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alexdok [17]
3 years ago
15

What is the meaning of life?

Physics
2 answers:
Harman [31]3 years ago
7 0

We all have our own purpose on the earth, because of course you aren't just born for no reason, everything happens for a reason, there's no such thing as coincidences. Everybody's reason is different. Some people were brought on the earth to make a change, some people to find love, others to do their dreams. However, everyone has their own reason, we just have to find it and achieve it. :)

Hope this helps, <3.

valentina_108 [34]3 years ago
3 0
1. Life is meaningless.
2. 42
You might be interested in
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
3 years ago
Which term is applied to an object through which light passes?
Elena L [17]

Answer:

D

Explanation:

transparent_objects that allows light to pass through and can you see through them

5 0
2 years ago
Read 2 more answers
What do you mean by booting ?whst are typez of booting​
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Answer:

plz mark brainlist

8 0
2 years ago
Does anyone know this
NARA [144]

1) The net force is 16 N to the right

2) The net force is 98 N to the left

3) The net  force is 0.5 N downward

4) The net force is 170 N to the right

5) The net force is 175 N to the  right

Explanation:

1)

To find the net force, we have to analyze all the forces acting on the box.

We have:

  • Force to the right: F_a = 20 N, the applied force
  • Force to the left: F_f = 4 N, the force of friction
  • Force to the bottom: F_g = 400 N, the weight of the box (the weight is always downward vertically)
  • Force to the top: F_N = 400 N. This is the normal force, which is the reaction force exerted by the table on the box: it points upward and counterbalances the weight of the box, preventing it from falling down)

Therefore, the horizontal net force is

F_x = F_a - F_f = 20 - 4 = 16 N (to the right)

While the vertical force is

F_y = F_N - F_g = 400 - 400 = 0

So the net force is 16 N to the right.

2)

In this case, we have the following forces:

  • F_g = 4 N downward, the weight of the ball
  • F_a = 100 N to the left, the force that kicks the ball
  • F_f = 2 N to the right, the force of friction
  • F_N = 4 N upward, the normal reaction exerted by the field on the ball

Therefore, the horizontal net force is

F_x =F_a - F_f = 100 -2 = 98 N (to the left)

While the vertical force is

F_y = F_g - F_N = 4 - 4 = 0 (downward)

And so, the net force is 98 N to the left.

3)

The force acting on the squirrel in this problem are:

  • F_g = 8 N downward, the weight of the squirrel
  • F_f = 7.5 N upward, the air resistance, acting upward

Both forces act vertically and there are no other forces acting in other directions, therefore the net force on the squirrel is simply equal to the net force on the vertical direction, which is:

F_y = F_g - F_f = 8 - 7.5 = 0.5 N

And since the weight is larger than the air resistance, the direction of the net force is downward.

4)

The forces acting on Monkey are:

  • F_1=95 N is the force applied to the right by Bunny
  • F_2 = 75 N is the force applied by Deer from the left (so, also on the right)
  • F_g = 50 N is the weight of Monkey, downward
  • F_N = 50 N is the normal reaction exerted by the surface, upward

So, the net force in the horizontal direction is

F_x = F_1 + F_2 = 95+75=170 N (to the right)

While the net force in the vertical direction is

F_y = F_N - F_g = 50 - 50 = 0

And therefore the net force is 170 N to the right

5)

The forces acting on Deer are:

  • F_a = 100 N + 100 N = 200 N to the right, the combined force applied by Bunny and Monkey
  • F_f = 25 N to the left, the force of friction
  • F_g = 150 N downward, the weight of the deer
  • F_N = 150 N upward, the normal reaction from the surface that balances the weight

So the net horizontal force is

F_x = F_a - F_f = 200 - 25 = 175 N to the right

While the net vertical force is

F_y = F_N - F_g = 150 - 150 = 0

So the net force is 175 N to the right.

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

3 0
2 years ago
A 13561 N car traveling at 51.1 km/h rounds
Minchanka [31]

Answer:

a) The centripetal acceleration of the car is 0.68 m/s²

b) The force that maintains circular motion is 940.03 N.

c) The minimum coefficient of static friction between the tires and the road is 0.069.

Explanation:

a) The centripetal acceleration of the car can be found using the following equation:

a_{c} = \frac{v^{2}}{r}

Where:

v: is the velocity of the car = 51.1 km/h

r: is the radius = 2.95x10² m

a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2}

Hence, the centripetal acceleration of the car is 0.68 m/s².

b) The force that maintains circular motion is the centripetal force:

F_{c} = ma_{c}

Where:

m: is the mass of the car

The mass is given by:

P = m*g

Where P is the weight of the car = 13561 N

m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg

Now, the centripetal force is:

F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N

Then, the force that maintains circular motion is 940.03 N.

c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:

F_{c} = F_{\mu}

F_{c} = \mu N = \mu P

\mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069

Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.

I hope it helps you!                

3 0
3 years ago
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