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zhuklara [117]
3 years ago
7

When the mallet hits the ball with an action force, the ball exerts a reaction 1 force on the mallet as explained by: 1) Newton'

s first law 2) Newton's second law 3) Newton's third law 4) all three of Newton's laws​
Physics
2 answers:
aliina [53]3 years ago
5 0

Answer:it should be Newton’s second law

Explanation: lmk if I’m wrong

alexandr402 [8]3 years ago
5 0

Answer:

It's Newton's third law. For every known reaction, there is a equal and opposite reaction.

Explanation:

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A piece of Nichrome wire has a radius of 7.1 x 10-4 m. It is used in a laboratory to make a heater that dissipates 3.20×102 W of
bekas [8.4K]

Answer:

the length of the Nichrome wire is <em>59.88 m</em>

<em></em>

Explanation:

The resistance of a wire is determined by:

P = V² / R                              or    R = V² / P

where

  • V is the voltage
  • R is the resistance
  • P is the power

We also know that

R = PL / A

where

  • A is the area and it is equal to πr²
  • P = 3.20 × 10² W

Therefore, the length of the Nichrome wire is

L  = RA / ρ

L = (V²/P)(πr²) / ρ

L = V²πr² /  ρ P

L = (110 V)²(π)(7.1 × 10⁻⁴ m)² / (3.2 × 10² W)(100 × 10⁻⁸ Ω·m)

<em>L = 59.88 m</em>

7 0
4 years ago
In the periodic, the elements are in order based on
Helga [31]

Answer: informative array

Explanation:

"The periodic table of elements arranges all of the known chemical elements in an informative array. Elements are arranged from left to right and top to bottom in order of increasing atomic number. Order generally coincides with increasing atomic mass. The rows are called periods."

https://www.livescience.com/25300-periodic-table.html

5 0
4 years ago
Producing a current by moving a wire through a magnetic field is called<br> induction
nlexa [21]

Answer:

Actually it's called electro-magnetic induction...

7 0
3 years ago
Read 2 more answers
A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 126 m/s from the top of a cliff
Nikolay [14]

Answer:

a) The initial total mechanical energy of the projectile is 498556.296 joules.

b) The work done on the projectile by air friction is 125960.4 joules.

c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

Explanation:

a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy (E) of the project is equal to the sum of gravitational potential energy (U_{g}) and translational kinetic energy (K), all measured in joules:

E = U_{g} + K (Eq. 1)

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2} (Eq. 1b)

Where:

m - Mass of the projectile, measured in kilograms.

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

y - Initial height of the projectile above ground, measured in meters.

v - Initial speed of the projectile, measured in meters per second.

If we know that m = 54\,kg, g = 9.807\,\frac{m}{s^{2}}, y = 132\,m and v = 126\,\frac{m}{s}, the initial mechanical energy of the earth-projectile system is:

E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}

E = 498556.296\,J

The initial total mechanical energy of the projectile is 498556.296 joules.

b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:

W_{loss} = E_{o}-E_{1} (Eq. 2)

Where:

E_{o} - Initial total mechanical energy, measured in joules.

E_{1} - FInal total mechanical energy, measured in joules.

W_{loss} - Work losses due to air friction, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

W_{loss} = E_{o}-K_{1}-U_{g,1}

W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1} (Eq. 2b)

Where:

m - Mass of the projectile, measured in kilograms.

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

y_{1} - Maximum height of the projectile above ground, measured in meters.

v_{1} - Current speed of the projectile, measured in meters per second.

If we know that E_{o} = 498556.296\,J, m = 54\,kg, g = 9.807\,\frac{m}{s^{2}}, y_{1} = 297\,m and v_{1} = 89.3\,\frac{m}{s}, the work losses due to air friction are:

W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)

W_{loss} = 125960.4\,J

The work done on the projectile by air friction is 125960.4 joules.

c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:

E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss} (Eq. 3)

K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}

Where:

E_{1} - Total mechanical energy of the projectile at maximum height, measured in joules.

U_{g,2} - Potential gravitational energy of the projectile, measured in joules.

K_{2} - Kinetic energy of the projectile, measured in joules.

W_{loss} - Work losses due to air friction during the upward movement, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss} (Eq. 3b)

m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}

v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}

v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}  }

If we know that E_{1} = 372595.896\,J, m = 54\,kg, g = 9.807\,\frac{m}{s^{2}}, y_{2} =0\,m and W_{loss} = 125960.4\,J, the final speed of the projectile is:

v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right)  }

v_{2} \approx 82.475\,\frac{m}{s}

The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

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