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2 years ago

15

what is the resistance of a car light bulb that conducts 0.025A current when connected to a 12V car accumulator? is the current

in the bulb steady?

1 answer:

Sophie [7]2 years ago

7 0

One form of Ohm's Law says . . . . . Resistance = Voltage / Current .

R = V / I

R = (12 v) / (0.025 A)

R = (12 / 0.025) (V/I)

<em>R = 480 Ohms</em>

I don't know if the current in the bulb is steady, because I don't know what a car's "accumulator" is. (Floogle isn't sure either.)

If you're referring to the car's battery, then the current is quite steady, because the battery is a purely DC storage container.

If you're referring to the car's "alternator" ... the thing that generates electrical energy in a car to keep the battery charged ... then the current is pulsating DC, because that's the form of the alternator's output.

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Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different direc

Advocard [28]

Answer:

the direction that should be walked by Ricardo to go directly to Jane is 23.52 m, 24° east of south

Explanation:

given information:

Ricardo walks 27.0 m in a direction 60.0 ∘ west of north, thus

A= 27

Ax = 27 sin 60 = - 23.4

Ay = 27 cos 60 = 13.5

Jane walks 16.0 m in a direction 30.0 ∘ south of west, so

B = 16

Bx = 16 cos 30 = -13.9

By = 16 sin 30 = -8

the direction that should be walked by Ricardo to go directly to Jane

R = √A²+B² - (2ABcos60)

= √27²+16² - (2(27)(16)(cos 60))

= 23.52 m

now we can use the sines law to find the angle

tan θ = $\frac{R_{y} }{R_{x} }$

= By - Ay/Bx -Ax

= (-8 - 13.5)/(-13.9 - (-23.4))

θ = 90 - (-8 - 13.5)/(-13.9 - (-23.4))

= 24° east of south

4 0

2 years ago

Nancy is pushing her empty grocery cart at a rate of 1.8 m/s. 30 seconds later,

Strike441 [17]

3.Es tarde y mi taxi no llega. Estoy ____.
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3 0

2 years ago

What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from

Strike441 [17]

To develop this problem we will apply the concepts related to the Doppler effect. The frequency of sound perceive by observer changes from source emitting the sound. The frequency received by observer $f_{obs}$ is more than the frequency emitted by the source. The expression to find the frequency received by the person is,

$f_{obs} = f_s (\frac{v_w}{v_w-v_s})$

$f_s$ = Frequency of the source

$v_w$ = Speed of sound

$v_s$ = Speed of source

The velocity of the ambulance is

$v_s = 119km/h (\frac{1000m}{1km})(\frac{1h}{3600s})$

$v_s = 30.55m/s$

Replacing at the expression to frequency of observer we have,

$f_{obs} = 800Hz(\frac{345m/s}{345m/s-30.55m/s})$

$f_{obs} = 878Hz$

Therefore the frequency receive by observer is 878Hz

8 0

2 years ago

Mt. Asama, Japan, is an active volcano complex. In 2009, an eruption threw solid volcanic rocks that landed far from the crater.

solong [7]

Answer:u=97.41m/s

Explanation:

Given

inclination $\theta =58.7^{\circ}C$

Horizontal distance travel by Particle $d=1200 m$

Vertical height $h=780 m$

Let u be the initial velocity

calculating vertical distance

$y=u\sin \theta +\frac{at^2}{2}$

$y=u\sin \theta t-\frac{gt^2}{2}$ -------1

Calculating horizontal distance

$x=u\cos \theta \times t+0$

$t=\frac{x}{u\cos \theta }$

put value of t in equation 1

$y=u\sin \theta \times \frac{x}{u\cos \theta }-\frac{g}{2}\times (\frac{x}{u\cos \theta })^2$

$y=x\tan \theta -\frac{gx^2}{2u^2\cos ^2\theta }$

$\frac{gx^2}{2u^2\cos ^2\theta }=x\tan \theta -y$

$u^2=\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}$

$u=\sqrt{\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}}$

at $y=-780\ m\ x=1200 m$

$u^2=\frac{18.154}{2753.65}\times 1200^2$

$u=97.41 m/s$

6 0

2 years ago

A 1100kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat,and trailer, if the car exerts

likoan [24]

Refer to the figure shown below.

m₁ = 1100 kg, the mass of the car
m₂ = 700 kg, the mass of the trailer and boat
F = 1900 N, the driving force acting on the car
N₁ = m₁g, the normal reaction on the car
N₂ = m₂g, the normal force on the trailer and boat
μN₁ and μN₂ are frictional forces, where = kinetic coefficient of friction
T = the force in the hitch between the car and trailer.

Part (a)
Let R₁ = the total force that resists the motion of the car, boat, and trailer.
Because the acceleration is 0.550m/s², therefore
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100+700 kg)*(0.55 m/s²) = (1900 - R₁) N
990 = 1900 - R
R₁ = 910 N

Answer: The resistive force is 910 N

Part (b)
80% of the resistive forces are experienced by the boat and trailer.
Let the resistive force be R₂.
Then
R₂ = 0.8*R₁ = 728 N
If the tension in the hitch between the car and the trailer is T, then
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²)
T - 728 = 385
T = 1113 N

Answer: The force in the hitch is 1113 N

3 0

2 years ago