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MAVERICK [17]
3 years ago
13

When two different species compete for resources, it is______ competition.

Physics
1 answer:
Katen [24]3 years ago
4 0

Answer:

it is a interspecific competition

Explanation:

i just took a test with this question

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The first law states that “objects at rest and objects in motion remain in motion in a straight line unless acted upon by an unbalanced force”. Keeping the ice smooth will make sure there is not friction, friction would slow the puck down
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Which is an example of transfroming potential energy to kinetic​
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Answer:

A ball being dropped to the ground

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Li and Raj are playing football. They take turns trying to run past each other with the football. Li weighs 125 pound and Raj we
Elenna [48]

Answer:

Li has less mass and therefore less inertia, so he can change his motion more easily than Raj.

Explanation:

Inertia describes the resistance of an object to any change in its state of motion, and it depends on the mass of the object only. In particular:

- if an object has a large inertia (large mass), then it is more difficult to change its state of motion (i.e. to put it in motion, or to slow it down, or to change its direction of motion)

- if an object has small inertia (small mass), then it is more easy to change its state of motion

In this problem, Li has less mass than Raj, so he has less inertia, therefore he can change his motion more easily than Raj.

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3 years ago
A 0.290 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 30.0 pC charge on it
mojhsa [17]

Answer:

Therefore,

The potential (in V) near its surface is 186.13 Volt.

Explanation:

Given:

Diameter of sphere,

d= 0.29 cm

radius=\dfrac{d}{2}=\dfrac{0.29}{2}=0.145\ cm

r = 0.145\ cm = 0.145\times 10^{-2}\ m

Charge ,

Q = 30.0\ pC=30\times 10^{-12}

To Find:

Electric potential , V = ?

Solution:

Electric Potential at point surface is given as,

V=\dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{Q}{r}

Where,  

V= Electric potential,  

ε0 = permeability free space = 8.85 × 10–12 F/m

Q = Charge  

r = Radius  

Substituting the values we get

V=\dfrac{1}{4\times 3.14\times 8.85\times 10^{-12}}\times \dfrac{30\times 10^{-12}}{0.145\times 10^{-2}}

V=\dfrac{30}{16.117\times 10^{-2}}=186.13\ Volt

Therefore,

The potential (in V) near its surface is 186.13 Volt.

3 0
4 years ago
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