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Klio2033 [76]
3 years ago
8

During the summer, when your lawn is growing quickly, estimate this speed, in m/s. Make this estimate from your experience notin

g, for instance, how often you mow the lawn and what length you trim.
Physics
1 answer:
Vsevolod [243]3 years ago
4 0

Answer: If T is the time you wait to trim the lawn since you last did it and B is the lenght at which you trim, then the speed will be:

v=\frac{B}{T}

Explanation:

Let's say T is the time you wait to trim the lawn since you last did it (1 month for example) and B is the lenght at which you trim (10cm for example).

If v is the speed at which the lawn grows, then the lenght of the lawn as a function of time will be:

b(t)=b_{0}+ vt

where b_{0} is an arbitrary initial lenght. Let's say b_{0}=0.

b(t)=vt

Then we have:

v=\frac{b(t)}{t}

So, at time T the lawn has grown a lenght B. And the speed is:

v=\frac{B}{T}

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If a car increases its velocity from 1m/s to 3.6km/hr in 5 seconds what’s its acceleration
EleoNora [17]

Answer:

applying 1st eq of motion vf=vi+at here we have to find a=vf-vi/t , a= 1-1/5 , a=0/5 then we got a=0 here(vf value 3.6km/h is converted in standard unit 3.6×1000/3600 so we get vf=1m/s²

7 0
3 years ago
Two train whistles, and , each have a frequency of 392 Hz. is stationary and is moving toward the right (away from ) at a speed
trasher [3.6K]

Answer:

Part a)

f = 371.1 Hz

Part b)

f = 417.7 Hz

Part c)

beat frequency = 46.6 Hz

Explanation:

Part a)

Due to doppler's Effect the frequency of the sound heard by the train which is moving away from the observer is given as

f_1 = f_0\frac{v + v_o}{v + v_s}

f_1 = 392(\frac{340 + 15}{340 + 35})

f_1 = 371.1 Hz

Part b)

Now from the second train which is approaching the person we can say

f_2 = f_0\frac{v - v_o}{v - v_s}

f_2 = 392(\frac{340 - 15}{340 - 35})

f_2 = 417.7 Hz

Part c)

As we know that beat frequency is the difference in the frequency from two sources

f_b = f_2 - f_1

f_b = 417.7 - 371.1 = 46.6 Hz

8 0
3 years ago
If a = 2.0 cm, b = 5.0 cm, and i = 20 a, what is the magnitude of the magnetic field at the point p?
attashe74 [19]

If a = 2.0 cm, b = 5.0 cm, and i = 20 a,  6.0 μt is the magnitude of the magnetic field at the point p, So the correct option is (a).

The magnetic influence on moving electric charges, electric currents, and magnetic materials is described by a magnetic field, which is a vector field. A force perpendicular to the charge's own velocity and the magnetic field acts on it when the charge is travelling through a magnetic field.

B_{1} = μ_{0} i \frac{\pi }{6} / 4\pi (a+b)

B_{2} = μ_{0} i \frac{\pi }{6} / 4\pi b

As, B_{1} is moving down and B_{2} is moving up so,

B_{2} - B_{1} = (μ_{0} i \frac{\pi }{6} / 4\pi b) -  [μ_{0} i \frac{\pi }{6} / 4\pi (a+b)]

B_{2} - B_{1} = μ_{0} i 24 / (\frac{1}{b} - \frac{1}{a+b} )

B_{2} - B_{1} = \frac{4\pi *10^{-7}*20 }{24} (\frac{1}{0.05} -\frac{1}{0.02})

B_{2} - B_{1} = 5.98×10^{-6} T ≈ 6μT

Therefore,  6.0 μt is the magnitude of the magnetic field .

Learn more about  magnetic field here;

brainly.com/question/23096032

#SPJ4

3 0
2 years ago
Show your work please ​
nekit [7.7K]

Answer:

V = U + at

Explanation:

<u>Given the following data;</u>

Initial velocity = 0 (since the stone is starting from rest).

Final velocity = 32 m/s

Acceleration = g = 10 m/s²

Time = 3.2 seconds

To show that the speed of the stone when it hits the ground is 32 m/s, we would use the first equation of motion;

V = U + at

Where;

  • V is the final velocity.
  • U is the initial velocity.
  • a is the acceleration.
  • t is the time measured in seconds.

Substituting into the formula, we have;

32 = 0 + 10*3.2

32 = 0 + 32

32 = 32

<em>Proven: 32 m/s = 32 m/s</em>

7 0
3 years ago
In what way is the liquid water different from the water vapor?
just olya [345]

Answer:

The difference is the way the molecules are arranged and the way they move

Explanation:

Water molecules in liquid water are constantly sliding past and bumping into each other; they keep moving from one place to another. The molecules of water in water vapor are far apart and moving freely.

3 0
4 years ago
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