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4 years ago

Jenny wanted to test whether worms prefer light or darkness best. After

Physics

1 answer:

myrzilka [38]4 years ago

3 0

Answer:

True

Explanation:

An independent variable is the variable that is changed or controlled in a scientific experiment to test the effects on the dependent variable. You are changing light or dark

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As waves get closer to a beach they:

hoa [83]

Answer:

As waves get closer to a beach they decrease in height.

Explanation:

As a wave crest approaches the shoreline, it is usual that one end of the line is closer to the shoreline than the other. The implication of this is that the energy in a wave is also spread over quite a larger area,this in turn reduces the height of the waves. In other words, refraction often makes waves smaller.

Waves are caused by wind. Wave height in the open ocean is determined by three factors. The greater the wind speed the larger the waves. The greater the duration of the wind (or storm) the larger the waves. The greater the fetch (area over which the wind is blowing - size of storm) the larger the waves.

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3 years ago

What is one way you can state how you feel?

dexar [7]

Words and more words

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3 years ago

A tank contains 150 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

nikitadnepr [17]

Answer:

The number A(t) of grams of salt in the tank at time t is $A(t) = 150 - 110 e^{-\frac{t}{50} }$

Explanation:

Knowing

$\frac{dA}{dt} = Rin - Rout$

First we have to find the Rin and Rout

Rin = (concentration of the salt inflow) * (input rate of brine)

Rin = 1 g/L * 3 L/min = 3 g/L

Rout = (concentration of the salt outflow) * (output rate of brine)

Rout = ( $\frac{A(t)}{150} g/L$ ) * (3 L/min) = $\frac{A(t)}{50} g/min$

Substituting this results

$\frac{dA}{dt}$ = 3 - $\frac{A(t)}{50}$ --> $\frac{dA}{dt} + \frac{1}{50} A(t) = 4$

Thus, integration factors is

$e^{ \int\limits^._. {\frac{1}{50} } \, dt } = e^{\frac{t}{50} }$

$e^{\frac{t}{50} } \frac{dA}{dt} + \frac{1}{50} e^{\frac{t}{50}$ A(t) = 4 $e^{\frac{t}{50} }$

$e^{\frac{t}{50} } A(t) = \int\limits^._. {3 e^{\frac{t}{50} } } \, dt\\ \\A(t) = 150 + c e^{-\frac{t}{50} }$

Applying the initial conditions

A(0) = 40

c = 150 - 40 = 110

Now, substitute this result in the solution to get

$A(t) = 150 - 110 e^{-\frac{t}{50} }$

8 0

4 years ago

Which one of the following best illustrates a primary reinforcer? A. Money B. The sight of a chocolate meringue pie C. A warm sm

konstantin123 [22]

Answer:

D . water when you're thirsty...

8 0

4 years ago

Read 2 more answers

As soon as a traffic light turns green, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.

Ganezh [65]

Answer: (a) The bicycle is ahead of the car for 4 s.

(b) The bicycle leads the car by the maximum distance of 55 m.

Explanation:

(a)

Use the equation of the motion to calculate the time taken by the car.

$v=u+at$

As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec2 .

Put u=0, v=22.3 m/sec and a=4.02 m/sec^2.

$22.3=0+4.02t$

$t=\frac{22.3}{4.02}$

t= 5.5 s

Use the equation of the motion to calculate the time taken by the bicycle.

$v=u+at_{1}$

As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.

Put u=0, v=8.94 m/sec and a=5.81 m/sec^2.

$8.94=0+5.81t_{1}$

$t_{1}=\frac{8.94}{5.81}[tex]t_{1}=1.5 s$

Calculate the time interval for which the bicycle is ahead of the car.

$t-t_{1}= 5.5 s - 1.5s$

$t-t_{1}= 4s$

Therefore, the bicycle is ahead of the car for 4 s.

(b)

Use the equation motion to calculate the distance covered by the car.

$S=ut+\frac{1}{2}at^{2}$

As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec^2 .

Put t= 5.5 s, u=0 s and a=4.02 m/sec^2.

$S=(0)t+\frac{1}{2}(4.02)(5.5)^{2}$

$S= 60.8 m$

Use the equation motion to calculate the distance covered by the bicycle.

$S_{1}=ut+\frac{1}{2}at^{2}$

As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.

Put t= 1.5 s, u=0 s and a=5.81 m/sec^2.

$S_{1}=(0)t+\frac{1}{2}(5.18)(1.5)^{2}$

$S_{1}= 5.8 m$

Calculate the maximum distance covered by the bicycle to lead the car.

$S-S_{1}=60.8-5.8=55m$

Therefore, the bicycle leads the car by the maximum distance of 55 m.

6 0

3 years ago