What is happening at the points on the graph where the slope is flat?

A 50 W light bulb is plugged into a standard

wolverine [178]

Answer:

$1.26

Explanation:

Power =energy/ time

energy =powerxtime

energy =50x31x24=37200

=37.2kwh

1kwh =3.39

37.2kwh=3.39x37.2=126.108cent

=$1.26

8 0

3 years ago

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

jeyben [28]

Answer:

$V_0$

Explanation:

Given that, the range covered by the sphere, $M$ , when released by the robot from the height, $H$ , with the horizontal speed $V_0$ is $D$ as shown in the figure.

The initial velocity in the vertical direction is $0$ .

Let $g$ be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. $V_0$ will remain constant throughout the projectile motion.

So, if the time of flight is $t$ , then

$D=V_0t\; \cdots (i)$

Now, from the equation of motion

$s=ut+\frac 1 2 at^2\;\cdots (ii)$

Where $s$ is the displacement is the direction of force, $u$ is the initial velocity, $a$ is the constant acceleration and $t$ is time.

Here, $s= -H, u=0,$ and $a=-g$ (negative sign is for taking the sigh convention positive in $+y$ direction as shown in the figure.)

So, from equation (ii),

$-H=0\times t +\frac 1 2 (-g)t^2$

$\Rightarrow H=\frac 1 2 gt^2$

$\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)$

Similarly, for the launched height $2H$ , the new time of flight, $t'$ , is

$t'=\sqrt {\frac {4H}{g}}$

From equation (iii), we have

$\Rightarrow t'=\sqrt 2 t\;\cdots (iv)$

Now, the spheres may be launched at speed $V_0$ or $2V_0$ .

Let, the distance covered in the $x-$ direction be $D_1$ for $V_0$ and $D_2$ for $2V_0$ , we have

$D_1=V_0t'$

$D_1=V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_1=\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_1=1.41 D$ (approximately)

This is in the $3$ points range as given in the figure.

Similarly, $D_2=2V_0t'$

$D_2=2V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_2=2\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_2=2.82 D$ (approximately)

This is out of range, so there is no point for $2V_0$ .

Hence, students must choose the speed $V_0$ to launch the sphere to get the maximum number of points.

7 0

3 years ago

The pictures shows what a student observed when he looked through a hand lens at a penny.

igomit [66]

The answer is A. Refracted

7 0

3 years ago

Read 2 more answers

Two objects separated by a distance r are each carrying a charge q The magnitude of the force exerted on the second object by th

Sindrei [870]

Answer:F=4F

Explanation: Columbs law states that The force between the two point charges is directly proportional to the product of charges and inversely proportional to the square of distance between them

Force between the two charges is given by

F=K*q1*q2/r^2

if one charge become 4 times, new force is,

F=4(K*q1*q2)/r^2

F=4F

Where q1 and q2 are the point charges

r is the distance between the two charges

K is a constant of proportion called electrostatic force

4 0

3 years ago

When preparing to strip insulation in order to splice conductors, a person should strip the insulation back far enough so that t

Fiesta28 [93]

Answer:

Closely fits into the connector.

Explanation:

It's one of the steps used for the splicing of aluminium conductors in the underground connections. Where we do the strip insulation to splice the conductors by using compression type connectors.

3 0

3 years ago