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3 years ago

In a chemical reaction blank are the substances left over

Physics

1 answer:

Zepler [3.9K]3 years ago

3 0

Answer: In a chemical reaction blank are the substances left over

Explanation: The substance left over after a reaction takes place

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Two particles are moving along the x axis. Particle 1 has a mass m₁ and a velocity v₁ = +4.7 m/s. Particle 2 has a mass m₂ and a

nirvana33 [79]

Answer:

m₁ / m₂ = 1.3

Explanation:

We can work this problem with the moment, the system is formed by the two particles

The moment is conserved, to simulate the system the particles initially move with a moment and suppose a shock where the particular that, without speed, this determines that if you center, you should be stationary, which creates a moment equal to zero

p₀o = m₁ v₁ + m₂ v₂

pf = 0

m₁ v₁ + m₂ v₂ = 0

m₁ / m₂ = -v₂ / v₁

m₁ / m₂= - (-6.2) / 4.7

m₁ / m₂ = 1.3

Another way to solve this exercise is to use the mass center relationship

Xcm = 1/M (m₁ x₁ + m₂ x₂)

We derive from time

Vcm = 1/M (m₁ v₁ + m₂v₂)

As they say the velocity of the center of zero masses

0 = 1/M (m₁ v₁ + m₂v₂)

m₁ v₁ + m₂v₂ = 0

m₁ / m₂ = -v₂ / v₁

m₁ / m₂ = 1.3

4 0

3 years ago

2. A girl whose mass is 55kg stands on a spring weighing machine inside a lift. When the lift starts to ascend, its acceleration

Annette [7]

Answer:

Explanation:

5 0

3 years ago

Science is the human endeavor to understand the natural world true or flase

hjlf

True yes TRUE
Science may also be defined as the study of surroundings

4 0

3 years ago

What is the magnitude of an electric field that balances the weight of a plastic sphere of mass 2.1 g that has been charged to -

Liula [17]

Answer:

Electric field, $E=6.86\times 10^6\ N/C$

Explanation:

It is given that,

Mass of sphere, m = 2.1 g = 0.0021 kg

Charge, $q=-3\ nC=-3\times 10^{-9}\ C$

We need to find the magnitude of electric field that balances the weight of a plastic spheres. So,

$ma=qE$

a = g

$E=\dfrac{mg}{q}$

$E=\dfrac{0.0021\ kg\times 9.8\ m/s^2}{-3\times 10^{-9}\ C}$

$E=6860000\ N/C$

$E=6.86\times 10^6\ N/C$

Hence, the magnitude of electric field that balances its weight is $6.86\times 10^6\ N/C$ . Hence, this is the required solution.

4 0

3 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

3 years ago