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lidiya [134]
3 years ago
11

3. A car does an emergency stop from 144 km\h in 4s What was its deceleration?

Physics
1 answer:
strojnjashka [21]3 years ago
5 0

Answer:

36?

Explanation:

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Which objects in space formed from the huge disk cisce and debris beyond the outer planets? Select two option
Dennis_Churaev [7]

Answer:

Comets

Explanation:

The Kuiper Belt is a collection of trans-Neptunian objects that consist of comets and other dwarf planets, including Pluto.

4 0
3 years ago
Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet
Valentin [98]

Answer:

v = 8.09   m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

       

starting point. Higher

         Em₀ = U = m gh

final point. To go down the slope

         Em_f = K = ½ m v²

The work of the friction force is

         W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

        N - W_y = 0

        N = W_y

X axis

        Wₓ - fr = ma

let's use trigonometry

        sin  θ = y / L

         sin θ = 11/110 = 0.1

         θ = sin⁻¹  0.1

          θ = 5.74º

         sin 5.74 = Wₓ / W

         cos 5.74 = W_y / W

         Wₓ = W sin 5.74

         W_y = W cos 5.74

the formula for the friction force is

         fr = μ N

         fr = μ W cos θ

Work is friction force is

         W_fr = - μ W L cos θ  

Let's use the relationship of work with energy

        W + ΔU = ΔK

         -μ mg L cos 5.74 + (mgh - 0) = 0  - ½ m v²

        v² = - 2 μ g L cos 5.74 +2 (gh)

        v² = 2gh - 2 μ gL cos 5.74

let's calculate

        v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

        v² = 215.6 -150.16

        v = √65.44

        v = 8.09   m/s

6 0
3 years ago
How does energy acquisition in the deep sea differ from energy acquisition near the ocean’s surface? a. Organisms in the deep se
Montano1993 [528]

that would be the energy from the core. because of the fact that ht eearths core is producing 45mill killawhats and giggawats


3 0
3 years ago
Read 2 more answers
The emt must assume that any unwitnessed water-related incident is accompanied by:________
natta225 [31]

The EMT must assume that any unwitnessed water-related incident is accompanied by potential spinal damage.

<h3>What is spinal damage?</h3>
  • Nerves or the spinal cord in any way damaged at the end of the spinal canal.
  • A rapid strike or cut to the spine can cause a traumatic spinal cord damage.
  • Below the damage site, a spinal cord injury frequently results in a lifelong loss of strength, feeling, and function.
  • A lot of people with spinal cord injuries may lead productive, independent lives with the help of rehabilitation and assistive technology.
  • Symptom-reducing medications and spinal stabilisation surgery are used as treatments.
  • Herniated discs are among the common injuries and diseases of the spine. Stenosis of the lower back and Scoliosis are others.
  • After taking part in a rehabilitation programme, over 80% of people with incomplete spinal cord injury (SCI) can walk again.

Learn more about spinal cord here:

brainly.com/question/23916836

#SPJ4

3 0
2 years ago
The kinetic energy of a proton and that of an a-particle are 4 eV and 1 eV,
madam [21]

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

5 0
3 years ago
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