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katen-ka-za [31]

3 years ago

11

Humans have existed for about 106 years, whereas the universe is about 1010 years old. If the age of the universe is defined as

1 "universe day," where a universe day consists of "universe seconds" as a normal day consists of normal seconds, how many universe seconds have humans existed?

1 answer:

son4ous [18]3 years ago

8 0

Answer:

8.64 seconds

Explanation:

According to the question

$10^{10}\ years=1\ day$

$1\ year=\dfrac{1}{10^{10}}\ day$

Humans have lived for $10^6$ years

In one universe day

$10^6\ years=10^6\times \dfrac{1}{10^{10}}=10^{-4}\ day$

In seconds

$10^{-4}\times 24\times 60\times 60=8.64\ seconds$

In one universe day humans have lived for 8.64 seconds

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A 2000 kg car moves along a horizontal road at speed vo

cluponka [151]

Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

$\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a$ (Eq. 1)

$\Sigma F_{y} = N-m\cdot g = 0$ (Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

$a = -\mu_{k}\cdot g$ (Eq. 3)

Where:

$a$ - Deceleration of the car, measured in meters per square second.

$\mu_{k}$ - Kinetic coefficient of friction, dimensionless.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $\mu_{k} = 0.0735$ and $g = 9.807\,\frac{m}{s^{2}}$ , then deceleration of the car is:

$a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )$

$a = -0.721\,\frac{m}{s^{2}}$

The stopping distance of the car ( $\Delta s$ ), measured in meters, is determined from the following kinematic expression:

$\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$ (Eq. 4)

Where:

$v_{o}$ - Initial speed of the car, measured in meters per second.

$v$ - Final speed of the car, measured in meters per second.

If we know that $v_{o} = 15.9\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $a = -0.721\,\frac{m}{s^{2}}$ , stopping distance of the car is:

$\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}$

$\Delta s = 175.319\,m$

The shortest possible stopping distance of the car is 175.319 meters.

8 0

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3. When you graph the motion of an object, you put ____ on the horizontal axis and ____ on the axis. a. speed, time b. dis

Papessa [141]

The answer is B. distance , Time

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Which of the following is the most reasonable weight in units of newtons for an average adult?

Oliga [24]

Answer: The reason weight for average adult is 530N

Explanation: The clearly see this answer, let's convert this weight to kg using F=mg

530=m×9.8

g is a constant of 9.8m/s2

We divide both sides by 9.8 to get

54kg then this weight is reasonable for an average adult.

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A squirrel drops an acorn onto the head of an unsuspecting dog. The acorn falls 4.0\,\text m4.0m4, point, 0, start text, m, end

kirill [66]

Answer:

0.903 seconds

Explanation:

To find how many seconds the acorn fall, we can use the formula for distance travelled with constant acceleration:

D = Vo*t + a*t^2/2,

where D is the distance travelled, Vo is the inicial speed, t is the time and a is the acceleration.

In our problem:

Vo = 0,

a = g = 9.81 m/s2,

D = 4 meters.

So, we can solve the equation to find the time:

4 = 0*t +9.81*t^2/2

4.905*t^2 = 4

t^2 = 4/4.905 = 0.8155

t = 0.903 seconds

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kobusy [5.1K]

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Explanation:

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