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2 years ago

how do the currents produced by a 1.5 v flashlight cell and a 12 v car battery compare if the resistance is the same

1 answer:

6 0

Current = (voltage) / (resistance)

I(flashlight) = (1.5 v) / (resistance)
----------------------------------------------
I(car battery) = (12 v) / (resistance)

If the resistance is the same in both cases, then

I(flashlight) 1.5
----------------- = ------
I(car battery) 12

The flashlight current is 1/8 (12.5%) as much as the car battery current.

The car battery current is 8 times as much as the flashlight current.

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For the first 10 seconds a squirrel runs 3 m/s to look for an acorn. The next 5 seconds he eats an acorn that he finds. Afterwar

Gala2k [10]

Distance covered by the squirrel to look for an acorn :

d = ( 3 m/s ) × 10 s = 30 m.

Time taken to eat an Acron is 5 seconds.

Time taken to cover distance of 30 m with 2 m/s speed is :

$T=\dfrac{30}{2}\ s= 15 \ s$

Therefore, total time take to get back to where he started is ( 10+5+15 ) = 30 s.

Hence, this is the required solution.

7 0

2 years ago

A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the vertical velocity of t

zhuklara [117]

Answer:

<em>You would use the kinematic formula:</em>

$\Delta y=V_{0y}\times t-g\times t^2/2$

Explanation:

The upwards vertical motion is ruled by the equation:

$y=y_0+V_{0y}\times t-g\times t^2/2$

Where:

$y \text{ is the position at the time }t:y=0.1m$

$y_0\text{ is the initial position: }y_0=0$

$t=2s$

$g\text{ is the gravitational acceleration: }\approx 9.8m/s^2$

$V_{0y}\text{ is the initial vertical velocity}$

Naming Δy = y - y₀, the equation becomes:

$\Delta y=V_{0y}\times t-g\times t^2/2$

Then, you just need to substitute with Δy = 0.1m, t = 2s, and g = 9.8m/s², ans solve for the intital vertical velocity.

7 0

2 years ago

Read 2 more answers

You have a resistor and a capacitor of unknown values. First, you charge the capacitor and discharge it through the resistor. By

Fittoniya [83]

Answer:

The frequency is $f = 0.221 \ Hz$

Explanation:

From the question we are told that

The time taken for it to decay to half its original size is $t = 3.40 \ ms = 3.40 *10^{-3} \ s$

Let the voltage of the capacitor when it is fully charged be $V_o$

Then the voltage of the capacitor at time t is said to be $V = \frac{V_o}{2}$

Now this voltage can be mathematical represented as

$V = V_o * e ^{-\frac{t}{RC} }$

Where RC is the time constant

substituting values

$\frac{V_o}{2} = V_o * e ^{-\frac{3.40 *10^{-3}}{RC} }$

$0.5 = e^{-\frac{3.40 *10^{-3}}{RC} }$

$- \frac{0.5}{RC} = ln (0.5)$

$-\frac{0.5}{RC} = -0.6931$

$RC = 0.721$

Generally the cross-over frequency for a low pass filter is mathematically represented as

$f = \frac{1}{2 \pi * RC }$

substituting values

$f = \frac{1}{2* 3.142 * 0.72 }$

$f = 0.221 \ Hz$

7 0

2 years ago

A mover pushes a 255 kg piano

faust18 [17]

Answer:

$0.495 ms^{-2}$

Explanation:

According to the newton's second law of motion we can apply F=ma hear

Force = mass * acceleration

(assume the piano is moving left side )

←F = ma

$F_(pull)+ F_(push)= M*a\\77.5 + 48.7 = 255 *a\\a = 0.495 ms^{-2}$

7 0

2 years ago

A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington

Roman55 [17]

Let $\theta$ be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

$\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

The current has velocity vector (relative to the Earth)

$\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath$

The swimmer's resultant velocity (her velocity relative to the Earth) is then

$\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}$

$\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath$

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

$\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ$

which is approximately 41º west of north.

6 0

2 years ago