how much current is in a circuit that includes a 9-volt battery and a bulb with a resistance of 3 ohms?

In an experiment, a shearwater (a seabird) was taken from its nest, flown a distance 5220 km away, and released. It found its wa

Korolek [52]

Answer:

4.6834625323 m/s

0 m/s

Explanation:

s = Displacement

t = Time

Velocity is given by

$v=\dfrac{s}{t}\\\Rightarrow v=\dfrac{5220000}{12.9\times 24\times 60\times 60}\\\Rightarrow v=4.6834625323\ m/s$

The bird's average velocity for the return flight is 4.6834625323 m/s

In the whole episode the bird went 5220 km away from its nest and came back. This means the displacement is zero.

Hence, the average velocity for the whole episode is 0 m/s

7 0

3 years ago

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5

Tanya [424]

Answer:

Reset

Explanation:

Digital methods are the methods that are uses methodological outlook to study societal change and cultural condition of online data. Reset is use to disguise data In digital methods. It is use to set again and conceal data by giving the data a different form. It restores the device to the original manufacture's settings.

8 0

3 years ago

A circuit contains two devices that are connected in parallel. If the resistance of one of these devices is 12 ohms and the resi

finlep [7]

<u>Answer</u>

3 Ohms

<u>Explanation</u>

when the resistors are in series, the resistance in the circuit increases. For example, if two resistors, R1 and R2 are in series, the combined resistance is R1+R2.

When connected in parallel, the total resistance is the reciprocal of (1/R1 + 1/R2)

In this case the resistors are in parallel.

Total resistance = (1/12 + 1/4)⁻¹

= (1/3)⁻¹

= 3 Ohms

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3 years ago

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A 5 kg wooden block sitson a flat straight-away12 meters fromthe bottom of an infinitely long ramp, which has an angle of 20 deg

saveliy_v [14]

Answer:

(a) 19.71801m/s Velocity just before going up the ramp.

(b) 74.56338m.

Explanation:

We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.

Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.

Following formulas will be used.

$x(t) = \frac{1}{2} t^2 = 12m =16.2m/s^2 t^2$

$F =ma$

$V(t) = V_{o} +at$

$x(t) = x_{0} +v_{0}t+\frac{1}{2}a t^2$

(a) Calculating velocity right before going up the ramp.

Wooden block is going on a straightaway and has net for on it.

$F_{n} =F-F_{s} = F-uF_{n} = 100N-0.4*9.8m/s^2*5kg$ = $81N$

and this force produces acceleration of

$a = \frac{F}{m}=\frac{81}{5} =16.2m/s^2 .$

With this acceleration, wooden block would reach at the foot of ramp in.

$x(t) = 12m = 16.2m/s^2*t^2$

$t = 1.217s$

and final velocity will be

$v(t) = v_{0}+at = 0+16.2m/s^2*1.2171s = 19.7180m/s.$

this velocity of wooden box just before going up the ramp.

(b) How far up the ramp will the wooden block go before stopping.

Ramp is at 20° relative to horizontal therefore velocity along the ramp that the wooden block would have will be.

$V= V_{h}cos(20) = 18.5288m/s$

and deceleration along the ramp is

$a = \frac{F_{s} }{m}$

Where $F_{s}$ force of friction along the inclined plane.

$F_s = uF_n = u*m*a$

$a = 9.8m/s^2*cos(20) = 9.2089m/s^2$

is a component of g along normal of the inclined plane.

$F_{s} = 0.25*5kg*9.2089m/s^2$

$= 11.5112N$

$a = \frac{11.5112N}{5kg} = 2.3022m/s^2$

And with this deceleration time needed to get wooded block to stop is.

$v(t) = v_o-at = 18.5288m/s-2.3022m/s^2*t = 0$

$t = \frac{18.5288m/s}{2.3022m/s^2} =8.04813s$

and in that time wooden block would travel

$x(8.04813s) = 18.52881m/s *8.04813s-\frac{1}{} 2.3022m/s^2*(8.0481)^2=74.56338m$

This is how up wooden box will go before coming to stop.

3 0

3 years ago

A 0.5-kilogram apple falls from a height of 2 meters to 1.50 meters. Ignoring frictional effects, what is the kinetic energy of

spin [16.1K]

The final kinetic energy of the ball is 2.45 J

Explanation:

We can solve this problem by using the law of conservation of energy.

In absence of frictional effect, the mechanical energy of the apple must be conserved during the fall. So we can write:

$U_i +K_i = U_f + K_f$

where :

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at the bottom

$K_f$ is the final kinetic energy, at the bottom

By explicing the potential energy, we can rewrite the equation as:

$mgh_i + K_i = mgh_f + K_f$

where:

m = 0.5 kg is the mass of the apple

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 2 m$ is the initial height

$h_f=1.50 m$ is the final height

The initial kinetic energy is zero, since the ball starts from rest:

$K_i = 0$

Therefore we can solve the equation for $K_f$ , the final kinetic energy of the ball:

$K_f = mg(h_i-h_f)=(0.5)(9.8)(2-1.50)=2.45 J$

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

3 0

3 years ago

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