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kkurt [141]
3 years ago
15

Invasive species are _______ species that have a ______ impact on a given ecosystem.

Physics
2 answers:
S_A_V [24]3 years ago
7 0
The answer is non-native, harmful.
Katyanochek1 [597]3 years ago
3 0

Answer:

non-native, harmful

Explanation:

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A hockey puck, mass of 0.115 kg, moving at 35 m/s strikes a ray thrown on the ice by a fan. The ray has a mass of 0.265 kg. The
emmainna [20.7K]
We can use conservation of momentum. MaVa + MbVb = (Ma+Mb)Vf.

Plugging in values:
(.115)(35) + (.265)0 = (.380)(Vf)

Now solve for Vf.
Vf= .115*35 / .380 = 10.6 m/s
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3 years ago
Which statement about a right triangle is NOT true?
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Answer:

The square of the hypotenuse is equal to the sum of the squares of the other two lengths.

Explanation:

only one that made sense

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An 8 g bullet leaves the muzzle of a rifle with
Elena-2011 [213]

Answer: 1872 N

Explanation:

This problem can be solved by using one of the Kinematics equations and Newton's second law of motion:

V^{2}=V_{o}^{2} + 2ad (1)

F=ma (2)

Where:

V=611.9 m/s is the bullet's final speed (when it leaves the muzzle)

V_{o}=0 is the bullet's initial speed (at rest)

a is the bullet's acceleration

d=0.8 m is the distance traveled by the bullet before leaving the muzzle

F is the force

m=8 g \frac{1 kg}{1000 g}=0.008 kg is the mass of the bullet

Knowing this, let's begin by isolating a from (1):

a=\frac{V^{2}}{2d} (3)

a=\frac{(611.9 m/s)^{2}}{2(0.8 m)} (4)

a=234013.5063 m/s^{2} \approx 2.34(10)^{5} m/s^{2} (5)

Substituting (5) in (2):

F=(0.008 kg)(2.34(10)^{5} m/s^{2}) (6)

Finally:

F=1872 N

4 0
3 years ago
Is the circuit you created a parallel, series or a series/parallel circuit? Support your answer with a description of this type
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It's a circuit wherein all the elements are in parallel and a resistance is fitted in series across which a voltmeter is connected to measure potential drop acrossR so as to find the current across the load that is bulb which connected to in series with the resistor
5 0
4 years ago
A fuzzy bunny sees a butterfly and chases it right off of a 20 m high cliff at 7m/s
bogdanovich [222]

A) 2.02 s

To find the time it takes for the bunny to reach the rocks below the cliff, we just need to analyze its vertical motion. This motion is a free fall motion, which is a uniformly accelerated motion. So we can use the suvat equation:

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

where

s is the vertical displacement

u is the initial vertical velocity

a is the acceleration

t is the time

For the bunny here, choosing downward as positive direction,

u_y = 0 (initial vertical velocity is zero)

s = 20 m

a=g=9.8 m/s^2 (acceleration of gravity)

And solving for t, we find the time of flight:

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s

B) 14.1 m

For this part, we need to consider the horizontal motion of the bunny.

The horizontal motion of the bunny is a uniform motion with constant velocity, which is the initial velocity of the bunny:

v_x = 7 m/s

Therefore the distance covered after time t is given by

d=v_x t

And substituting the time at which the bunny hits the ground,

t = 2.02 s

We find how far the bunny went from the cliff:

d=(7)(2.02)=14.1 m

C) 21.0 m/s at 70.5^{\circ} below the horizontal

The horizontal component of the velocity is constant during the entire motion, so it will still be the same when the bunny hits the ground:

v_x = 7 m/s

Instead, the vertical velocity is given by

v_y = u_y +at

And substituting t = 2.02 s, we find the vertical velocity at the moment of impact:

v_y = 0+(9.8)(2.02)=19.8 m/s

So, the magnitude of the final velocity is

v=\sqrt{v_x^2+v_y^2}=\sqrt{7^2+(19.8)^2}=21.0 m/s

And the angle is given by

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{19.8}{7})=70.5^{\circ}

below the horizontal

5 0
3 years ago
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