How many sig figs does 0.009 have?

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.