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maw [93]
3 years ago
9

How many sig figs does 0.009 have?

Physics
1 answer:
VLD [36.1K]3 years ago
4 0
There are 6 sig figs
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What type of motion is shown with this graph? (5 points)
Vladimir [108]
It should be Constant speed. The line goes straight & doesn’t change within the graph.
7 0
3 years ago
Find the position of the center of mass of the system of the sun and Jupiter? (Since Jupiter is more massive than the rest of th
8090 [49]

Answer:

r_{cm} = 0.074 m from the position of the center of the Sun

Explanation:

As we know that mass of Sun and Jupiter is given as

M_s = 1.98 \times 10^{30} kg

M_j = 1.89 \times 10^{27} kg

distance between Sun and Jupiter is given as

r = 7.78 \times 10^{11} m

now let the position of Sun is origin and position of Jupiter is given at the position same as the distance between them

so we will have

r_{cm} = \frac{M_s r_1 + M_j r_2}{M_s + M_j}

r_{cm} = \frac{1.98 \times 10^{30} (0) + (1.89 \times 10^{27})(7.78 \times 10^{11})}{1.98 \times 10^{30} + 1.89 \times 10^{27}}

r_{cm} = 0.074 m from the position of the center of the Sun

3 0
3 years ago
Assume that the average mass of each of the approximately 1 billion people in China is 55 kg.Assume that they all gather in one
Harman [31]

Answer: 5.9(10)^{-8} m

Explanation:

The equation to calculate the center of mass C_{M} of a particle system is:

C_{M}=\frac{m_{1}r_{1}+m_{1}r_{1}+...+m_{n}r_{n}}{m_{1}+m_{2}+...+m_{n}}

In this case we can arrange for one dimension, assuming the geometric center of the Earth and the ladder are on a line, and assuming original center of mass located at the Earth's geometric center:

C_{M}=\frac{m_{E}(0 m) + m_{p} r_{E-p}}{m_{E}+m_{p}}

Where:

m_{E}=5.9(10)^{24} kg is the mass of the Earth

m_{p}=55(10)^{9} kg is the mass of 1 billion people

r_{E}=6371000 m is the radius of the Earth

r_{E-p}=6371000 m- 2m=6370998 m is the distance between the center of the Earth and the position of the people (2 m above the Earth's surface)

C_{M}=\frac{m_{p}55(10)^{9} kg (6370998 m)}{5.9(10)^{24} kg+55(10)^{9} kg}

C_{M}=5.9(10)^{-8} m This is the displacement of Earth's center of mass from the original center.

8 0
3 years ago
A kayaker needs to paddle north across a 100-m-wide harbor. The tide is going out, creating a tidal current that flows to the ea
Savatey [412]

Answer:

41.81^{\circ}

Explanation:

The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.

Let Vector \overrightarrow{OA} is the tidal current velocity as shown in the diagram.

In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, \vec {R} must be in the north direction.

Let \overrightarrow{AB} is the speed of the kayaker having angle \theta measured north of east as shown in the figure.

For the resultant velocity in the north direction, the tail of the vector \overrightarrow {OA} and head of the vector \overrightarrow{AB} must lie on the north-south line.

Now, for this condition, from the triangle OAB

|\overrightarrow{AB}|\sin \theta=|\overrightarrow{OA}|

\Rightarrow \sin\theta=\frac{|\overrightarrow{OA}|}{|\overrightarrow{AB}|}=\frac 2 3

\Rightarrow \theta=\sin^{-1}\frac23

\Rightarrow \theta=41.81^{\circ}

Hence, the kayaker must paddle in the direction of 41.81^{\circ}  in the north of east direction.

3 0
3 years ago
What would be the orbital speed and period of a satellite in orbit 0.82 ✕ 108 m above the earth?
Ivenika [448]
The answer should be 88.56
3 0
3 years ago
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