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3 years ago

Arrange the events of the big bang in sequential order

Physics

1 answer:

ankoles [38]3 years ago

8 0

One - Cosmic "egg"

Two - Expansion

-no explosion

-expansion causes cooling

Three - Subatomic particles form quarks/electrons

Four - More cooling

Protons and neutrons form

Five - Formation of atoms

Six - Gravity causes atoms to form clouds of gas which forms stars

Seven - Stars group together to form galaxies

hope that helped somehow...

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A 0.10kg hockey puck decreases it’s speed from 40m/s to 0m/s in 0.025s. Determine the force that it experiences

BigorU [14]

Vf = vi + at
0m/s = 40m/s + a(0.025s)
a = -1600m/s^2

Fnet = ma
Fnet = (0.10kg)(-1600m/s^2)
Fnet = -160 N

hope that helps

6 0

3 years ago

Help me please pleasee

klio [65]

Answer:18/2

Explanation:

its half

6 0

3 years ago

How do I do these? My teacher didn’t show us how.

melisa1 [442]

Explanation:

Displacement is simply the change in position. So in the first part of problem 1, looking at the graph between 0 s and 2 s, the position changes from 0 m to -4 m. So the displacement is:

Δx = -4 m − 0 m

Δx = -4 m

Between 2 s and 4 s, the position stays at -4 m. The displacement is:

Δx = -4 m − (-4 m)

Δx = 0 m

Finally, between 4 s and 6 s, the position goes from -4 m to 6 m. The displacement is:

Δx = 6 m − (-4 m)

Δx = 10 m

The net displacement is the change in position from 0 s to 6 s:

Δx = 6 m − 0 m

Δx = 6 m

In the second part of problem 1, we have a velocity vs time graph.

Car 1 starts with 0 velocity and ends with a velocity of 6 m/s, so it is accelerating and constantly moving to the right.

Car 2 starts with a velocity of -6 m/s and ends with a velocity of 6 m/s. It is also accelerating, but first it is moving to the left, comes to a stop at t = 3 s, then moves to the right.

Car 3 starts with a velocity of 2 m/s and ends with a velocity of 2 m/s. So it is moving constantly to the right, but never speeds up or slows down.

We want to know when two of the cars meet. Unfortunately, this isn't as easy as looking for where the lines cross on the graph. We need to calculate their displacements. We can do this by finding the area under the graph (assuming all the cars start from the same point).

Let's start with Car 2. Half of the area is below the x-axis, and half is above. Without doing calculations, we can say the total displacement for this car is 0. This means it ends back up where it started, and that it never meets either of the other cars, both of which have positive displacements.

So we know Car 1 and Car 3 meet, we just have to find where and when. For Car 1, the area under the curve is a triangle. So its displacement is:

Δx = ½ t v(t)

where t is the time and v(t) is the velocity of Car 1 at that time. Since the line has a slope of 1 and y intercept of 0, we know v(t) = t. So:

Δx = ½ t²

Now look at Car 3. The area under the curve is a rectangle. So its displacement is:

Δx = 2t

When the two cars have the same displacement:

½ t² = 2t

t² = 4t

t² − 4t = 0

t (t − 4) = 0

t = 0, 4

t = 0 refers to the time when both cars are at the starting point, so t = 4 is the answer we're looking for. Where are the cars at this time? Simply plug in t = 4 into either of the equations we found:

Δx = 8

So Cars 1 and 3 meet at 4 s and 8 m.

7 0

3 years ago

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

3 years ago

What are the strengths and weaknesses of the four methods of waste management?<br>

Nitella [24]

Answer & Explanation: Waste management are all activities and actions required to manage waste from its inception to its final disposal. There are several methods of managing waste with its strengths and weaknesses. The strengths include;

* It creates employment

* It keeps the environment clean

* The practice is highly lucrative

* It saves the earth and conserves energy

The weaknesses of the methods of waste management includes;

* The sites are often dangerous

* The process is mostly

* There is a need for global buy-in

* The resultant product had a short life

4 0

3 years ago