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2 years ago

1. The geologic time scale divides time into years and centuries, true or false

Physics

1 answer:

hodyreva [135]2 years ago

5 0

Answer:

The answer is "False"

Explanation:

The geologic time scale is the "schedule" for occasions in Earth history. It partitions time into named units of unique time called in descending order of duration "eons, eras, periods, epochs, and ages". The specification of those geologic time units depends on stratigraphy, which is the relationship and order of rock layers. The fossil structures that happen in the stones, nonetheless, give the central methods for setting up a geologic time scale, with the circumstance of the development and vanishing of far and wide species from the fossil record being used to outline the beginnings and endings of ages,, periods, and different stretches.

Geologic time is the broad time period involved by the geologic history of Earth. Formal geologic time starts toward the beginning of the Archean Eon (4.0 billion to 2.5 billion years back) and proceeds to the current day.

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anice is watching her granddaughter drive a Barbie Jeep with a 6 V battery and an electric motor with 5 ohm of resistance. How m

Scrat [10]

Answer:

Power required will be 7.2 watt

Explanation:

We have given battery of Barble jeep is 6 volt

So potential difference V = 6 volt

Resistance of electric motor R = 5 ohm '

We have to find the power motor using to drive the jeep

Power is given by $P=\frac{V^2}{R}$ , here V is voltage and R is resistance

So $P=\frac{V^2}{R}=\frac{6^2}{5}=7.2watt$

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3 years ago

Read 2 more answers

An electron and a proton are both released from rest, midway between the plates of a charged parallel-plate capacitor. The only

topjm [15]

Answer:

Explanation:

Let the potential difference between the middle point and one of the plate be ΔV .

electric potential energy will be lost and it will be converted into kinetic energy .

Electrical potential energy lost = Vq , where q is charge on charge particle .

For proton

ΔV× q = 1/2 M V² ( kinetic energy of proton )

where M is mass and V be final velocity of proton .

For electron

ΔV× q = 1/2 m v² ( kinetic energy of electron )

where m is mass and v be final velocity of electron . Charges on proton and electron are same in magnitude .

As LHS of both the equation are same , RHS will also be same . That means the kinetic energy of both proton and electron will be same

1/2 M V² = 1/2 m v²

(V / v )² = ( m / M )

(V / v ) = √ ( m / M )

In other words , their velocities are inversely proportional to square root of their masses .

4 0

3 years ago

A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before

Darina [25.2K]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}$

$\\$

☯ As we know that,

$\\$

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }$

$\\$

$\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}$

$\\$

$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }$

$\\$

$\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }$

$\\$

$\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }$

$\\$

$\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }$

$\\$

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}$

$\\$

$\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }$

$\\$

$\dashrightarrow\:\: \sf{ s = 0.04 \: m }$

$\\$

☯ For left penetration (s₂)

$\\$

$\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }$

$\\$

$\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}$

$\\$

$\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}$

$\\$

$\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\$

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2 years ago

Which state of matter has atoms that are spread out and bouncy?

NikAS [45]

The stage where atoms are spread out and bouncy is the gas stage.

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In half wave rectifier circuit the diode and load resistance are connected in ...to ac power source

Fudgin [204]

Working of a Half wave rectifier
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