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iren [92.7K]
2 years ago
7

How long must a tow truck apply a force of 600 N to increase the speed of a 1,500 kg car at rest to 2 m/s?

Physics
1 answer:
slega [8]2 years ago
7 0

The time the truck must apply the given force to increase its speed to given value is 5 s.

The given parameters;

  • <em>applied force, F = 600 N</em>
  • <em>mass of the truck, m = 1,500 kg</em>
  • <em>speed of the truck, v = 2 m/s</em>

The force applied to the truck is determined by Newton's second law of motion; <em>which states that the force applied to an object is directly proportional to the product of mass and acceleration of the object.</em>

F = ma

F = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{1500 \times 2}{600} \\\\t = 5 \ s

Thus, the time the truck must apply the given force to increase its speed to given value is 5 s.

Learn more here:brainly.com/question/1988795

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The kinetic energy of a proton and that of an a-particle are 4 eV and 1 eV,
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Answer:

  • <u><em>(b) 1:1</em></u>

Explanation:

<u></u>

<u>1. Formulae:</u>

  • E = hf  
  • E = h.v/λ
  • λ = h/(mv)
  • E = (1/2)mv²

Where:

  • E = kinetic energy of the particle
  • λ = de-Broglie wavelength
  • m = mass of the particle
  • v = speed of the particle
  • h = Planck constant

<u><em>2. Reasoning</em></u>

An alha particle contains 2 neutrons and 2 protons, thus its mass number is 4.

A proton has mass number 1.

Thus, the relative masses of an alpha particle and a proton are:

       \dfrac{m_\alpha}{m_p}=4

For the kinetic energies you find:

          \dfrac{E_\alpha}{E_p}=\dfrac{m_\alpha \times v_\alpha^2}{m_p\times v_p^2}

            \dfrac{1eV}{4eV}=\dfrac{4\times v_\alpha^2}{1\times v_p^2}\\\\\\\dfrac{v_p^2}{v_\alpha^2}=16\\\\\\\dfrac{v_p}{v_\alpha}=4

Thus:

           \dfrac{m_\alpha}{m_p}=4=\dfrac{v_p}{v__\alpha}

          m_\alpha v_\alpha=m_pv_p

From de-Broglie equation, λ = h/(mv)  

       \dfrac{\lambda_p}{\lambda_\alpha}=\dfrac{m_\lambda v_\lambda}{m_pv_p}=\dfrac{1}{1}=1:1

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we know a = F/ M

Explanation:

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What are the advantages and disadvantages of series and parallel circuits?
DerKrebs [107]
Parallel circuit

Advandages: 1. Every unit that is connected in a parallel circuit gets equal amount of voltage.

2. It becomes easy to connect or disconnect a new element without affecting the working of other elements.

3. If any fault happened to the circuit, then also the current is able to pass through the circuit through different paths.

Disadvantages: 1. It requires the use of lot of wires.

2. We cannot increase or multiply the voltage in a parallel circuit.
3. Parallel connection fails at the time when it is required to pass exactly same amount of current through the units.

series circuit

Advantages: 1. Series circuits do not overheat easily. This makes them very useful in the case of something that might be around a potentially flammable source, like dry plants or cloth.

2. Series circuits are easy to learn and to make. Their simple design is easy to understand, and this means that it’s simple to conduct repairs .

3. we can add more power devices, they have a higher output in terms of voltage .

4. The current that flows in a series circuit has to flow through every component in the circuit. Therefore, all of the components in a series connection carry the same current.

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2. As the number of components in a circuit increases ,greater will be the circuit resistance.
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What is the distance from axis about which a uniform, balsa-wood sphere will have the same moment of inertia as does a thin-wall
andrey2020 [161]

Answer:

D_{s} ≈ 2.1 R

Explanation:

The moment of inertia of the bodies can be calculated by the equation

     I = ∫ r² dm

For bodies with symmetry this tabulated, the moment of inertia of the center of mass

Sphere               Is_{cm} = 2/5 M R²

Spherical shell   Ic_{cm} = 2/3 M R²

The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass

    I = I_{cm} + M D²

Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation

Let's start with the spherical shell, axis is along a diameter

     D = 2R

    Ic = Ic_{cm} + M D²

    Ic = 2/3 MR² + M (2R)²

    Ic = M R² (2/3 + 4)

    Ic = 14/3 M R²

The sphere

    Is =Is_{cm} + M [D_{s}²

    Is = Ic

    2/5 MR² + M D_{s}² = 14/3 MR²

    D_{s}² = R² (14/3 - 2/5)

    D_{s} = √ (R² (64/15)

    D_{s} = 2,066 R

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3 years ago
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