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4 years ago

6

A child goes down slide starting from rest if the length of the slide is to him and it takes a child three seconds to go down th

e slide what is the child’s acceleration

1 answer:

AlexFokin [52]4 years ago

4 0

Answer:

I. dont. know. sorry. I. need. points

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A 15.0-mW laser puts out a narrow beam 2.00 mm indiameter.

UkoKoshka [18]

Explanation:

It is known that formula for energy per unit area per unit time is as follows.

S = $\frac{P}{A}$

= $\frac{P}{\pi \times r^{2}} = \frac{4P}{\pi d^{2}}$

$c \epsilon_{o} E^{2} = \frac{4P}{\pi d^{2}}$

E = $\sqrt{\frac{4P}{\pi d^{2} c \epsilon_{o}}}$

Now, putting the values we will calculate the electric field as follows.

E = $\sqrt{\frac{4P}{\pi d^{2} c \epsilon_{o}}}$

= $\sqrt{\frac{4 \times 15 \times 10^{-3} W}{3.14 \times (2mm \times \frac{1m}{1000 mm})^{2} \times 3 \times 10^{8} \times 8.85 \times 10^{-12}}}$

= 1341.03 V/m

Now, we will calculate the average magnetic field as follows.

B = $\frac{E}{c}$

= $\frac{1341.03 V/m}{3 \times 10^{8} m/s}$

= $4.47 \times 10^{-6}$ T

= $4.47 \mu T$

Thus, we can conclude that the average (rms) value of the magnetic field is $4.47 \mu T$ .

3 0

3 years ago

400 gallons of pesticide is accidentally spilled into a lake and uniformly mixes with the water. The volume of the lake includin

svet-max [94.6K]

Answer:

t=27725.8minutes

Explanation:

To determinate use the law the change of volume in determinate time is derivate so:

$\frac{dV}{dt}=flowin-flowout$

The flow in is zero and the flow out is modeling with V as the volume out in the volume of the lake in the reason of change.

$\frac{dV}{dt}=0-(\frac{V}{10x10^{8}})*5000\\\frac{dV}{dt}=-5x10^{-5}*V \\dV*V=-5x10^{-5}*dt$

Integrate to get the volume function

$\int\limits{\frac{1}{V}}\, dv=\int\limits{-5x10^{-5} }\, dt\\ ln(V)=-5x10^{-5}*t\\ln*e(V)=C*e^{-5x10^{-5}*t}\\$

The find the constant C in the time t=0 the volume is knowing so:

$t=0s\\V=400\\V=C*e^{0}\\C=400$

$V=400*e^{-5x10^{-5}*t}$

The volume is 1 part per million so:

$\frac{V}{10x^{8}}=\frac{1}{10x^{6}}\\ V=100$

$V=400*e^{-5x10^{-5}*t}\\100=400*e^{-5x10^{-5}*t}\\\frac{100}{400}=e^{-5x10^{-5}*t}\\ln*\frac{1}{4}=ln*e^{-5x10^{-5}*t}\\-ln(4)=-5x10^{-5}*t\\ t=\frac{-ln(4)}{-5x10^{-5}}=27725.88\\t=27725.8minutes$

8 0

3 years ago

NEED HELP ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

mote1985 [20]

Answer:<em> tried my best to answer your ?s:</em>

<em>B</em>

<em>A</em>

<em>B</em>

<em>D</em>

<em>A</em>

<em>A</em>

<em>B?</em>

<em>Hope this helps!!</em>

3 0

3 years ago

Using the midpoint and the distance formulas, calculate he coordinate of the midpoint and the length of the segment.

arsen [322]

Answer:

Explanation:

Formula to calculate the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is,

d = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Therefore, length of CD having coordinates C(3,1) and D(3, 3),

CD = $\sqrt{(3-3)^2+(1-3)^2}$

= 2 units

Formula to find the midpoint of CD is,

M = $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

= $(\frac{3+3}{2},\frac{1+3}{2})$

= (3, 2)

Therefore, length of CD = 2 units and (3, 2) are the coordinates of the midpoint of CD.

7 0

3 years ago

Can a moving object have gravitational energy?

Evgesh-ka [11]

Answer:

It is worth noting that the higher the gravitational energy of an object moving downwards, the lower the kinetic energy, and the lower the kinetic energy of an object moving upwards, the higher its gravitational energy.

Gravitational potential energy is acquired by an object when it has been moved against a gravitational field. For example, an object raised above the surface of the Earth will gain energy, which is released if the object is allowed to fall back to the ground.

5 0

3 years ago