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Viktor [21]
3 years ago
8

Arace car starts from rest and accelerates uniformly to a speed of 40 m/s in 8.

Physics
1 answer:
9966 [12]3 years ago
8 0

Answer:

160 m

Explanation:

For computing the distance first we have to determine the acceleration which is shown below:

v = u + at

where,

v = 0 = initial velocity

u = 40 = uniformly to a speed

t = 8 = time

So, the a is

= 40 ÷ 8

= 5 m/s

Now we have to use the distance formula i.e

d = ut + 1 ÷ 2 at^2

= 1 ÷2 × 5 × 8^2

= 160 m

Basically we have to applied the above formulas so that the distance could come

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A 37-cm-long wire of linear density 18 g/m vibrating at its second mode, excites the third vibrational mode of a tube of length
lora16 [44]

Answer:

T = 4.42 10⁴ N

Explanation:

this is a problem of standing waves, let's start with the open tube, to calculate the wavelength

        λ = 4L / n                 n = 1, 3, 5, ...    (2n-1)

How the third resonance is excited

       m = 3

       L = 192 cm = 1.92 m

       λ = 4 1.92 / 3

       λ = 2.56 m

As in the resonant processes, the frequency is maintained until you look for the frequency in this tube, with the speed ratio

      v = λ f

      f = v / λ

      f = 343 / 2.56

      f = 133.98 Hz

       

Now he works with the rope, which oscillates in its second mode m = 2 and has a length of L = 37 cm = 0.37 m

The expression for standing waves on a string is

           λ = 2L / n

           λ = 2 0.37 / 2

           λ = 0.37 m

The speed of the wave is

          v = λ f

As we have some resonance processes between the string and the tube the frequency is the same

          v = 0.37 133.98

          v = 49.57 m / s

Let's use the relationship of the speed of the wave with the properties of the string

              v = √ T /μ

              T = v² μ

              T = 49.57²   18

              T = 4.42 10⁴ N

4 0
3 years ago
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. 30
schepotkina [342]

Answer:

Explanation:

Length if the bar is 1m=100cm

The tip of the bar serves as fulcrum

A force of 20N (upward) is applied at the tip of the other end. Then, the force is 100cm from the fulcrum

The crate lid is 2cm from the fulcrum, let the force (downward) acting on the crate be F.

Using moment

Sum of the moments of all forces about any point in the plane must be zero.

Let take moment about the fulcrum

100×20-F×2=0

2000-2F=0

2F=2000

Then, F=1000N

The force acting in the crate lid is 1000N

Option D is correct

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