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yanalaym [24]
3 years ago
10

What two particles in an atom are equal in number

Physics
1 answer:
Ainat [17]3 years ago
8 0
The two particles that are equal in a atom is the proton and electron.
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Plz tll answer of this . pleaseeeeeeeeeeee. Q 6,7,8<br>​
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3 years ago
Jane has a mass of 55 kg and his body covers 700 nails with a surface area of 1.00 mm,
Oksana_A [137]

We have,

  • Jane mass is 55 kg
  • His body covered with 700 nails all of them having a surface area of 1.00 mm² each = 700 × 1 = 700 mm² = 700/1000000 = 7/10000

We know that,

  • Pressure = Force/Area

Let's calculate force as we already have area;

  • F = ma
  • F = 55 × 9.8 { Acceleration due to gravity }
  • F = 539 N

Now, if should she would be on 700 nails then pressure will be;

  • P = F/A
  • P = 539/7 × 10000
  • P = 5390000/7
  • P = 770,000 Pascal

And if should would be on a 1 nail only,

  • P = F/A
  • P = 539/1 × 1000000
  • P = 539000000 Pascal

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6 0
2 years ago
What happens when bromine reacts with carbon?
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D. Electrons are shared between the bromine atoms and carbon atoms
4 0
3 years ago
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A 4 kg bird is flying with a velocity of 4 m/s. What is its kinetic energy?
Maslowich
32 kg m/s would be the kinetic energy.
3 0
3 years ago
A rotating flywheel has moment of inertia 18.0 kg⋅m^2 for an axis along the axle about which the wheel is rotating. Initially th
timama [110]

Answer:

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

Explanation:

By the Principle of Energy Conservation and the Work-Energy Theorem we know that flywheel slow down due to the action of non-conservative forces (i.e. friction), the energy losses are equal to the change in the rotational kinetic energy. That is:

\Delta E = K_{1}-K_{2} (1)

Where:

\Delta E - Energy losses, measured in joules.

K_{1}, K_{2} - Initial and final rotational kinetic energies, measured in joules.

By definition of rotational kinetic energy, we expand the equation above:

\Delta E = \frac{1}{2}\cdot I\cdot (\omega_{1}^{2}-\omega_{2}^{2}) (2)

Where:

I - Moment of inertia of the flywheel, measured in kilograms per square meter.

\omega_{1}, \omega_{2} - Initial and final angular speed, measured in radians per second.

If we know that K_{1} = 30\,J, K_{2} = 15\,J and I = 18\,kg\cdot m^{2}, then the initial angular speed is:

K_{1} = \frac{1}{2}\cdot I \cdot \omega_{1}^{2} (3)

\omega_{1}=\sqrt{\frac{2\cdot K_{1}}{I} }

\omega_{1} = \sqrt{\frac{2\cdot (30\,J)}{18\,kg\cdot m^{2}} }

\omega_{1} \approx 1.825\,\frac{rad}{s}

\omega_{1}\approx 0.291\,\frac{rev}{s}

K_{2} = \frac{1}{2}\cdot I \cdot \omega_{2}^{2} (4)

\omega_{2}=\sqrt{\frac{2\cdot K_{2}}{I} }

\omega_{2} = \sqrt{\frac{2\cdot (15\,J)}{18\,kg\cdot m^{2}} }

\omega_{2} \approx 1.291\,\frac{rad}{s}

\omega_{2} \approx 0.205\,\frac{rev}{s}

Under the assumption that flywheel is decelerating uniformly, we get that the time taken for the flywheel to slowdown is:

t = \frac{\omega_{2}-\omega_{1}}{\alpha} (5)

If we know that \omega_{1}\approx 0.291\,\frac{rev}{s}, \omega_{2} \approx 0.205\,\frac{rev}{s} and \alpha = -0.200\,\frac{rev}{s^{2}}, then the time needed is:

t = \frac{0.205\,\frac{rev}{s}-0.291\,\frac{rev}{s}}{-0.200\,\frac{rev}{s^{2}} }

t = 0.43\,s

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

6 0
3 years ago
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