac{W}{d} = \frac{4,475 J}{26.0m} =  172.1 N

4 0
3 years ago
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