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KiRa [710]
3 years ago
13

Two planetoids with radii R1 and R2 and local accelerations g1 and g2 are separated by a centre-to-centre distance D, and are me

mbers of a simple two-body system. A rocket is located on planetoid 1 and is scheduled to launch from this planetoid to a point between the planetoids where the net gravitational force on the rocket by the two planetoids is zero. At what distance from the centre of planetoid 1 is the zero gravitational point? Choose R1 = 1400m, R2 = 1000m, g1 = 7.5m/s2, g2 = 5.3m/s2 and D = 4800m. [5 points]
Physics
1 answer:
balu736 [363]3 years ago
5 0

Answer:

Therefore, the gravitational zero points between two planetoids lie at a distance of 3000 m from the center of planetoid 1.

Explanation:

From Newton’s gravitation formula, the expression of the mass (M) of the planet of radius R is given as,

F_{G} = mg_{1}\\ \left ( \frac{GMm}{{R_{1}}^{2}} \right )=mg_{1}\\\\M= \frac{gR^{2}}{G}\rightarrow \left ( 1 \right )

Let's take x to be the distance of the zero gravitational points from the center of the planetoid 1.

Thus, the distance of the zero gravitational points from the center of the planetoid 2 is (D-x).

At zero gravitational point, the gravitational force between the planets and the rocket must be equal.

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A space vehicle accelerates uniformly from 85 m/s at t = 0 to 164 m/s at t = 10.0 s .How far did it move between t = 2.0 s and t
creativ13 [48]

First, we have a change in the velocity from 85 to 164 m/s in 10 sec.

Then, we calculate the <u>acceleration </u>as:

a=\frac{v_{f}-v_{i} }{t} =\frac{164-85}{10}=7.9 m/s^2

Hence we need to calculate the velocity of the space vehicle at t = 2 sec using the first equation of motion:

v_{f}=v_{i}+at=85+7.9*2=100.8m/s

Then, using the second equation of motion to calculate the distance:

d=v_{i}  t+\frac{1}{2}at^2

d=100.8*2+\frac{1}{2}*7.9*(2)^2=217.4m

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4 years ago
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julia-pushkina [17]

Answer:

The horizontal distance covered by the firework will be \frac{1876.8}{g}

Explanation:

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v_{y}=u_{y}+(-g) t

u_{y}= u\sin \phi

and v_{y}=0

therefore \frac{u\sin \phi }{g} =t

writing equation of motion in horizontal direction:

s_{x}=u_{x}t

u_{x} = u\cos \phi

therefore the equation becomes s_{x}=\frac{u^{2}   \sin \phi  \cos \phi}{g}

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DiKsa [7]

Answer:

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\theta=75^o

Explanation:

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\vec F_n=m\vec a

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N

The components of the first force are

\vec F_1=

\vec F_1=\ N

The components of the second force are

\vec F_2=

\vec F_2=\ N

The net force is

\vec F_n=

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The magnitude of the net force is

|F_n|=\sqrt{14.64^2+54.64^2}

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The acceleration has a magnitude of

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\displaystyle |a|=\frac{56.57}{20}

|a|=2.83\ m/s^2

The direction of the acceleration is the same as the net force:

\displaystyle tan\theta=\frac{54.64}{14.64}

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Answer:

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