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3 years ago

If the voltage impressed across a circuit is held constant while the resistance is halved, what change occur?

Physics

1 answer:

egoroff_w [7]3 years ago

3 0

Answer:

The current doubles

Explanation:

In a circuit, Ohm's law gives the relationship between voltage, current and resistance:

$V=RI$

where

V is the voltage

R is the resistance

I is the current

In this problem,

V is held constant

R is halved: $R'=\frac{R}{2}$

Therefore, the new current is

$I'=\frac{V}{R'}=\frac{V}{R/2}=2\frac{V}{R}=2I$

So, the current doubles.

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3 years ago

A skateboarder rolls off a 2.5 m high bridge into the river. If the skateboarder was originally moving at 7.0 m/s, how much time

saul85 [17]

Answer:

t = 0.714 s and x = 5.0 m

Explanation:

This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed

vₓ = 7.0 m / s

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y = y₀ + v_{oy} t - ½ g t²

the initial vertical speed is zero and when it reaches the river its height is zero

0 = y₀ + 0 - ½ g t²

t = $\sqrt{\frac{2y_o}{g} }$

t = ra 2 2.5 / 9.8

t = 0.714 s

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3 years ago

A train that has wheels with a diameter of 91.44 cm (36 inches used for 100 ton capacity cars) slows down from 82.5 km/h to 32.5

podryga [215]

Answer:

The answer is below

Explanation:

The initial velocity = u = 82.5 km/h = 22.92 m/s, the final velocity = 32.5 km/h = 9.03 m/s, diameter = 91.55 cm = 0.9144 cm

radius (r) = diameter / 2 = 0.9144 / 2= 0.4572 m

a) Initial angular velocity ( $\omega_o$ ) = u /r = 22.92 / 0.4572 = 50.13 rad/s, final velocity (ω) = v / r = 9.03 / 0.4592 = 19.67 rad / s

θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

angular acceleration (α) is:

$\omega^2=\omega_o^2+2\alpha \theta\\\\19.67^2-50.13^2=2\alpha(272.9)\\\\19.67^2=50.13^2+2\alpha(272.9)\\\\2\alpha(272.9)=-2126.108\\\\\alpha=-3.89\ rad/s^2\\\\$

$\omega=\omega_o+\alpha t\\\\19.67=50.13+(-3.89t)\\\\3.89t=50.13-19..67\\\\3.89t=30.46\\\\t=7.83\ s$

c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

a) When it stops, the final angular velocity is 0. Hence:

$\omega^2=\omega_o^2+2\alpha \theta\\\\0=50.13^2+2(-3.89)\theta\\\\2(3.89)\theta=50.13^2\\\\2(3.89)\theta=2513\\\\\theta=323\ rad\\\\revolutions=\frac{\theta}{2\pi r}=\frac{323}{2\pi(0.4572)} =112.4\ rev$

θ = 323 rad

4 0

3 years ago