All categories

2 years ago

Which circuit would have the most electrical power?

Physics

2 answers:

Sauron [17]2 years ago

7 0

Answer:C

Explanation:if you look at it closely it has 10 and 10 is bigger

myrzilka [38]2 years ago

3 0

Answer:

Explanation:

You might be interested in

Interpret the following graphs by answering the questions below.

Furkat [3]

Answer:

Nodes = 5 and antinodes = 4

Explanation:

Nodes are the points of zero amplitude and appear to be fixed. On the other hand, antinodes are points on a stationary wave that oscillates with maximum amplitude.

In this given standing wave, there are 5 points where the amplitude is 0. So, there are 5 nodes. Also, there are 4 points where the amplitude is maximum.

So, there are 5 nodes and 4 antinodes.

6 0

2 years ago

How far did the object travel by the end of eight seconds, according to the graph above?

Ulleksa [173]

Answer:

The object traveled 4 cm by the end of eight seconds. Correct: A)

Explanation:

Speed vs Time Graph

In a speed-time graph, speed is plotted on the vertical axis and time is plotted on the horizontal axis. If the graph is a horizontal line, the speed is constant, if the line is sloped up, the speed is increasing and the acceleration is positive and constant, and if the line is sloped down, the speed is decreasing and the acceleration is negative and constant.

The distance traveled by the object can be found by calculating the area under the graph and above the x-axis.

The graph provided shows two different zones: the first 4 seconds, the speed is constant at 1 cm/s, and the last 4 seconds, the speed is zero, i.e. the object is not moving.

The area behind the first zone is a rectangle of height 1 cm/s and base 4 sec, thus the distance is 1 * 4 = 4 centimeters.

The second zone corresponds to an object at rest, thus no distance is traveled.

The object traveled 4 cm by the end of eight seconds.

A) Correct. As shown above

B) The distance traveled is 4 cm. Incorrect

C) The distance traveled is 4 cm. Incorrect

D) The distance traveled is 4 cm. Incorrect

6 0

3 years ago

A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s

Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

Given:

Time of flight = t = 20 s
Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

Assume:

Initial velocity of the projectile = u
R = Range of the projectile during the time of flight
H = maximum height of the projectile
D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

Initial horizontal velocity of the projectile = $u\cos \theta$
Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$