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3 years ago

Does the universe need explaining or can it just 'be' there?

1 answer:

4 0

Recently, We've been discussing this specific topic in my RS class. So, there are many theories to explain the existence of the world such as the Big Bang, Evolution or the Creation story.

Although these are three different ideas, they can still be related or interconnected with one another in some way. For example, George Lemaintre's Big Bang explanation can be linked with the Creation story in ways such as God being present at the Big Bang or even God being the one to allow the Big Bang occur.

Furthermore, God might have created us as apes or monkeys then we evolved to become humans.

In conclusion, the universe couldn't have come to pass just like that, they're must be Scientific or Theological truths to back this up. The only thing is that people have their unique beliefs, according to their religion, culture and childhood background.

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A bicyclist starts from rest and accelerates at a rate of 2.3 m/s^2 until it reaches a speed of 23 m/s. It then slows down at a

Rama09 [41]

Answer:

33 seconds.

Explanation:

The equation for speed with constant acceleration at time t its:

$V(t) \ = \ V_0 \ + \ a \ t$

where $V_0$ is the initial speed, and a its the acceleration.

<h3>First half of the problem</h3>

Starting at rest, the initial speed will be zero, so

$V_0 = 0$

the final speed is

$V(t_{f1}) = 23 \frac{m}{s}$

and the acceleration is

$a = 2.3 \frac{m}{s^2}$ .

Taking all this together, we got

$V(t_{f1}) = 23 \frac{m}{s} = 0 + 2.3 \ \frac{m}{s^2} t_{f1}$

$23 \frac{m}{s} = 2.3 \ \frac{m}{s^2} t_{f1}$

$\frac{23 \frac{m}{s}}{2.3 \ \frac{m}{s^2}} = t_{f1}$

$10 s = t_{f1}$

So, for the first half of the problem we got a time of 10 seconds.

<h3>Second half of the problem</h3>

Now, the initial speed will be

$V_0 = 23 \frac{m}{s}$ ,

the acceleration

$a=-1.0 \frac{m}{s^2}$ ,

with a minus sign cause its slowing down, the final speed will be

$V(t_{f2}) = 0$

Taking all together:

$V(t_{f2}) = 0 = 23 \frac{m}{s} - 1.0 \frac{m}{s^2} t_{f2}$

$23 \frac{m}{s} = 1.0 \frac{m}{s^2} t_{f2}$

$\frac{23 \frac{m}{s}}{1.0 \frac{m}{s^2}} = t_{f2}$

$23 s = t_{f2}$

So, for the first half of the problem we got a time of 23 seconds.

<h3>Total time</h3>

$t_total = t_{f1} + t_{f2} = 33 \ s$

5 0

3 years ago

The atomic number of an element is equal to the number of neutrons in the element?

tia_tia [17]

Answer: No

Explanation:

The number of neutrons is equal to the difference between the mass number of the atom (M) and the atomic number (Z).

7 0

3 years ago

A string trimmer is a tool for cutting grass and weeds; it utilizes a length of nylon "string" that rotates about an axis perpen

Kaylis [27]

Answer:

r = 22.3 cm

Explanation:

given,

angular speed = 45 rev/s

tangential speed = 63 m/s

length of the rotating string

angular velocity =

ω = 2 π N

ω = 2 π x 45

ω = 282.74 rad/s

we know,

v = r ω

$r = \dfrac{v}{\omega}$

$r = \dfrac{63}{282.74}$

$r =0.223\ m$

r = 22.3 cm

3 0

4 years ago

(5) A 4 kg. object rests on a flat, horizontal surface with a static

baherus [9]

Answer:

F = 9.81 [N]

Explanation:

To solve this problem we must use Newton's third le which tells us that the sum of forces on a body that remains static must be equal to one resulting from these forces in the opposite direction.

Let's perform a summation of forces on the vertical axis-y to determine the normal force N.

∑F = 0 (axis-y)

$N - m*g = 0$

where:

m = mass = 4 [kg]

g = gravity acceleration = 9.81 [m/s²]

$N - (4*9.81)=0\\N = 39.24 [N]$

Now we know that the frictional force can be calculated using the following equation.

f = μ*N

where:

f = friction force [N]

μ = friction coefficient = 0.25

N = normal force = 39.24 [N]

Now replacing:

$f = 0.25*39.24\\f = 9.81[N]$

Then we perform a sum of forces on the X-axis equal to zero. This sum of forces allows us to determine the minimum force to be able to move the object in a horizontal direction.

∑F = 0 (axis-x)

$F-f=0\\F-9.81=0\\F= 9.81[N]$

If the coefficient was smaller, a smaller force (F) would be needed to start the movement, this can be easily seen by replacing the value of 0.25, by smaller values, such as 0.1 or 0.05.

If the coefficient were larger, a larger force would be needed.

3 0

3 years ago

Please help , will surely mark As brainlest

Oliga [24]

The north arrow of the compasses will all point to the south since opposite poles attract. I have put an example picture below.

For the horse shoe magnet the magnetic field line will also go from north to south. I have put a picture of this below

8 0

3 years ago