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NISA [10]
3 years ago
13

A student creates a model of a closed ecosystem by filling a glass tank half full with water then adding 10 snails and two small

aquatic plants the next day all the snails are dead what is the most likely cause of their death
Physics
2 answers:
worty [1.4K]3 years ago
8 0

In any closed ecosystem, if we put some snails and some plants in them, the snails would survive for a while but next day they will die, this is because of the reason that the snails can survive for a while in closed ecosystem but the waste products which takes place in the ecosystem due to both the plants as well as the snails affects the snails due to which they are likely to die soon.

GenaCL600 [577]3 years ago
4 0
The reason why the snails are dead because the snails lack oxygen for it is a model of closed ecosystem. Another reason to this is that they lack  abiotic factors that they need in surviving or as they live for the place that they are living lacks a lot and are not enough for them to last or survive. Lastly, the water could have not been a contributor for them to live for they may not be able to live in a water environment and could only dwell in land.
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What characteristics of a force can be measured?
Helga [31]
Direction and Strength can be measured of force 
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3 years ago
A muon is traveling at 0.988
AlexFokin [52]

As per Einstein's relation of relativity

m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}

here we know that

m_0 = 207 m_e = 207 \times 9.11 \times 10^{-31} kg

m_0 = 1.88 \times 10^{-28} kg

now here we know that

v = 0.988 c

now from above equation mass of the muon is given as

m = \frac{1.88 \times 10^{-28}}{\sqrt{1 - 0.988^2}}

m = 1.22 \times 10^{-27} kg

now for the momentum of muon we can use

P = mv

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6 0
4 years ago
A child on a sled slides (starting from rest) down an icy slope that makes an angle of 15◦ with the horizontal. After sliding 20
statuscvo [17]

Answer:

A) v₁ = 10.1 m/s t₁= 4.0 s

B) x₂= 17.2 m

C) v₂=7.1 m/s

D) x₂=7.5 m

Explanation:

A)

  • Assuming no friction, total mechanical energy must keep constant, so the following is always true:

       \Delta K + \Delta U = (K_{f} - K_{o}) +( U_{f} - U_{o}) = 0  (1)

  • Choosing the ground level as our zero reference level, Uf =0.
  • Since the child starts from rest, K₀ = 0.
  • From (1), ΔU becomes:
  • \Delta U = 0- m*g*h = -m*g*h (2)  
  • In the same way, ΔK becomes:
  • \Delta K = \frac{1}{2}*m*v_{1}^{2}  (3)      
  • Replacing (2) and (3) in (1), and simplifying, we get:

       \frac{1}{2}*v_{1}^{2}  = g*h  (4)

  • In order to find v₁, we need first to find h, the height of the slide.
  • From the definition of sine of an angle, taking the slide as a right triangle, we can find the height h, knowing the distance that the child slides down the slope, x₁, as follows:

       h = x_{1} * sin \theta_{1} = 20.0 m * sin 15 = 5.2 m (5)

       Replacing (5) in (4) and solving for v₁, we get:

      v_{1} = \sqrt{2*g*h} = \sqrt{2*9.8m/s2*5.2m} = 10.1 m/s  (6)

  • As this speed is achieved when all the energy is kinetic, i.e. at the bottom of the first slide, this is the answer we were looking for.
  • Now, in order to finish A) we need to find the time that the child used to reach to that point, since she started to slide at the its top.
  • We can do this in more than one way, but a very simple one is using kinematic equations.
  • If we assume that the acceleration is constant (which is true due the child is only accelerated by gravity), we can use the following equation:

       v_{1}^{2} - v_{o}^{2} = 2*a* x_{1}  (7)

  • Since v₀ = 0 (the child starts from rest) we can solve for a:

       a = \frac{v_{1}^{2}}{2*x_{1} } = \frac{(10.1m/s)^{2}}{2* 20.0m} = 2.6 m/s2  (8)

  • Since v₀ = 0, applying the definition of acceleration, if we choose t₀=0, we can find t as follows:

       t_{1} =\frac{v_{1} }{a} =\frac{10.1m/s}{2.6m/s2} = 4.0 s  (9)

B)

  • Since we know the initial speed for this part, the acceleration, and the time, we can use the kinematic equation for displacement, as follows:

       x_{2} = v_{1} * t_{2} + \frac{1}{2} *a_{2}*t_{2}^{2}  (10)

  • Replacing the values of v₁ = 10.1 m/s, t₂= 2.0s and a₂=-1.5m/s2 in (10):

       x_{2} = 10.1m/s * 2.0s + \frac{1}{2} *(-1.5m/s2)*(2.0s)^{2}  = 17.2 m (11)

C)  

  • From (6) and (8), applying the definition for acceleration, we can find the speed of the child whem she started up the second slope, as follows:

       v_{2} = v_{1} + a_{2} *t_{2} = 10.1m/s - 1.5m/s2*2.0s = 7.1 m/s (12)

D)

  • Assuming no friction, all the kinetic energy when she started to go up the second slope, becomes gravitational potential energy when she reaches to the maximum height (her speed becomes zero at that point), so we can write the following equation:

       \frac{1}{2}*v_{2}^{2}  = g*h_{2}   (13)

  • Replacing from (12) in (13), we can solve for h₂:

       h_{2} =\frac{v_{2} ^{2}}{2*g} = \frac{(7.1m/s) ^{2}}{2*9.8m/s2} = 2.57 m  (14)

  • Since we know that the slide makes an angle of 20º with the horizontal, we can find the distance traveled up the slope applying the definition of sine of an angle, as follows:

       x_{3} = \frac{h_{2} }{sin 20} = \frac{2.57m}{0.342} = 7.5 m (15)

4 0
3 years ago
A point charge Q moves on the x-axis in the positive direction with a speed of 370 m/s. A point P is on the y-axis at y = +80 mm
mixas84 [53]

Answer:

+58

Explanation:

Cause I intelligent

3 0
3 years ago
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