All categories

3 years ago

Circuit that has only one path for the current to follow

Physics

1 answer:

Vadim26 [7]3 years ago

4 0

Answer:

Series circuit

Explanation:

The type of circuit can be divided into series and parallel. In the series circuit, there is only one loop/path. If the path is cut or stopped, the current of the circuit will be stopped completely. A parallel circuit has more than one loops. This means the circuit has alternative paths that it can travel in case that one path is blocked or cut. This alternative path makes the parallel circuit have a lower resistance compared to the series.

You might be interested in

Radiation is the main way of transferring heat in a vacuum b solid c gas d liquid

drek231 [11]

Answer:

gas

................

Explanation:

hope it helps u

3 0

3 years ago

If a pitcher throws a baseball 10 m in 30 seconds - what would be the average speed?

allsm [11]

Answer:

0.333 m/s

Explanation:

avg speed= (total distance)/(total time)

10m/30s

0.333 m/s

7 0

3 years ago

Read 2 more answers

What is the instantaneous speed of the object after the five seconds?

OverLord2011 [107]

Answer:

12.5 m/s

Explanation:

In a acceleration time graph the area under the curve gives the change in velocity of the object. Here object starts at rest and therefore initial velocity is 0. After 5 seconds acceleration is 5m/s2.

change in velocity=area under the curve

change in velocity= 0.5*acceleration* change in time

v-0=0.5*5*5

v=12.5 m/s

7 0

4 years ago

Physics student is dropped. If they reach the floor at a speed of 3.2 m/s, from what height did they fall?

Mekhanik [1.2K]

Answer:

0.52 m

Explanation:

The motion of the studnet is an accelerated motion, with constant acceleration $g=9.8 m/s^2$ (acceleration of gravity) toward the ground. We can find the distance covered by the student (which is equal to the height from which he falls) by using the SUVAT equation:

$v^2 -u^2 = 2ad$

where

v = 3.2 m/s is the final speed

u = 0 is the initial speed

a = 9.8 m/s^2 is the acceleration

d is the distance covered

By re-arranging the equation, we find:

$d=\frac{v^2-u^2}{2a}=\frac{(3.2 m/s)^2-0}{2(9.8 m/s^2)}=0.52 m$

5 0

4 years ago

Point A and B located at 4 meters and 9 meters from a source of the sound. If IA and IB are intensity at point A and point B, th

elena-14-01-66 [18.8K]

The intensity ratio at point A and B will be 81:16.

<u>Explanation:</u>

Sound waves are known to get faded with increase in the distance. This is because, the intensity of the sound is inversely proportional to the square of the distance of the source from the observer. So, if an observer is standing greater distance from the source of the sound, he/she will find difficulty in hearing the sound.

So, as the distance between the source and observer increases, the intensity of the sound wave decreases.

$I = \frac{1}{r^{2} }$

As here two points A and B are located at 4 m and 9 m distance from the source, then the intensity of sound at A and B will be inversely proportional to their respective square of the distance as shown below.

$I_{A} =\frac{1}{r_{A}^{2} } = \frac{1}{4 \times 4}=\frac{1}{16}$

Similarly,

$I_{B} =\frac{1}{r_{B}^{2} } = \frac{1}{9 \times 9}=\frac{1}{81}$

So, the ratio of intensity at point A and B will be

$\frac{I_{A} }{I_{B} } = \frac{81}{16} =81:16$

Thus, the intensity ratio at point A and B will be 81:16.

7 0

3 years ago