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ELEN [110]
4 years ago
15

A race car starts from rest on a circular track of radius 478 m. The car’s speed increases at the constant rate of 0.747 m/s 2 .

At the point where the magnitudes of the centripetal and tangential accelerations are equal, find the speed of the race car
Physics
1 answer:
Keith_Richards [23]4 years ago
3 0

Answer:

18.896ms-1

Explanation:

The tangential acceleration of the car= 0.747ms-2

The centripetal acceleration is given by v^2/r

0.747= v^2/r

V= √0.747*478= 18.896ms-1

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A proton moves with a speed of 1.00 x 106 m/s perpendicular to a magnetic field, B. As a result, the proton moves in circle of r
Softa [21]

Answer:

0.0109 m ≈ 10.9 mm

Explanation:

proton speed = 1 * 10^6 m/s

radius in which the proton moves = 20 m

<u>determine the radius of the circle in which an electron would move </u>

we will apply the formula for calculating the centripetal force for both proton and electron ( Lorentz force formula)

For proton :

Mp*V^2 / rp  = qp *VB   ∴  rp = Mp*V / qP*B    ---------- ( 1 )

For electron:

re = Me*V/ qE * B -------- ( 2 )

Next: take the ratio of equations 1 and 2

re / rp = Me / Mp                                 ( note: qE = qP = 1.6 * 10^-19 C )

∴ re ( radius of the electron orbit )

= ( Me / Mp ) rp

= ( 9.1 * 10^-31 / 1.67 * 10^-27 ) 20

= ( 5.45 * 10^-4 ) * 20

= 0.0109 m ≈ 10.9 mm

5 0
3 years ago
A proton enters a uniform magnetic field of strength 1 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’
scZoUnD [109]

A charged particle moving in a magnetic field experiences a force equal to:

\vec{F}=q\vec{v}\times \vec{B}

Thus, the magnitude of the force that the proton experiences is given by:

F=qvBsin\theta

The magnetic field is perpendicular to the proton's velocity, therefore, we have \theta=90^\circ. Replacing the given values, we obtain:

F=1.6*10^{-19}C(300\frac{m}{s})(1T)sin(90^\circ)\\F=4.8*10^{-17}N

3 0
3 years ago
Consider the incomplete equation below mc020-1.jpg If this equation was completed, which statement would it best support? Nuclea
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<span>Nuclear fission produces elements that are heavier than helium.

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Which type of fusion occurs in a high-mass star near the end stages of its life cycle?
uysha [10]
The answer is "iron fusion".

In fact, initially all stars burn hydrogen through nuclear fusion. As they run out of hydrogen, they start to burn the next heavier element, which is helium. The very massive stars continue this cycle, and when they run out of helium they start to burn the heavier elements until reaching iron. This element represents the end of the chain, because nuclear fusion of iron does not release energy, but it absorbs energy. This means that the star can't produce energy anymore and eventually it collapses.
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4 years ago
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What is the relationship between refraction and the speed of light?
tankabanditka [31]

Answer:

n=\frac{c}{v}

Explanation:

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When a light travel form denser to rare medium then the light ray bend away from the normal.

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8 0
3 years ago
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