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belka [17]
3 years ago
13

1 Point

Physics
1 answer:
Margarita [4]3 years ago
7 0

Answer: C

Combustion

Explanation:

Oxygen support combustion in a chemical reaction. Oxygen support burning.

In the absence of oxygen, combustion reaction can never take place. And when the supply of oxygen is being cut off, combustion reaction will automatically stop.

The correct option is C which is combustion

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Two small particles of mass m1 and mass m2 attract each other with a force that varies inversely with the cube of their separati
Naya [18.7K]

Answer:

r_{cm}\ =\ \dfrac{m_2d\ +\ m_1v_0 (t_1\ -\ t_0)}{m_1\ +\ m_2}

Explanation:

Given,

  • mass of the first particle = m_1
  • velocity of the first particle = v_o
  • mass of the second particle = m_2
  • velocity of the second particle = v_2 = 0
  • Time interval = (t_1\ -\ t_o)

Let v_{cm} be the initial velocity of the center of mass of the system of particle at time t_o

\therefore v_{cm}\ =\ \dfrac{m_1v_1\ +\ m_2v_2}{m_1\ +\ m_2}\\\Rightarrow v_{cm}\ =\ \dfrac{m_1v_0}{m_1\ +\ m_2}

Assuming that the first particle is at origin, distance of the second particle from the origin is 'd'

  • x_1\ =\ 0
  • x_2\ =\ d

Center of mass of the system of particles

x_{cm}\ =\ \dfrac{m_1x_1\ +\ m_2x_2}{m_1\ +\ m_2}\\\Rightarrow x_{cm}\ =\ \dfrac{m_2d}{m_1\ +\ m_2}\\

Hence, at time t_0, the center of mass of the system is at x_0\ =\ \dfrac{m_2d}{m_1\ +\ m_2} at an initial speed of v_{cm}

Both the particles are assumed to be the point masses, therefore at the time t_1 the center of mass is at the position of the second particle which should be equal to the total distance traveled by the first particle because the second particle is at rest.

Let r_{cm} be the distance traveled by the center of mass of the system of particles in the time interval (t_1\ +\ t_0)

From the kinematics,

s\ =\ x_0\ +\ vt\\\Rightarrow r_{cm}\ =\ x_{cm}\ +\ v_{cm}{t_1\ -\ t_0}\\\Rightarrow r_{cm}\ =\ \dfrac{m_2d}{m_1\ +\ m_2}\ +\ \left ( \dfrac{m_1v_0}{m_1\ +\ m_2}\ \right )\times (t_1\ -\ t_0)\\\Rightarow r_{cm}\ =\ \dfrac{m_2d\ +\ m_1v_0 (t_1\ -\ t_0)}{m_1\ +\ m_2}

Hence, this is the required distance traveled by the first mass to collide with the second mass which is at rest.

5 0
3 years ago
PLEASE HELP WHICH STATEMENTS ARE CORRECT DO NOT GUESS
Anni [7]
THe first one and the third one!!
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3 years ago
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Explain what must be true of component waves for reinforcement or interference to occur.
alisha [4.7K]

The component of waves must have same frequency and phase.


when the component of waves vibrate at the same rate and attain maximum point at the same time, reinforcement of the waves amplitude occur to cause a constructive interference.However, when the two waves are out of phase where one is at minimum when the other is at maximum a destructive interference happens.
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Exercising in hot weather can cause what side effects ?
trapecia [35]

Answer:

dehydration, lots of sweating, heat stroke, heat exhaustion: nausea, dizziness, vomitting, diahreaa, headache

Explanation:

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A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed
Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

\int EdS=\frac{Q_{in}}{\epsilon_o}   (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum =  8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

\int EdS=ES=E(4\pi r^2)   (2)

Qin is calculate by using the charge density:

Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho  (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}

Next, you use the results of (3), (2) and (1):

E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}

hence, the electric field is 1580594.95 N/C

7 0
3 years ago
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