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Luden [163]
3 years ago
8

For an object to orbit another object

Physics
1 answer:
STatiana [176]3 years ago
8 0
B. Inertia and gravity must be balanced.

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Which of the followijg is an example of kinetic energy
Ipatiy [6.2K]

Answer:

D

Explanation:

D. A swing moving back and forth

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3 years ago
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The ball and socket joint in the hip is similar to the ball and socket joint of the shoulder but is designed more for __________
Viktor [21]

Answer:

a. power

b. precision

Explanation:

The hip is a spherical joint that allows the upper leg to move from front to back and from side to side. The largest joint that supports weight in the body, the hip joint is surrounded by strong ligaments and muscles.

It differs mainly with the circular articulation of the shoulder since it allows it to perform movement of greater power and with greater precision

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3 years ago
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Calculate the speed of the ball, vo in m/s, just after the launch. A bowling ball of mass m = 1.5 kg is launched from a spring c
klemol [59]

Answer:

v_0=17.3m/s

Explanation:

In this problem we have three important moments; the instant in which the ball is released (1), the instant in which the ball starts to fly freely (2) and the instant in which has its maximum height (3). From the conservation of mechanical energy, the total energy in each moment has to be the same. In (1), it is only elastic potential energy; in (2) and (3) are both gravitational potential energy and kinetic energy. Writing this and substituting by known values, we obtain:

E_1=E_2=E_3\\\\U_e_1=U_g_2+K_2=U_g_3+K_3\\\\\frac{1}{2}kd^2=mg(d\sin\theta)+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}m(v_0\cos\theta)^2

Since we only care about the velocity v_0, we can keep only the second and third parts of the equation and solve:

mgd\sin\theta+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}mv_0^2\cos^2\theta\\\\\frac{1}{2}mv_0^2(1-\cos^2\theta)=mg(h-d\sin\theta)\\\\v_0=\sqrt{\frac{2g(h-d\sin\theta)}{1-\cos^2\theta}}\\\\v_0=\sqrt{\frac{2(9.8m/s^2)(4.4m-(0.21m)\sin32\°)}{1-\cos^232\°}}\\\\v_0=17.3m/s

So, the speed of the ball just after the launch is 17.3m/s.

4 0
3 years ago
Energy that comes from the heat of the Earth's core is known as:
cestrela7 [59]
Geothermal energy is earths natural heat
3 0
2 years ago
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Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass
Arlecino [84]

Answer: The rank of the items is E,B,F,A,D,C..

Explanation:      

In the given problem, The satellites fire rockets that provide the force needed to maintain a circular orbit around the space station. The centripetal force is acting to keep up the continuous circular motion.

The expression for the centripetal force is as follows;

F=\frac{m\times v^{2}}{r}

Here, m is the mass of the object, v is the velocity and r is the radius.

In the given problem, the orbital period is constant.

r\propto v

Thus,

F\propto m\times r

In the given problem,  each satellite has mass m and radius of orbit L.

r=L+R

Calculate the net force acting on each satellite from their rockets for the part A.

mr=m(L+R)

Put m= 200 kg, R=6378 km and L=5000 m.

mr=200(5000+6378)

mr=1.28\times 10^{9}kg\times m

Calculate the net force acting on each satellite from their rockets for the part B.

mr=m(L+R)

Put m= 400 kg, R=6378 km and L=2500 m.

mr=400(2500+6378)

mr=2.55\times 10^{9}kg\times m  

Calculate the net force acting on each satellite from their rockets for the part C.

mr=m(L+R)

Put m= 100 kg, R=6378 km and L=2500 m.

mr=100(2500+6378)

mr=6.38\times 10^{8}kg\times m

Calculate the net force acting on each satellite from their rockets for the part D.

mr=m(L+R)

Put m= 100 kg, R=6378 km and L=10000 m.

mr=100(10000+6378)

mr=6.39\times 10^{8}kg\times m

Calculate the net force acting on each satellite from their rockets for the part E.

mr=m(L+R)

Put m= 800 kg, R=6378 km and L=5000 m.

mr=800(5000+6378)

mr=5.11\times 10^{9}kg\times m

Calculate the net force acting on each satellite from their rockets for the part F.

mr=m(L+R)

Put m= 300 kg, R=6378 km and L=7500 m.

mr=300(7500+6378)

mr=1.92\times 10^{9}kg\times m

Therefore, the rank of the items is E,B,F,A,D,C.

5 0
3 years ago
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