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Nuetrik [128]
2 years ago
11

Introduced species often thrive and multiply in an environment very different from their original one. Why are they often able t

o do so well in their new habitat?
Physics
1 answer:
Anestetic [448]2 years ago
8 0
They begin to adapt into their new location. They then end up having adaptations to help them survive.
You might be interested in
A weight lifter applies an upward force of 1100 N while lowering a dumbbell
Paul [167]

Answer:

A

Explanation:

work = force \times distance

work = 1100 \times 0.5

= 550 \: j

hope it helped a lot

pls mark brainliest with due respect .

6 0
3 years ago
What is the resultant of a and b if a = 3i 3j and b = 3i − 3j?
ss7ja [257]
 <span>a+b= ? 
3i +3j + (3i -3j) = ? 
3i + 3j + 3i -3j =? 
= 6i + 0j</span>
3 0
3 years ago
Read 2 more answers
The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 13.1
ahrayia [7]

Answer:

The temperature is  T  = 168.44 \ K

Explanation:

From the question ewe are told that

   The rate of heat transferred is    P  = 13.1 \ W

     The surface area is  A = 1.55 \ m^2

      The emissivity of its surface is  e = 0.287

Generally, the rate of heat transfer is mathematically represented as

           H  =  A e \sigma  T^{4}

=>         T  =  \sqrt[4]{\frac{P}{e* \sigma } }

where  \sigma is the Boltzmann constant with value  \sigma  = 5.67*10^{-8} \ W\cdot  m^{-2} \cdot  K^{-4}.

substituting value  

             T  =  \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }

            T  = 168.44 \ K

7 0
3 years ago
(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

3 0
3 years ago
Read 2 more answers
The diagram shows a person using a piece of gym equipment to lift weights.
ki77a [65]

Answer:

C. The lower legs are levers, and the knees are fulcrums. The ankles hold the loads.

Explanation:

3 0
3 years ago
Read 2 more answers
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