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zaharov [31]
3 years ago
12

Which of the following best describes a coastal plain?

Physics
2 answers:
evablogger [386]3 years ago
5 0
Area near a sea having flat land and low relief
Katarina [22]3 years ago
4 0

<em>Hello,</em>

<em>The correct answer is....</em>

<em>B</em><em>) Area near a sea having flat land and low relief.</em>

<em>Hope this helps!!!! :)</em>

<em>**(</em><em>Vanessa</em><em>)**</em>

<em>**Can you please mark me Brainliest answer!!!!! Thanks!!!!! :)</em>

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A cd player uses a ___ to convert info to an electrical signal
Alex_Xolod [135]
C.) Laser. the light from the laser reflects off the shiny surface as the CD rotates
7 0
3 years ago
Achilles and the tortoise are having a race. The tortoise can run 1 mile (or whatever the Hellenic equivalent of this would be)
fenix001 [56]

Answer:

Surely Achilles will catch the Tortoise, in 400 seconds

Explanation:

The problem itself reduces the interval of time many times, almost reaching zero. However, if we assume the interval constant, then it is clear that in two hours Achilles already has surpassed the Tortoise (20 miles while the Tortoise only 3).

To calculate the time, we use kinematic expression for constant speed:

x_{final}=x_{initial}+t_{tor}v_{tor}=1+t_{tor}\\x_{final}=x_{initial}+t_{ach}v_{ach}=10t_{ach}

The moment that Achilles catch the tortoise is found by setting the same final position for both (and same time as well, since both start at the same time):

1+t=10t\\t=1/9 hour=0.11 hours

7 0
3 years ago
Suppose a particle moves along a straight line with velocity v(t)=t2e−2tv(t)=t2e−2t meters per second after t seconds. How many
dimulka [17.4K]

Explanation:

It is given that,

Velocity of the particle moving in straight line is :

v(t)=t^2e^{-2t}\ m/s

We need to find the distance (x)  traveled by the particle during the first t seconds. It is given by :

x=\int\limits {v.dt}

x=\int\limits {t^2e^{-2t}dt}

Using by parts integration, we get the value of x as :

x=\dfrac{-(2t^2+2t+1)e^{-2t}}{4}\ meters

Hence, this is the required solution.

6 0
3 years ago
1.- Un barco recorre la distancia que separa Gran Canaria de Tenerife (90 km) en 6 horas. ¿Cuál es
bezimeni [28]

Answer:

The speed is 15 km/h or 4.16 m/s.

Explanation:

A boat travels the distance that separates Gran Canaria from Tenerife (90 km) in 6 hours. Which  the speed of the boat in km / h? And in m / s?

Given that,

Distance, d = 90 km = 90000 m

Time, t = 6 hours = 21600 s

Speed = distance/time

v=\dfrac{90\ km}{6\ h}\\\\=15\ km/h

or

v=\dfrac{90000\ m}{21600\ s}\\\\=4.16\ m/s

So, the required speed is 15 km/h or 4.16 m/s.

7 0
3 years ago
5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc
ludmilkaskok [199]

Answer:

Δθ₁ =  172.5 rev

Δθ₁h =  43.1 rev

Δθ₂ =   920 rev

Δθ₂h = 690 rev

Explanation:

  • Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

       \Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2}  (1)  

  • Now, we need first to find the value of  the angular acceleration, that we can get from the following expression:

       \omega_{f1}  = \omega_{o} + \alpha * \Delta t  (2)

  • Since the machine starts from rest, ω₀ = 0.
  • We know the value of ωf₁ (the operating speed) in rev/min.
  • Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

       3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)

  • Replacing by the givens in (2):

       57.5 rev/sec = 0 + \alpha * 6 s  (4)

  • Solving for α:

       \alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)

  • Replacing (5) and Δt in (1), we get:

       \Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev  (6)

  • in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

       \Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev  (7)

  • In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

       \Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2}  (8)

  • First of all, we need to find the value of the angular acceleration during the second period.
  • We can use again (2) replacing by the givens:
  • ωf =0 (the machine finally comes to an stop)
  • ω₀ = ωf₁ = 57.5 rev/sec
  • Δt = 32 s

       0 = 57.5 rev/sec + \alpha * 32 s  (9)

  • Solving for α in (9), we get:

       \alpha_{2}  =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)

  • Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

        \Delta \theta_{2}  = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)

  • In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
  • \Delta \theta_{2h}  = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)
7 0
2 years ago
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