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Dvinal [7]
3 years ago
10

A block of wood 3.0 cm on each side has a mass of 27g. what is the density of this block?

Physics
1 answer:
schepotkina [342]3 years ago
8 0
The block of wood is 3cm on each side so it is a cube. The volume of a cube is given by s^3. So the volume of this block is 3cm x 3cm x3 cm = 27 cm^3. density = mass/volume =27 g / 27 cm^3 = 1 g/cm^3
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A stone with a weight of 5.29 N is launched vertically from ground level with an initial speed of 26.0 m/s, and the air drag on
makkiz [27]

Answer:

a) 19.4 m/s

b) 19 m/s

Explanation:

a) In the given question,

the potential energy at the initial point = Ui = 0

the potential energy at the final point = Uf = mgh

the kinetic energy at the initial point = Ki = 1/2 mv₀².

the kinetic energy at the final point = Kf = 0

work done by air= Ea= fh =  0.262 N

Now, using the law of conservation of energy

initial energy= final energy

Ki +Ui = Kf + Uf +Ea

1/2 mv₀² + 0 = 0 + mgh + fh

1/2 mv₀² = mgh + fh

h = v₀²/ 2g (1 +f/w)

calculate m

m= w/g = 5.29 /9.8

= 0.54 kg

h = 20 ²/ (2 x9.80) x (1 0.265/5.29)

h = 19.4 m.

b) 1/2 mv² + 2fh = 1/2 mv₀²

Vg = 19 m/s

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3 years ago
the atoms in a sample are close together but can slide past one another. as the atoms lose energy, they move slower. the atoms b
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The change of state that is occurring is from the liquid state to the solid state.
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A technique in which the muscles are stretched by an outside force is called _____.
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3 years ago
In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st
bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

  • Total time will be the sum of the partial times between stations plus the time stopped at the stations.
  • Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
  • At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
  • In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

       v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s  (1)

  • Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

       t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)

  • Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

       x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m  (3)

  • In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

       t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)

  • We can find the distance traveled while the train was decelerating as follows:

       x_{3} = (v_{1} * t_{3})   + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m  (5)

  • Finally, we need to know the time traveled at constant speed.
  • So, we need to find first the distance traveled at the constant speed of 26.4m/s.
  • This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
  • x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s   (7)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
  • Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
  • tm = t*5 = 131.6 * 5 = 658.2 s (9)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
  • Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

  • Using all the same premises that for a) we know that the only  difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
  • Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
  • We can find the distance traveled at constant speed, rewriting (6) as follows:
  • x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s   (12)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
  • Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
  • tm = t*3 = 207.4 * 3 = 622.2 s (14)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
  • Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)
7 0
3 years ago
How long does it take for a train to increase its velocity from 10m/s to 40m/s if it accelerates at 3 m/s
DIA [1.3K]

Answer:

Explanation:

Givens

Vi = 10 m/s

Vf = 40 m/s

a = 3 m/s^2

Formula

a = (vf - vi) /t              Substitute the givens into this formuls

Solution

3 = (40 - 10) / t          Multiply both sides by t

3*t = t(40 - 10)/t        Combine. Cancel t's on the right

3*t = 30                     Divide by 3

3t/3 = 30 / 3

Answer: t = 10 seconds.

6 0
2 years ago
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