Celina has a water sample that’s contaminated with salt and microorganisms. Which method should she use to purify the water?

A slinky is stretched across a classroom to 9 meters. A compression travels along the slinky at a velocity of 2 m s . How long d

Dmitrij [34]

It really doesn't matter whether it's a compression wave in a Slinky®,
or a trained frog, or a model airplane, or a baby chicken.

Time = distance / speed

Time = (9 m) / (2 m/s) = 4.5 seconds

8 0

4 years ago

Read 2 more answers

A current of 1.40 a flows in a wire. how many electrons are flowing past any point in the wire per second?

ikadub [295]

The current is the quantity of charge Q flowing through a certain point of the wire in a time interval of $\Delta t$ :
$I= \frac{Q}{\Delta t}$

So, by using this relationship and I=1.40 A, we can find the charge passing any point in the wire in 1 second:
$Q=I \Delta t=(1.40 A)(1.0 s)=1.40 C$

To find how many electrons corresponds to this charge, we should divide this value by the charge of a single electron ( $q=1.6 \cdot 10^{-19}C$ ):
$N= \frac{Q}{q}= \frac{1.40 C}{1.6 \cdot 10^{-19} C}=8.75 \cdot 10^{18}$ electrons

3 0

4 years ago

Draw a diagram to show the life-cycle of a silk moth.

Advocard [28]

$\huge{ \underline{ \underline{ \green{ \bf{Answer:}}}}}$

The female silk moth lays eggs. When the eggs hatch a tiny Caterpillar comes out, This is the Silkworm or Larva. The silkworm feeds on leaves and grass. When it is ready to enter the next stage in its lifecycle, it first weaves a net around itself. It then secrets a fibre made up of proteins which hardens on exposure to air. This is the Silk fibre. It covers itself completely with this fibre to form a cocoon at this stage the Larva is called Pupa. The Pupa grows and changes inside the cocoon. A few weeks later the cocoon opens and an adult moth come out.

The Silk yarn is obtained from the cocoon of the silkworm. The breeding and management of silkworms is known as Sericulture.

━━━━━━━━━━━━━━━━━━━━

6 0

3 years ago

A water balloon slingshot launches its projectiles essentially from ground level at a speed of 27.5 m/s . (You can ignore air re

RoseWind [281]

A ) A range:
R = v² sin 2 α / g
The maximum range is when sin 2 α = 1, or when: 2 α = 90°
α = 45°
B ) R = 27.5² · 1 / 9.81 = 77 m
C ) Time:
T = 2 v sin α / g = 2 · 27.5 · sin 45° / 9.81 =
= 2 · 27.5 · 0.7 / 9.81 = 3.92 s

3 0

3 years ago

A 95kg fullback (football player for those not into sports) moving south with a speed of 5.0 m/s has a perfectly inelastic colli

Lunna [17]

Answer:

a. $v=3.11mls$ , $29.4^{0}$

b. $K.E =-697.8J$

Explanation:

To calculate the values in the question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 $p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\$

Momentum of player 2 $p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\$

Hence the total momentum $p_{12}=p_{1}+p_{2}\\$

Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate

Hence $p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\$

$(95+90)v=475i+270j\\$

$v=2.57i+1.45j\\$

solving for the resultant velocity, we have

$v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls$

To calculate the direction of movement, we have

$\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\ \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}$

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

$K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\$

And the final kinetic energy after collision is

$K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\ K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J$

The decrease in Kinetic energy is

$K.E =K.E_{final}- K.E_{initial}=894.7-1592.5$

$K.E =-697.8J$

The negative sign indicate a decrease in Kinetic energy

4 0

3 years ago