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myrzilka [38]
2 years ago
5

When will the solar system die

Physics
2 answers:
scoundrel [369]2 years ago
8 0
No the system will never doe that's just like people say in the world is going to end
Alenkinab [10]2 years ago
3 0
I would say our solar sustem will die when the sun dies or a black hole sucks it up.
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Why are viruses hard to fight
ki77a [65]

Answer:Compared to other pathogens, such as bacteria, viruses are minuscule. And because they have none of the hallmarks of living things — a metabolism or the ability to reproduce on their own, for example — they are harder to target with drugs.

Explanation:

7 0
3 years ago
Two blocks of masses 3.0 kg and 5.0 kg are connected by a spring and rest on a frictionless surface. They are given velocities t
miskamm [114]

Answer:

-0.7 m/sec

Explanation:

Mass of first block = m1 =3.0 kg

Mass of second block = m2= 5.0 kg

Velocity of first block = V1= 1.2 m/s

Velocity of second block = V2 = ?

Momentum of Center of mass MVcom  is sum of both blocks momentum and is given by

MVcom= m1v1+m2v2

Where

M= mass of center of mass

Vcom= Velocity of center of mass=0 m/s (because center of mass is at rest , so Vcom = 0 m.sec)

Putting values, we get;

0= 3×1.2+5v2

==> v2=  3.6/5= - 0.7 m/s

-ve sign indicates that block 2 is moving in opposite direction of block 1

3 0
3 years ago
When a wave has λ=3 m and f=15 Hz, what is the speed of the wave?
Tamiku [17]

Wave speed = (wavelength) x (frequency)

Wave speed = (3 m) x (15 Hz)

<em>Wave speed = 45 m/s</em>

5 0
3 years ago
An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com
solong [7]

Answer:

(a) 3.81\times 10^5\ Pa

(b) 4.19\times 1065\ Pa

Explanation:

<u>Given:</u>

  • T_1 = The first temperature of air inside the tire = 10^\circ C =(273+10)\ K =283\ K
  • T_2 = The second temperature of air inside the tire = 46^\circ C =(273+46)\ K= 319\ K
  • T_3 = The third temperature of air inside the tire = 85^\circ C =(273+85)\ K=358 \ K
  • V_1 = The first volume of air inside the tire
  • V_2 = The second volume of air inside the tire = 30\% V_1 = 0.3V_1
  • V_3 = The third volume of air inside the tire = 2\%V_2+V_2= 102\%V_2=1.02V_2
  • P_1 = The first pressure of air inside the tire = 1.01325\times 10^5\ Pa

<u>Assume:</u>

  • P_2 = The second pressure of air inside the tire
  • P_3 = The third pressure of air inside the tire
  • n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)

Part (a):

Using the above equation for this part of compression in the air, we have

\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa

Hence, the pressure in the tire after the compression is 3.81\times 10^5\ Pa.

Part (b):

Again using the equation for this part for the air, we have

\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa

Hence, the pressure in the tire after the car i driven at high speed is 4.19\times 10^5\ Pa.

8 0
2 years ago
A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.
GarryVolchara [31]

a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is 8.75 m/s^2 in the forward direction

Explanation:

a)

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

- The applied force, F = 50 N, pushing forward in the horizontal direction

- The frictional force, F_f = 15 N, pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)

- The weight of the book, W=mg, where m is the mass of the book and g=9.8 m/s^2 is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:

W=(4)(9.8)=39.2 N

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

b)

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

N-W=ma

where a is the acceleration; however, since N = W, this becomes

a=0

And since the book is initially at rest on the desk, this means that there is no motion.

c)

We said there are two forces acting in the horizontal direction:

- The applied force, F = 50 N, forward

- The frictional force, F_f = 15 N, backward

Since they act along the same line, we can calculate their resultant as

\sum F = F - F_f = 50 - 15 = 35 N

and therefore the net force is 35 N in the forward direction.

d)

The net force is obtained as the resultant  of the net forces in the horizontal and vertical direction. However, we have:

- The net force in the horizontal direction is 35 N

- The net force in the vertical direction is zero, because the weight is balanced by the normal reaction

Therefore, this means that the total net force acting on the book is just the net force acting on the horizontal direction, so 35 N forward.

e)

The acceleration of the book can be calculated by using Newton's second law:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

Here we have:

\sum F = 35 N (in the forward direction)

m = 4 kg

Therefore, the acceleration is

a=\frac{\sum F}{m}=\frac{35}{4}=8.75 m/s^2 (forward)

Learn more about forces, weight and Newton's second law:

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#LearnwithBrainly

8 0
3 years ago
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