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3 years ago

+ 1. Energy due to motion is _______ energy

Physics

2 answers:

Novay_Z [31]3 years ago

4 0

Answer:

Kinetic energy

Explanation:

The energy associated with an object's motion is called kinetic energy. A speeding bullet, a walking person, and electromagnetic radiation like light all have kinetic energy.

love history [14]3 years ago

4 0

Kinetic energy. For example, a walking person.

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Caves are a type of island ecosystem according to the theory of island biogeography. Please select the best answer from the choi

tatuchka [14]

Answer:

True

Explanation:

A different type of habitat exists in the caves. There exist no green plants/producers but still various life forms evolve with in the caves. The major source of nutrients with in the caves is water and intrusion of other organism that led to deposition of food in the form of debris/guano etc. These food deposits are intake by the organisms living in caves leading to start of food cycle. This is very much similar to any other food cycle

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4 years ago

A ball is falling from the second floor balcony to the floor below

aleksandrvk [35]

Answer:

KE increases and PE decreases

Explanation:

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3 years ago

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A car is traveling at a speed of 20m/s and has a mass of 1200kg how much kinetic energy does it have

Oxana [17]

Answer:

Explanation:

The formula for Kinetic Energy is

$KE=\frac{1}{2}mv^2$ . Filling in:

$KE=\frac{1}{2}(1200)(20)^2$ It looks like we only need 1 significant digit here but I'll give you 2 and you can round how you want.

KE = 2.4 × 10⁵ J

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3 years ago

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

<u>Given:</u>

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$