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Vladimir79 [104]
3 years ago
5

If we assume that, in the African savanna, the predator population remains constant, what other, nonmobile, factor would influen

ce the antelope population?
Physics
2 answers:
musickatia [10]3 years ago
7 0
On usa test prep the answer is c
Phantasy [73]3 years ago
6 0

Answer:

C)  

available grass

Explanation:

just did quiz on usatestprep

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Explain why it is easier to climb a mountain on a zigzag path rather than one straight up the side.
GaryK [48]
Although a zig zag pattern going up a mountain means you walk further, the incline of the slope is a lot less so you don't have to work as hard.
8 0
3 years ago
What is the speed of a truck that travels 10 i’m in 10 minutes ?
fiasKO [112]

10km/10min is a legitimate speed. So is meters/sec, km/hour (kph), etc.  

Kph is very common for vehicles:  

10 km/10 min (60 min/hr) = 60 kph

7 0
3 years ago
Why are latitude and longitude included on maps
Veseljchak [2.6K]

Answer:

to locate places on earth

3 0
2 years ago
A cyclotron designed to accelerate protons has a magnetic field of magnitude 0.15 T over a region of radius 7.4 m. The charge on
klio [65]

Explanation:

It is given that,

Magnetic field, B = 0.15 T

Charge on a proton, q=1.60218\times 10^{-19}\ C

Mass of a proton, m=1.67262 \times 10^{-27}\ kg

The cyclotron frequency is given by :

f=\dfrac{qB}{2\pi m}

f=\dfrac{1.60218\times 10^{-19}\ C\times 0.15\ T}{2\pi \times 1.67262 \times 10^{-27}\ kg}

f = 2286785.40 Hz

or

\omega=14368296.44\ rad/s

\omega=1.43\times 10^7 rad/s

Hence, this is the required solution.

8 0
3 years ago
You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state
posledela

Answer:

a)  y₂ = 49.1 m ,    t = 1.02 s , b)   y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

            v_{y}² = v_{oy}² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

           y-yo = v_{oy}² /2 g
            y- 0 = 10.0²/2 9.8
            y - 0 = 5.10 m
            
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
             y₂ = 5.1 + 44
             y₂ = 49.1 m
Let's use the other equation to find the time
              [tex]v_{y} = v_{oy} - g t

              t = v_{oy} / g

              t = 10 / 9.8

              t = 1.02 s

b) the maximum height

            y- 44.0 = v_{y}² / 2 g

            y - 44.0 = 5.1

            y = 5.1 +44.0

            y = 49.1 m

The time is the same because it does not depend on the initial height

              t = 1.02 s

7 0
3 years ago
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