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fgiga [73]
3 years ago
5

What would cause rain showers to continue over a particular area for several days?

Physics
2 answers:
fiasKO [112]3 years ago
8 0
<h2>Answer:<em><u> a high-pressure system</u></em></h2>

Aloiza [94]3 years ago
5 0
I think is a high-pressure system because it is only in one particular area.
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In an effort to protect a rhino, volunteers are following its steps with air monitoring and ground cameras. The rhino starts on
finlep [7]

Answer:

Explanation:

We shall represent each displacement by vectors . i will represent east , -i west , j north and - j south .

Rhino walks 1.5 km west on day 1.

D₁ = - 1.5 i

The rhino then moves 0.7 km on a straight line in a direction of 15o east of north toward ground camera B

D₂ = .7 sin15 i + .7cos15 j

On the second day, the rhino moves 2.5 km directly south

D₃ = - 2.5 j

D = D₁ + D₂ + D₃

= - 1.5 i + .7 sin15 i + .7cos15 j - 2.5 j

= - 1.5 i + .181 i + .676 j - 2.5 j

= - 1.32 i - 1.824 j

magnitude  of total displacement

= √ (1.32² +1.824²

= 2.25 km

For direction we shall calculate slope with x axis

Tanθ = - 1.824 / - 1.32

= 54°

So rhino will be towards 54° south of west as both x and y coordinates are negative.

6 0
3 years ago
If the tire reaches a temperature of 48 ∘c, what fraction of the original air must be removed if the original pressure of 250 kp
andrey2020 [161]
Assuming air as ideal gas and amount of air in no of moles is known then by gas law,

PV= nRT

Pressure is constant

P* (change in volume) = nR* (change in temperature)
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3 years ago
The weakest type of friction that occurs between solid surfaces is _____ friction
Evgen [1.6K]

Answer:

Static, sliding, and rolling friction occur between solid surfaces. Static friction is strongest, followed by sliding friction, and then rolling friction, which is weakest. Fluid friction occurs in fluids, which are liquids or gases.

Explanation:

4 0
3 years ago
Read 2 more answers
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
2 years ago
large truck exerts 7000 n of force on a piston with an area of 0.4m squared and the is ton can only support 16,000 pa of pressur
sasho [114]
7000 N / 0,4m^2  = 17 500 Pa.
6 0
3 years ago
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