What is the ultimate energy for all life on earth

If there are 3 resistors of .5 ohms, 1 ohms, and 1 ohms connected in parallel then what is the equivalent resistance

Ede4ka [16]

Answer:0.45ohms

Explanation:

Let R be there equivalent resistance

1/R=1/r+1/r+1/r

1/R=1/5+1/1+1/1

1/R=1/5+2

1/R=(1+10)/5

1/R=11/5

Cross multiplying we get

11R=5

Divide both sides by 11

11R ➗ 11=5 ➗ 11

R=0.45ohms

3 0

3 years ago

Should we continue to rely on fossils fuels, such as coal, oil, and natural gas, as energy sources? Or, should we invest more in

Dennis_Churaev [7]

Answer:

i`d say energy

Explanation:

7 0

3 years ago

The average value of the load between A and B is 6.0 N. The spring has an unstretched length

Katyanochek1 [597]

Answer:

sorry just need the points

8 0

3 years ago

the velocity (v) of a tranvas wave in a string is related to tension (T) the mass (M) and the length (L) of de string by the rel

Lana71 [14]

1

Explanation:

1

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6 0

3 years ago

In the Hunger Games movie, Katniss Everdeen fires a 0.0200-kg arrow from ground level to pierce an apple up on a stage. The spri

egoroff_w [7]

Answer:

a) $v=99.8584\ m.s^{-1}$

b) $v'=99.366\ m.s^{-1}$

Explanation:

Given:

mass of the arrow, $m=0.02\ kg$

stiffness constant of the bow, $k=330\ N,m^{-1}$

distance of pulling back the arrow on the bow from its mean position, $\Delta x=0.55\ m$

height of the apple targeted, $h=5\ m$

<u>Force on the arrow due to the stiffness of the bow:</u>

$F=k.\Delta x$

$F=330\times 0.55$

$F=181.5\ N$

Now the acceleration of the arrow upwards:

$a=\frac{F}{m}$

$a=\frac{181.5}{0.02}$

$a=9075\ m.s^{-2}$

a) For the course of motion when the arrow leaves the bow after the stretch is relaxed we consider that the arrow left the bow after its string goes to the mean position. During this phase the arrow also faces gravity in the downward direction.

Using the equation of motion:

$v^2=u^2+2(a-g).\Delta x$

where:

$v=$ velocity with which the arrow leaves the bow

$u=$ initial velocity of the arrow after it left

$v^2=0^2+2\times (9075-9.81)\times 0.55$

$v=99.8584\ m.s^{-1}$

b) Now when the arrow travels up then it is under a constant gravitational force acting opposite to the motion.

<u>Using eq. of motion:</u>

$v'^2=v^2-2\times g.h$

where:

$v'=$ final velocity when the arrow hits the target

$v=$ initial velocity after the arrow has been launched

$v'^2=99.8584^2-2\times 9.81\times 5$

$v'=99.366\ m.s^{-1}$

3 0

3 years ago