What type of light interaction is the picture below an example of?

Jack and Jill exercise in a 25.0 m long swimming pool. Jack swims 9 lengths of the pool in 156.9 s ( 2 min and 36.9 s ) , wherea

Leya [2.2K]

Answer:

Jill average velocity is 0

Jack average velocity is 0.159337

Jill average speed = 1.593372

Jack average speed = 1.434034

Explanation:

given data

long swimming pool = 25.0 m

9 lengths of the pool = 156.9 s ( 2 min and 36.9 s )

10 lengths = same time interval

to find out

average velocity and average speed

solution

we know that average velocity that is express as

average velocity = $\frac{displacement}{time}$ .....................1

Jill come back where she start

so here velocity will be = 0

and

Jack ends up on the other end of pool

so average velocity = $\frac{25}{156.9}$

average velocity = 0.159337

now we get here average speed that is express as

average speed = $\frac{distance}{time}$ .............2

jack speed = 9 × $\frac{25}{156.9}$

jack speed = 1.434034

and

Jill speed = 10 × $\frac{25}{156.9}$

Jill speed = 1.593372

5 0

4 years ago

At which SQR3 step should a student highlight important terms?

slega [8]

I think its c revising

7 0

4 years ago

A motorcycle is capable of accelerating at 1.5 m/s2. Starting from rest, how far can it travel in 0.5 seconds?

Helen [10]

Answer:
Here,
Initial velocity(u)=0 m/s
acceleration(a)=1.5m/s
time(t)=0.5s
Now,
distance covered(s)=ut+(1/2)at^2
=0*0.5+(1/2)*1.5*0.5*0.5
=0+(1/2)*0.375
=0+0.1875
=0.1875(nearly 0.19)
Hence,0.19 is correct answer.

6 0

4 years ago

A 15-µF capacitor and a 25-µF capacitor are connected in parallel, and charged to a potential difference of 60 V. How much energ

Alborosie

Answer:

Energy stored, E = 0.072 J

Explanation:

Given that,

Capacitance, $C_1=15\ \mu F$

Capacitance, $C_2=25\ \mu F$

These two capacitor are connected in parallel, and charged to a potential difference of, V = 60 volts

We know that in parallel combination of capacitor, the equivalent capacitance is given by :

$C=C_1+C_2\\\\C=(15+25)\ \mu F\\\\C=40\times 10^{-6}\ F$

The energy stored in the capacitor is given by :

$E=\dfrac{1}{2}CV^2\\\\E=\dfrac{1}{2}\times 40\times 10^{-6}\times (60)^2\\\\E=0.072\ J$

So, the energy stored in the capacitor in this capacitor combination is 0.072 J.

4 0

3 years ago

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30° angle. The block’s initial speed is 10 m/s. What vert

Romashka [77]

Answer:

Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s

Explanation:

Given Data:

V is the overall velocity vector, $Vi$ and $Ui$ are its initial vertical and horizontal components

$R = 10 m/s\\ Projection Angle (theta) = 30 degrees\\Vi = 10*sin(30) = 5 m/s\\Ui = 10*cos(30) = 8.66 m/s$

To find:

Max Height $h$ achieved

Calculation:

1) Using the $3^{rd}$ equation of motion, we know

$2*a*s = Vf^{2} - Vi^{2}$

2) In terms of gravity $g$ height $h$ and the vertical component of Velocity $Vf , Vi$ .

3) As $Vf = 0$ as at maximum height the vertical component of velocity is zero maximum height achieved

$2*g*h = Vf^{2} -Vi^{2}$

putting values

4) $h = 0.784 m/s$

5) As for the speed when it reaches back its starting point, it will have a speed similar to its launching speed, the reason being the absence of air friction (Air drag)

3 0

4 years ago