The swimmer swims 50 meters to one end of the pool but has to swim back therefore you double 50 which would be 100 meters. Then you have to multiply 100 by 4 since the swimmer did it 4 times.

7 0

3 years ago

How fast does light year travel

KIM [24]

"Light year" is a distance, not a speed. It's the distance light travels in one year, at the speed of 299,792,458 meters per second.

5 0

3 years ago

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Suppose you're performing experiments in science class in which you start with 70 bacteria and the amount of bacteria triples ev

barxatty [35]

1h----------------> 70x3=210 bacteria
2h-----------------> 210*3=630 bactaeria
let be y the number of bacteria at the t=0h
it is y=70 3^0
for t= 1h
y=70*3^1=210
for t=2h
y=70*3^2=630

so we can write y=70*3^x, where x is the number of hour

5 0

3 years ago

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A 2 kg ball is moving 3 m/s when it starts rolling up a hill.

AURORKA [14]

Answer:

the height reached is = 0.458 [m]

Explanation:

We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:

$Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]$

Replacing the values on the equation we have:

$Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\$

This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.

$Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\$

Now we have:

$h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]$

In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]

6 0

3 years ago