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Kisachek [45]
3 years ago
12

. A little car has a maximum acceleration of 2.57 m/s2. What is the new maximum acceleration of the little car if it tows anothe

r car that has the same mass?
Physics
1 answer:
valkas [14]3 years ago
8 0

Answer:

a'=1.285\ m/s^2

Explanation:

Let m be the mass of a little car and m' be the mass of another car.

We know that,

Force = mass × acceleration

ATQ,

m × a = 2m × a'

a = 2 × a'

a'=\dfrac{a}{2}\\\\a'=\dfrac{2.57}{2}\\\\a'=1.285\ m/s^2

So, the acceleration of another little car is equal to 1.285\ m/s^2.

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5. A 500 kg satellite is in a circular orbit at an altitude of 500 km above the Earth's surface. Because of air friction, the sa
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Answer:

the energy transformed into internal energy is E fr = 1.57*10¹⁰ J

Explanation:

From energy conservation , the internal energy gained by friction should be equal to the loss of total energy of the satellite

Since the total energy of the satellite can be decomposed into kinetic and potential energy

E fr = ΔE = E₂-E₁

E₁=  K₁+V₁ = 1/2*m*v₁² + m*g*h₁

E₂ = K₂+V₂=  1/2*m*v₂² + m*g*h₂

first, we can choose our reference state so that h₂=0 and h₁=h=500 km

second , we can calculate the approximate the inicial velocity as the velocity required for a stable circular orbit

g = v₁²/(h+R) → v₁² = g*(h+R)

as the velocity diminishes, h diminishes, falling into the earth

assuming the radius of the Earth as R= 6371 km , then

v₁² = g*(h+R) = 9.8 m/s² * (500 km+ 6371 km) *1000 m/km = 6.73 * 10⁷ (m/s)²

replacing values

E₁ = 1/2*m*v₁² + m*g*h₁ = 1/2* 500kg *6.73 * 10⁷ (m/s)² + 500kg* 9.8m/s² * 500 km = 1.67*10¹⁰ J

E₂= 1/2*m*v₂² + m*g*h₂ =  1/2* 500kg *(2000 m/s)² + 0 = 1*10⁹ J

therefore

E fr = ΔE = E₂-E₁ = 1.67*10¹⁰ J - 1*10⁹ J = 1.57*10¹⁰ J  

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