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3 years ago

Do we live in a simulation?

Physics

2 answers:

WINSTONCH [101]3 years ago

4 0

Answer:

There is a 50-50 chance that we do live in a simulation.

Explanation:

Lilit [14]3 years ago

3 0

Answer:

well no, but yes

Explanation:

depends on your religion tbh

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I couldn't answer this question may someone please help me

Neko [114]

Distance (s) =20 m

Initial velocity =0 m/s

g=10 m/s²

By using the formula:

v²= u²+ 2as

v²=0²+ 2×10×15

v²= 300

V= 10root over 3

$10 \sqrt{3}$

For calculating time,

v= u+ at

10 root over 3 = 0+ 10t

10 root over= 10t

t = 1 root over 3

t = 1.73 s

$1 \sqrt{3}$

The the stone will hit the ground with a velocity of 10 root over 3 m/s in 1 root over 3 sec.

8 0

3 years ago

Conservation of Energy-not sure how to do problem-confused on how to find the speed and how to figure out energy bar graph

nika2105 [10]

Given data

*The given mass of the rock is m = 2 kg

*The given potential energy is U_p = 407 J

(a)

The diagram of the energy bar graph is drawn below

(b)

If an object is at rest and has potential energy, once it starts to fall from its rest state then this potential energy is completely transferred to kinetic energy. This means that the magnitude of the kinetic energy is equal to the potential energy of the object.

The change in kinetic energy of the rock while falling to the ground is given as

$\begin{gathered} U_k=U_p \\ =407\text{ J} \end{gathered}$

(c)

The formula for the speed of the block is given as

$\begin{gathered} U_k=\frac{1}{2}mv^2 \\ v=\sqrt[]{\frac{2U_k}{m}} \end{gathered}$

Substitute the known values in the above expression as

$\begin{gathered} v=\sqrt[]{\frac{2\times407}{2}} \\ =20.17\text{ m/s} \end{gathered}$

Hence, the speed of the object is v = 20.17 m/s

6 0

1 year ago

A summary short about fire saftey

pentagon [3]

Answer:

Fire safety is the set of practices intended to reduce the destruction caused by fire. Fire safety measures include those that are intended to prevent ignition of an uncontrolled fire, and those that are used to limit the development and effects of a fire after it starts.

Explanation:

3 0

3 years ago

A new ride being built at an amusement park includes a vertical drop of 71.6 meters. Starting from rest the ride vertically drop

AnnyKZ [126]

Answer:

We can conclude 2.3x10^7 J was converted to thermal energy

Explanation:

<u>Energy Conservation </u>

According to the law of conservation of energy, the total energy of an isolated system must be constant. If some kind of energy is 'lost', we know it was transformed into another type.

Let's check the conditions of the problem. The ride vertically drops a distance of 71.6 m starting from rest, and at the bottom of the drop, its speed is 10 m/s. Knowing the mass of the cart plus passengers is 3.5X10^4 kg, we compute the total energy at the top of the drop.

$E=U+K$

Where E is the total energy (which must be conserved) at the top of the drop, U is the gravitational potential energy and K is the kinetic energy. We use the equations for each:

$U=m.g.h$

$\displaystyle K=\frac{mv^2}{2}$

$\displaystyle E=m.g.h+\frac{mv^2}{2}$

At the top, the speed is 0, thus

$\displaystyle E=m.g.h=(35000)(9.8)(71.6)=2.5X10^7\ J$

Now we compute the 'total' energy at the bottom (quoted because we know there is some mechanical energy loss in the drop)

$\displaystyle E'=m.g.h'+\frac{mv'^2}{2}$

This time h'=0 and v=10 m/s, thus

$\displaystyle E'=\frac{(35000)10^2}{2}=1.8x10^6\ J$

The mechanical energy at the top and the bottom are not the same, thus we can know part of it was converted to heat or thermal energy. We compute the difference

$2.5x10^7\ J-1.8x10^6\ J=2.3x10^7\ J$