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3 years ago

Describe two possibilities of what will happen to the universe in the future.

1 answer:

4 0

In our current view of the Universe there are 2 possible futures. One possiblity is that the Universe will come to an end in the opposite of a Big Bang called The Big Crunch. The other possibility is that we live in an Eternal Universe that will never come to end.

i hope this helps :)

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What is a centripetal acceleration of a point on a bicycle wheel of a radius of 0.70 m when a bike is moving 8.0 m/s

Furkat [3]

Answer:

The acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

Explanation:

The centripetal acceleration acts on a body when it is performing a circular motion.

Here, a point on the bicycle is performing circular motion as the rotation of the wheel produces a circular motion.

The centripetal acceleration of a point moving with a velocity $v$ and at a distance of $r$ from the axis of rotation is given as:

$a=\frac{v^2}{r}$

Here, $v=8\ m/s,r=0.70\ m$

∴ $a=\frac{8}{0.70}=11.43\ m/s^2$

Therefore, the acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

3 0

3 years ago

Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with

Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

Part A)

Let's use the parabolic motion equation to solve it. Let's define the variables:

y(i) is the initial height, it is a constant.
y(f) is the final height, in our case is 0
v(i) is the initial velocity (v(i)=16 m/s)
θ1 is the first angle, 30°
θ2 is the first angle, -30°

For the first stone

$y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}$

$0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}$

$0=y_{i1}+8t_{1}-4.905*t_{1}^{2}$ (1)

For the second stone

$0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}$

$0=y_{i2}-8t_{2}-4.905t_{2}^{2}$ (2)

If we solve the equation (1) we will have:

$t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can do the same procedure for the equation (2)

$t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

Part B)

We can use the equation of the horizontal position here.

First stone

$x_{f1}=x_{i1}+vcos(30)t_{1}$