An increase of 50 dB increases the sound intensity by a factor of

Which term best describes bats' natural habitats?

galina1969 [7]

Arboreal

Explanation:

The best term that describes the habitat of bats is arboreal.

The habitat of an organism is its dwelling place in nature. This encompasses the role it plays in the ecosystem.

Bats spends most of their life time hanging down on trees.
They are blind flying mammals.
Animals that lives in trees and trunks are said to be arboreal.
Other examples are koalas, primates, sloths, spider monkeys, leopards, chameleons etc.
These animals are adapted for life in tree trunks.
The have modified body parts that makes them thrive in such habitats.

learn more:

Interactions in the ecosystem brainly.com/question/2321688

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4 0

3 years ago

What is the mass of a 2 kg object on the Earth and on the moon?

zubka84 [21]

Answer:

Same

Explanation:

Mass is the quantity of matter in a certain object.

WHEREVER you take a 2kg object, the mass will remain 2kg. All that changes is the Weight ..Weight the force which the centre of a Planet uses to pull everything towards itself.

On earth, it is 9.81 whereas on the Moon it is 1.6

7 0

3 years ago

A train is travelling along a straight track at constant velocity from Western Station to Eastern station. The mile markers incr

S_A_V [24]

Answer:

Explanation:

Displacement of train = 60 - 25 = 35 mile

= 35 x 1.6 = 56 km

duration of time = 45 - 15 = 30 minutes

= 30 x 60 = 1800 s

velocity of train = displacement / time

= 56 / 1800 = .03111 km /s

= 31.111 m / s

3 0

3 years ago

A fighter jet is launched from an aircraft carrier with the aid of its own engines and a steam-powered catapult. The thrust of i

kirza4 [7]

Answer:

2.5 x 10⁷ J

Explanation:

F = thrust of the engine = 2.3 x 10⁵ N

d = distance traveled = 87 m

Work done by the engine is given as

W = F d = (2.3 x 10⁵) (87) = 200.1 x 10⁵ J

W' = Net work done

W'' = work done by catapult

KE₀ = initial kinetic energy = 0 J

KE = final kinetic energy = 4.5 x 10⁷ J

Net work done is given as

W' = KE - KE₀

W' = 4.5 x 10⁷ J

We know that

W' = W + W''

4.5 x 10⁷ = 2.001 x 10⁷ + W''

W'' = 2.5 x 10⁷ J

8 0

3 years ago

Two electric charges qa = 1. 0 μc and qb = - 2. 0 μc are located 0. 50 m apart. how much work is needed to move the charges apar

melisa1 [442]

The total work done of 0.018 joules is needed to move the charges apart and double the distance between them.

We have two electric charges q(A) = 1μc and q(B) = -2μc kept at a distance 0.5 meter apart.

We have to calculate much work is needed to move the charges apart and double the distance between them.

<h3>What s the formula to calculate the Potential Energy of a system of two charges (say 'q' and 'Q') separated by a distance 'r' ?</h3>

The potential energy of the system of two charges separated by a distance is given by -

$U = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}$

In order to solve this question, it is important to remember the work - energy theorem which states -

"The change in the energy of the body is equal to work done on it"

Hence, using this work -energy theorem in the question given to us we get -

$U_{f} -U_{i} =W_{net}$

In our case -

$U_{f} = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r}\\\\U_{i} = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W=\frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r} - \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W = \frac{qQ}{4\pi\epsilon_{o}r} (\frac{1}{2} -1)\\\\W = 9\times 10^{9}\times \frac{1 \times 10^{-6} \times 2\times10^{-6} }{0.5} \times \frac{-1}{2}$

W = 0.018 joules

Hence, the total work done should be 0.018 joules.

To solve more question on potential energy, visit the link below -

brainly.com/question/15014856

#SPJ4

4 0

2 years ago