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Ulleksa [173]
3 years ago
6

Which is an example of an abiotic factor in a pound environment

Physics
2 answers:
zvonat [6]3 years ago
4 0
A. The water

Biotic factors mean living this such as frogs, grasshoppers, snakes, etc.

Aboriginal factors means non-living such as water, temperature, sunlight, etc.
Abiotic factors are the non-living parts of the environment that have a major influence on living organisms which makes water an abiotic factor.

Hope this helped, if it did please like (if you want to :)
ella [17]3 years ago
4 0

Answer: the water

Explanation: its not living

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WARRIOR [948]

Answer:

Explanation:distance-time =speed

a,V =s/t

V=2m/2s

V=1m/s

b,v=s/t

V=80m/40s

V= 2m/s

The average speed is 2m/s

8 0
2 years ago
Problem 1 Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed
ANTONII [103]

Answer:

markers are 29.76 m far apart in the laboratory

Explanation:

Given the data in the question;

speed of particle = 0.624c

lifetime = 159 ns = 1.59 × 10⁻⁷ s

we know that; c is speed of light which is equal to 3 × 10⁸ m/s

we know that

distance = vt

or s = ut

so we substitute

distance = 0.624c × 1.59 × 10⁻⁷ s

distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s

distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s

distance =  29.76 m

Therefore, markers are 29.76 m far apart in the laboratory

3 0
2 years ago
Imagine the ball on the left is given a nonzero initial speed in the horizontal direction, while the ball on the right continues
Llana [10]

Let say the height of two balls from the ground is H

now we can use kinematics

s = v_i * t + \frac{1}{2} at^2

now we have

H = \frac{1}{2}gt^2

t = \sqrt{\frac{2H}{g}}

now in the same time ball on the left will cover the horizontal distance between them

v_x = \frac{d}{ t}[/tex[tex]v_x = \frac{3}{\sqrt{\frac{2H}{g}}}

<em>so above is the horizontal speed of the left ball</em>

8 0
3 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
2 years ago
An 85 kg object is moving at a constant speed of 15 m/s in a circular path, which has a radius of 20 meters. What centripetal fo
elena55 [62]
Mass, m = 85 kg, velocity = 15 m/s, radius = 20m.

Centripetal force = mv²/r

F = 85*15²/20 =  85*15*15/20 = 956.25 N
3 0
3 years ago
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