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3 years ago

Which of the following would create voltage in a coil of wire?

Physics

1 answer:

zubka84 [21]3 years ago

8 0

Its d all the above your welcome

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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]

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

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

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

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

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

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

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

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

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☯ As we know that,

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

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

$\\$

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

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

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

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

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

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

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

$\\$

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

$\\$

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

$\\$

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

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

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

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

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☯ For left penetration (s₂)

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

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

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

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

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

4 0

2 years ago

Which of the following biomes might be found at a latitude of 33° South?

cupoosta [38]

The answer is going be desert.

5 0

3 years ago

An athlete kicks a soccer ball that starts at rest so that it leaves their foot with a speed of 10m/s from the top o f a rectang

kirza4 [7]

Answer:

$a=500m/s^2$

Explanation:

We need only to apply the definition of acceleration, which is:

$a=\frac{v_f-v_i}{t_f-t_i}$

In our case the final velocity is $v_f=10m/s$ , the initial velocity is $v_i=0m/s$ since it departs from rest, the final time is $t_f=0.02s$ and the initial time we are considering is $t_i=0s$

So for our values we have:

$a=\frac{10m/s-0m/s}{0.02s-0s}=500m/s^2$

3 0

3 years ago

How to find average speed in physics

Triss [41]

Answer: you divide total distance by time. To get the time, divide total distance by speed. To get distance, multiply speed times the amount of time.

Explanation:

I hope this helps

7 0

2 years ago

A bug splats against the windshield of a car traveling at high speeds down a backcountry road. Which statement correctly compare

zvonat [6]

Answer:

C. The bug's change in momentum is equal to the car's change in momentum.

Explanation:

As we know by Newton's 2nd law

$F = \frac{\Delta P}{\Delta t}$

here we have also know that when car hits the bug then force applied by wind shield on the bug is same as the force applied by the bug on the car's wind shield as per Newton's III law

$F_{12} = F_{21}$

so we know that

$\frac{\Delta P_{12}}{\Delta t} = \frac{\Delta P_{21}}{\Delta t}$

so we have

$\Delta P_{12} = \Delta P_{21}$

so correct answer will be

C. The bug's change in momentum is equal to the car's change in momentum.

6 0

3 years ago