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dybincka [34]
2 years ago
11

How large is the tension in a rope that is being used to accelerate a 100 kg box upward at 2m/s2?

Physics
1 answer:
Flauer [41]2 years ago
5 0
  • m=100kg
  • Acceleration=2m/s^2

\\ \sf\Rrightarrow T=F

\\ \sf\Rrightarrow T=ma

\\ \sf\Rrightarrow T=100(2)

\\ \sf\Rrightarrow T=200N

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What is the ionisation energy of an atom
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its the minimum amount of energy required to remove the most loosely bound electron

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An electron accelerates through a 12.5 V potential difference, starting from rest, and then collides with a hydrogen atom, excit
Yanka [14]

Answer:

Initial state    Final state

     3           ⇒        2

     3           ⇒        1

     2          ⇒         1

Explanation:

For this exercise we must use Bohr's atomic model

         E = - 13.606 / n²

where is the value of 13.606 eV is the energy of the ground state and n is the integer.

The energy acquired by the electron in units of electron volt (eV)

          E = e V

          E = 12.5 eV

all this energy is used to transfer an electron from the ground state to an excited state

        ΔE = 13.6060 (1 / n₀² - 1 / n²)

the ground state has n₀ = 1

       ΔE = 13.606 (1 -  1/n²)

        1 /n² = 1 - ΔE/13,606

         1 / n² = 1 - 12.5 / 13.606

         1 / n² = 0.08129

          n = √(1 / 0.08129)

          n = 3.5

 since n is an integer, maximun is

         n = 3

because it cannot give more energy than the electron has

From this level there can be transition to reach the base state.

 

Initial state    Final state

     3           ⇒        2

     3           ⇒        1

     2          ⇒         1

8 0
3 years ago
Freight car A with a gross weight of 200,000 lbs is moving along the horizontal track in a switching yard at 4 mi/hr. Freight ca
zhenek [66]

Answer: a) 4.7 mi/hr.  b) 86,500 lbs. mi²/Hr²

Explanation:

As in any collision, under the assumption that no external forces exist during the very small collision time, momentum must be conserved.

If the collision is fully inelastic, both masses continue coupled each other as a single mass, with a single speed.

So, we can write the following:

p₁ = p₂ ⇒m₁.v₁ + m₂.v₂ = (m₁ + m₂). vf

Replacing by the values, and solving for vf, we get:

vf = (200,000 lbs. 4 mi/hr + 100,000 lbs. 6 mi/hr) / 300,000 lbs = 4.7 mi/hr

If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

After the collision, total kinetic energy is as follows:

K₂ = 1/2 ((m₁ + m₂) vf²) = 3,313,500 lbs. mi²/hr²

So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

DE = K₁ - K₂ = 86,500 lbs. mi² / hr²

This loss is due to the impact, and is represented by the work done by friction forces (internal) during the impact.

8 0
3 years ago
Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad
love history [14]

Answer:

a) The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}, b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}

Where:

a_{r} - Magnitude of the radial acceleration, measured in meters per square second.

a_{t} - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

\| \vec a \| = a_{t}

\| \vec a \| = r \cdot \alpha

Where:

\alpha - Angular acceleration, measured in radians per square second.

r - Radius of rotation (Radius of a tire), measured in meters.

Given that \alpha = 15\,\frac{rad}{s^{2}} and r = 0.31\,m. The linear acceleration experimented by the car is:

\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)

\| \vec a \| = 4.65\,\frac{m}{s^{2}}

The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}.

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

Where:

\theta - Final angular position, measured in radians.

\theta_{o} - Initial angular position, measured in radians.

\omega_{o} - Initial angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

If \theta_{o} = 0\,rad, \omega_{o} = 0\,\frac{rad}{s}, \alpha = 15\,\frac{rad}{s^{2}}, the final angular position is:

\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}

\theta = 46.875\,rad

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

\theta = 7.46\,rev

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0
3 years ago
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