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Crazy boy [7]
3 years ago
12

If a car has an initial velocity of 20 m/s and accelerates at 2.0 m/s2 for 100 m, what is its final velocity?

Physics
1 answer:
rodikova [14]3 years ago
3 0

Answer:

28.28 ms-1

Explanation:

v² = u² + 2as

v² = 20² + 2×2×100

v²= 800

v = 28.28 ms-1

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R S ( M ) = 2 G M c 2 , where G is the gravitational constant and c is the speed of light. It is okay if you do not follow the d
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The provided question's answer is "Schwarzschild radius".

The conversion factor between mass and energy is the speed of light squared.

GM/r stands for gravitational potential energy, also known as energy per unit mass.

GM/rc² then has "mass per unit mass" units. In other words, as mass/mass splits out in a dimensional analysis, "dimensionless per unit."

The derivation yields a formula for time or space coordinate ratios requiring sqrt(1 - 2GM/rc²). This number becomes 0 when r=2GM/c2, or the formula becomes infinite if in the denominator. However, there is no justification for using c² as a conversion factor there. Consider the initial expression sqrt(1 - 2GM/rc²).

Assume that m is used as the test particle's mass instead of 1. Then you have sqrt(m - 2GMm/rc² and mass units. This expression denotes that the rest energy of the test mass m you introduced into the gravitational field is "gone" at that radius.

The 2 would be absent if the gravitational field were Newtonian. However, at the event horizon, Einstein gravity is slightly stronger than Newton gravity, resulting in the factor 2 in qualitative terms.

So, the given equation is of Schwarzschild radius.

Learn more about Schwarzschild radius here:

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3 0
2 years ago
You step onto a hot beach with your bare feet. A nerve impulse, generated in your foot, travels through your nervous system at a
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3 years ago
Horses can run at speeds of 16 meters per second. If a horse running at full
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3 years ago
How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for a
Luba_88 [7]

Answer:

6

Explanation:

We are given that

\theta=2.12^{\circ}

Slid width,a=0.110 mm=0.11\times 10^{-3} m

1mm=10^{-3} m

Wavelength,\lambda=582 nm=582\times 10^{-9} m

1nm=10^{-9} m

We have to find the number of diffraction maxima are contained in a region of the Fraunhofer single-slit pattern.

asin\theta=\frac{2N+1}{2}\lambda

Using the formula

0.11\times 10^{-3}sin(2.12)=\frac{2N+1}{2}(582\times 10^{-9})

2N+1=\frac{0.11\times 10^{-3}sin(2.12)\times 2}{582\times 10^{-9}}

2N+1=13.98

2N=13.98-1=12.98

N=\frac{12.98}{2}\approx 6

Hence, 6 diffraction maxima are contained in a region of the Fraunhofer single-slit pattern

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3 years ago
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