All categories

3 years ago

A 4 V power source is connected in parallel to four resistors: 1, 2, 2, and 3. Find the total current.

Physics

1 answer:

Keith_Richards [23]3 years ago

4 0

Answer:

Explanation:

Equivalent resistance is 1 / ((1/1) + (1/2) + (1/2) + (1/3)) = 3/7 Ω

I = V/R = 4(7/3) = 28/3 = 9.3 A

You might be interested in

One horsepower is a unit of power equal to 746 W. How much energy can a 150 horsepower engine transform in 10.0 seconds

11111nata11111 [884]

1,119,000 basically 746 times 150 then multiply that answer by 10

7 0

3 years ago

Which is the only mechanism that can introduce new alleles in a gene pool? genetic drift random mating mutation

Vikentia [17]

Mutation because they change the genes

8 0

3 years ago

What is a force which resists all motion

yaroslaw [1]

The answer is friction

4 0

4 years ago

Read 2 more answers

A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught

meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$ (Eq. 1)

Where:

$y_{o}$ - Initial height of the softball, measured in meters.

$y$ - Final height of the softball, measured in meters.

$v_{o}$ - Initial velocity of the softball, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $y = y_{o}$ , $t = 3.56\,s$ and $g = -9.807\,\frac{m}{s^{2}}$ , the initial velocity of the softball is:

$v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0$

$3\cdot v_{o} -44.132\,m= 0$

$v_{o} = 14.711\,\frac{m}{s}$

The initial velocity of the softball is 14.711 meters per second.

8 0

4 years ago

Why is the curve between 1950 and 1980 relatively flat and centered around zero degrees difference from the baseline? (Hint: how

zimovet [89]

Look at the title of the graph, in small print under it.

Each point is "compared to 1950-1980 baseline". So the set of data for those years is being compared to itself. No wonder it matches up pretty close !

3 0

4 years ago