Which statement best explains the relationship between current, voltage, and resistance?

Physics

1 answer:

nalin [4]3 years ago

4 0

Explanation:

Ohm's law gives the relationship between current, voltage, and resistance. Its mathematical form is given by :

V = IR

I is current and R is resistance

Resistance resists the flow of electric current in a circuit. When the amount of applied voltage is increased, it will not change the resistance. It increases the current. Hence, the correct option is (C).

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The ball clears by 11.79 meters Let's first determine the horizontal and vertical velocities of the ball. h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s Now determine how many seconds it will take for the ball to get to the goal. t = 36.0 m / 15.04 m/s = 2.394 s The height the ball will be at time T is h = vT - 1/2 A T^2 where h = height of ball v = initial vertical velocity T = time A = acceleration due to gravity So plugging into the formula the known values h = vT - 1/2 A T^2 h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2 h = 42.92 m - 4.9 m/s^2 * 5.731 s^2 h = 42.92 m - 28.0819 m h = 14.84 m Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m

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The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$