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3 years ago

If current I=2A and resistance R= 5 ohms what is the potential difference (voltage) V?

Physics

1 answer:

damaskus [11]3 years ago

6 0

Answer:

10 V

Explanation:

From Ohm's law, V=IR where V is the potential difference voltage, I is the current and R is resistance. Substituting 2A for current and 5 Ohms for resistance then V=2*5= 10 V

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2. The longest snake ever found was a python that was 10.0 m long. Sup-

Akimi4 [234]

Answer:

See below

Explanation:

Hypotenuse is snake length 10 m

y coordinate = 10 sin 60 = 8.7 m

x coordinate = 10 cos 60 = 5m

8 0

2 years ago

A car drives on a circular road of radius R. The distance driven by the car is given by = + [where a and b are constants, and t

pishuonlain [190]

Answer:

The question is incomplete, the complete question is "A car drives on a circular road of radius R. The distance driven by the car is given by d(t)= at^3+bt [where a and b are constants, and t in seconds will give d in meters]. In terms of a, b, and R, and when t = 3 seconds, find an expression for the magnitudes of (i) the tangential acceleration aTAN, and (ii) the radial acceleration aRAD3"

answers:

a. $18a m/s^{2}$

b. $a_{rad}=\frac{(27a +b)^{2}}{R}$

Explanation:

First let state the mathematical expression for the tangential acceleration and the radial acceleration.

a. tangential acceleration is express as

$a_{tan}=\frac{d|v|}{dt} \\$

since the distance is expressed as

$d=at^{3}+bt$

the derivative is the velocity, hence

$V(t)=\frac{dd(t)}{dt}\\V(t)=3at^{2}+b\\$

hence when we take the drivative of the velocity we arrive at

$a_{tan}=\frac{dv(t)}{dt}\\ a_{tan}=6at\\t=3 \\we have \\a_{tan}=18a m/s^2$

b. the expression for the radial acceleration is expressed as

$a_{rad}=\frac{v^{2}}{r}\\a_{rad}=\frac{(3at^{2} +b)^{2}}{R}\\t=3\\a_{rad}=\frac{(27a +b)^{2}}{R}$

4 0

3 years ago

A trapeze artist, with swing, weighs 800 N; he is momentarily heldto one side by his partner using a horizontal force so that th

Tamiku [17]

Answer:

461.88 N

Explanation:

$F_{g}$ = Weight of the swing = 800 N

$T$ = Tension force in the rope

$F$ = Horizontal force being applied by the partner

Using equilibrium of force in vertical direction using the force diagram, we get

$T Cos30 = F_{g}\\T Cos30 = 800\\T = \frac{800}{Cos30} \\\\T = 923.76 N$

Using equilibrium of force in horizontal direction using the force diagram, we get

$F = T Sin30\\F = (923.76) (0.5)\\F = 461.88 N$

6 0

3 years ago

What is the speed of a bobsled whose distance-time graph indicates that it traveled 114m in 30s? m/s

Kitty [74]

3.8 m/s
--------------

5 0

3 years ago

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

3 years ago