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adell [148]
3 years ago
9

Which gas is responsible for polarizination defect?​

Physics
2 answers:
mestny [16]3 years ago
5 0

Explanation:

During chemical reaction in a cell, hydrogen gas is produced and it accumulates, it is known as polarization.

Alex_Xolod [135]3 years ago
3 0

Answer:

<h3>Hydrogen gas is responsible for polarizination defect.</h3>

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A proton moves at constant velocity in the direction, through a region in which there is an electric field and a magnetic field.
dimaraw [331]

Answer:

F_{net}=0

Explanation:

It is given that, a proton moves at constant velocity, through a region in which there is an electric field and a magnetic field such that,

The electric field is, E = 800 V/m

Magnetic field, B = 0.25 T

We know that the net force in the region of magnetic and electric field is given by Lorentz forces. But here, the proton moves with constant velocity. So, the net force acting on it is 0.

i.e.

F_{net}=0

Hence, this is the required solution.

7 0
3 years ago
What do you mean by velocity ratio of a wheel and axle​
IgorC [24]

Answer:

Explanation:

In wheel and axle. …with the system is the velocity ratio, or the ratio of the velocity (VF) with which the operator pulls the rope at F to the velocity at which the weight W is raised (VW). This ratio is equal to twice the radius of the large drum divided by the difference…

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3 years ago
A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (v), measured in meters per second, by using this kinematic equation:

v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s} (1)

Where:

a - Acceleration, measured in meters per square second.

\Delta s - Travelled distance, measured in meters.

v_{o} - Initial velocity, measured in meters per second.

If we know that v_{o} = 0\,\frac{m}{s}, a = 2.35\,\frac{m}{s^{2}} and \Delta s = 595\,m, the final velocity of the rocket is:

v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}

v\approx 52.882\,\frac{m}{s}

The time associated with this launch (t), measured in seconds, is:

t = \frac{v-v_{o}}{a}

t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }

t = 22.503\,s

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o}) (2)

Where:

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

If we know that v_{o} = 52.882\,\frac{m}{s}, v = 0\,\frac{m}{s}, a = -9.807\,\frac{m}{s^{2}} and s_{o} = 595\,m, then the maximum height reached by the rocket is:

v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})

s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}

s = 737.577\,m

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (2)

Where:

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

v_{o} - Initial speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that s_{o} = 595\,m, v_{o} = 52.882\,\frac{m}{s}, s = 0\,m and a = -9.807\,\frac{m}{s^{2}}, then the time needed by the rocket is:

0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}

-4.904\cdot t^{2}+52.882\cdot t +595 = 0

Then, we solve this polynomial by Quadratic Formula:

t_{1}\approx 17.655\,s, t_{2} \approx -6.872\,s

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0
2 years ago
An imaginary line perpendicular to a reflecting surface is called ____refrac_____.
umka2103 [35]
Not totally sure but i would say a normal? its not refraction or incidence if its perpendicular and i dont think its a mirror if its an imaginary line so yeah normal (normals are always perpendicular to their surface too i think so)
4 0
3 years ago
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