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2 years ago

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If a current of 2.0 A is flowing from point a to point b, the potential difference between Vb- Va (in V) is:

1 answer:

Aliun [14]2 years ago

5 0

Answer:

$\huge\color{skyblue}\boxed{\colorbox{black}{Answer ☘}}$

$total \: resistance \: ( R_{t} ) = (3 + 1)ohm \\ (since \: the \: resistors \: are \: connected \\ in \: series)$

$current \: flowing \: through \: circuit(I) = 2A \\ \\ now... \: by \: ohms \: law \\ V = IR\\ V = (2)(4) \\ = > 8v$

<h3>therefore , option (b) is correct!!</h3>

hope helpful~

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A sample contains 10g of radioactive isotope. How much radioactive isotope will remain in the sample after 3 half-lives?

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Answer:

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Explanation:

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A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.10 m/s

yarga [219]

Answer:

$I=336.6kgm/s$

Explanation:

The equation for the linear impulse is as follows:

$I=F\Delta t$

where $I$ is impulse, $F$ is the force, and $\Delta t$ is the change in time.

The force, according to Newton's second law:

$F=ma$

and since $a=\frac{v_{f}-v_{i}}{\Delta t}$

the force will be:

$F=m(\frac{v_{f}-v_{i}}{\Delta t})$

replacing in the equation for impulse:

$I=m(\frac{v_{f}-v_{i}}{\Delta t})(\Delta t)$

we see that $\Delta t$ is canceled, so

$I=m(v_{f}-v_{i})$

And according to the problem $v_{i}=0m/s$ , $v_{f}=5.10m/s$ and the mass of the passenger is $m=66kg$ . Thus:

$I=(66kg)(5.10m/s-0m/s)$

$I=(66kg)(5.10m/s)$

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3 years ago

An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the

garik1379 [7]

(a) 6.04 rev/s

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$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the ball from the centre of the circle

In situation 1), we have

$\omega=8.13 rev/s \cdot 2\pi = 51.0 rad/s$

r = 0.600 m

So the speed of the ball is

$v=(51.0 rad/s)(0.600 m)=30.6 m/s$

In situation 2), we have

$\omega=6.04 rev/s \cdot 2\pi = 37.9 rad/s$

r = 0.900 m

So the speed of the ball is

$v=(37.9 rad/s)(0.900 m)=34.1 m/s$

So, the ball has greater speed when rotating at 6.04 rev/s.

(b) $1561 m/s^2$

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$a=\frac{v^2}{r}$

where

v is the speed

r is the distance of the ball from the centre of the trajectory

For situation 1),

v = 30.6 m/s

r = 0.600 m

So the centripetal acceleration is

$a=\frac{(30.6 m/s)^2}{0.600 m}=1561 m/s^2$

(c) $1292 m/s^2$

For situation 2 we have

v = 34.1 m/s

r = 0.900 m

So the centripetal acceleration is

$a=\frac{v^2}{r}=\frac{(34.1 m/s)^2}{0.900 m}=1292 m/s^2$

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3 years ago

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Lera25 [3.4K]

Answer:

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Explanation:

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