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raketka [301]
3 years ago
8

An asteroid revolves around the Sun with a mean orbital radius twice that of Earth’s. Predict the period of the asteroid in Eart

h years.
help
Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
5 0

Answer:

8 years

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between square of the orbital period and the cube of the orbital radius of any object revolving around the Sun is constant:

\frac{T_e^2}{r_e^3}=\frac{T_a^2}{r_a^3}

where:

T_e = 1 y is the orbital period of Earth

r_e is the orbital radius of Earth

r_a=2 r_e is the orbital radius of the asteroid (which is twice that of Earth)

T_a = ? is the orbital period of the asteroid

Substituting and re-arranging the equation, we find

T_a^2 = \frac{r_a^3}{r_e^3} T_e^2 = \frac{(2 r_e)^2}{r_e^3} T_e^2=8 \frac{r_e^3}{r_e^3} T_e^2=8 T_e^2 = 8 (1y)^2=8 y

so, the orbital period of the asteroid will be 8 years.

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An automobile engine delivers 47.4 hp. how much time will it take for the engine to do 6.82 × 105 j of work? one horsepower is e
kogti [31]
1 horsepower is equal to 746 W, so the power of the engine is
P=47.4 hp \cdot 746  \frac{W}{hp}=35360 W
The power is also defined as the energy E per unit of time t:
P= \frac{E}{t}
Where the energy corresponds to the work done by the engine, which is E=6.82 \cdot 10^5 J. Re-arranging the formula, we can calculate the time t needed to do this amount of work:
t =  \frac{E}{P}= \frac{6.82 \cdot 10^5 J}{35360 W}=19.3 s
8 0
3 years ago
Robert hook used an early microscope to observed a cork sample. How did this help contribute to cell theory?
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Robert Hooke used an early microscope to observe a cork sample. How did this help contribute to cell theory? It helped to show that cells contain water. ... It helped to show that some cells are visible to the naked eye.

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3 0
3 years ago
Find the tension in the two ropes that are holding the 4.2 kg object in place.
Blababa [14]

Answer:

The tension in the two ropes are;

T1 = 23.37N T2 = 35.47N

Explanation:

Given mass of the object to be 4.2kg, the weight acting on the bag will be W= mass × acceleration due to gravity

W = 4.2×10 = 42N

The tension acting on the bag plus the weight are three forces acting on the bag. We need to find tension in the two ropes that will keep the object in equilibrium.

Using triangular law of force and sine rule to get the tension we have;

If rope 1 is at 57.6° with respect to the vertical and rope 2 is at 33.8° with respect to the vertical, our sine rule formula will give;

T1/sin33.8° = T2/sin57.6° = 42/sin{180-(33.8°+57.6°)}

T1/sin33.8° = T2/sin57.6° = 42/sin88.6°

From the equality;

T1/sin33.8° = 42/sin88.6°

T1 = sin33.8°×42/sin88.6°

T1 = 23.37N

To get T2,

T2/sin57.6°= 42/sin88.6°

T2 = sin57.6°×42/sin88.6°

T2 = 35.47N

Note: Check attachment for diagram.

7 0
3 years ago
A toy car having mass m = 1.50 kg collides inelastically with a toy train of mass M = 3.60 kg. Before the collision, the toy tra
baherus [9]

Answer:

  v_{f} = 3,126 m / s

Explanation:

In a crash exercise the moment is conserved, for this a system formed by all the bodies before and after the crash is defined, so that the forces involved have been internalized.

the car has a mass of m = 1.50 kg a speed of v1 = 4.758 m / s and the mass of the train is M = 3.60 kg and its speed v2 = 2.45 m / s

Before the crash

    p₀ = m v₁₀ + M v₂₀

After the inelastic shock

    p_{f}= m v_{1f} + M v_{2f}

    p₀ =  p_{f}

    m v₀ + M v₂₀ = m v_{1f}  + M v_{2f}

We cleared the end of the train

     M v_{2f} = m (v₁₀ - v1f) + M v₂₀

Let's calculate

     3.60 v2f = 1.50 (2.15-4.75) + 3.60 2.45

     v_{2f}  = (-3.9 + 8.82) /3.60

      v_{2f}  = 1.36 m / s

As we can see, this speed is lower than the speed of the car, so the two bodies are joined

set speed must be

      m v₁₀ + M v₂₀ = (m + M) v_{f}

      v_{f}  = (m v₁₀ + M v₂₀) / (m + M)

      v_{f}  = (1.50 4.75 + 3.60 2.45) /(1.50 + 3.60)

      v_{f} = 3,126 m / s

4 0
3 years ago
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