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Alla [95]
3 years ago
7

Driving your car with a constant speed of 11 m/s, you encounter a hump in the road with a circular cross section. If the radius

of curvature of the hump is 54 m, what is the apparent weight of a 80 kg person in your car as you pass over the top of the hump?
Physics
1 answer:
Simora [160]3 years ago
6 0

Answer:604.74  N

Explanation:

Given

Speed of car=11 m/s

radius of curvature is 54 m

mass of person=80 kg

At Top

mg-N=\frac{mv^2}{r}

where N=normal reaction

N=mg-\frac{mv^2}{r}

N=80\time 9.8-\frac{80\times 11^2}{54}

N=604.74 N

Apparent weight will be equal to Normal reaction

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Answer: KE = 62.5J

Explanation:

Given that

Mass of object = 5kg

kinetic energy KE = ?

velocity of object = 5m/s

Since kinetic energy is the energy possessed by a moving object, and it depends on the mass (m) of the object and the velocity (v) by which it moves. Therefore, the object has kinetic energy.

i.e K.E = 1/2mv^2

KE = 1/2 x 5kg x (5m/s)^2

KE = 0.5 x 5 x 25

KE = 62.5J

Thus, the object has 62.5 joules of kinetic energy.

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The speed of light in a vacuum is approximately 2.99 × 108 m/s, and the speed of light through a piece of glass is approximately
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Option (b) is correct.The index of refraction for the glass is 1.52

Explanation:

velocity of light in vacuum= C= 2.99 x 10⁸m/s

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Thus the refractive index of glass is 1.52

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Two black holes (the remains of exploded stars), separated by a distance of
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The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

The given parameters;

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  • <em>gravitational force between the two black holes, F = 6.9 x 10²⁵ N.</em>
  • <em>combined mass of the two black holes = 5.20 x 10³⁰ kg</em>

The product of the two masses is calculated from Newton's law of universal gravitational as follows;

F = \frac{Gm_1m_2}{r^2} \\\\m_1m_2 = \frac{Fr^2}{G} \\\\m_1m_2 = \frac{(6.9\times 10^{25}) \times (1.5\times 10^{12})^2}{6.67\times 10^{-11}} \\\\m_1m_2 = 2.328 \times 10^{60} \ kg^2

The sum of the two masses is given as;

m₁ + m₂ = 5.2 x 10³⁰ kg

m₂ = 5.2 x 10³⁰ kg - m₁

The first mass is calculated as follows;

m₁(5.2 x 10³⁰ - m₁) = 2.328 x 10⁶⁰

5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰

m₁² - 5.2 x 10³⁰m₁  + 2.328 x 10⁶⁰ = 0

<em>solve the quadratic equation using formula method</em>;

a = 1, b =-  5.2 x 10³⁰, c = 2.328 x 10⁶⁰

m_1 = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\m_1 = \frac{-(-5.2\times 10^{20})  \ \ +/- \ \ \sqrt{(-5.2\times 10^{20})^2 - 4(1\times 2.328\times 10^{60})} }{2(1)} \\\\m_1 = 4.7 \times 10^{30} \ kg \ \ or \ \ 4.9 \times 10^{29} \ kg

The second mass is calculated as follows;

m₂ = 5.2 x 10³⁰ kg - m₁

m₂ = 5.2 x 10³⁰ kg  -  4.7 x 10³⁰ kg

m₂ = 5 x 10²⁹ kg

or

m₂ = 5.2 x 10³⁰ kg  -  4.9 x 10²⁹ kg

m₂ = 4.7 x 10³⁰ kg

Thus, the largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

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