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3 years ago

What value is needed to create a motion diagram? a. position c. time interval b. velocity d. both a and c

1 answer:

5 0

For classical kinematics, you usually graph position versus time, meaning D. This would be a graph where time is on the x-axis and position (can also be treated as displacement if you consider the net distance) is on the y axis.

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Many counties in Florida missed many school days in the fall of 2004 due to hurricanes that year. A solution for how to make up

Ivenika [448]

Answer:

108 extended days

Explanation:

Regular school hours a day = 6 hr

No. of school days to make up by extending the regular hours = 3 days

Amount of time added to the regular hours of school = 10 min

No. of extended school days to make up the 3 school days by following the above mentioned criteria be x.

Time of school hours in 3 days = $3\times 6= 18\,hr\,\,\,\, ;or\,\,\, (18\times 60)\,minutes$

$\therefore x=\frac{18\times 60}{10}$

$x=108\,\,days$ are required to make up 3 days of school having 6 hours of regular timing with 10 minutes of add-on time each day.

4 0

3 years ago

In his novel From the Earth to the Moon (1866), Jules Verne describes a spaceship that is blasted out of 12,000 yards/s. the Col

saw5 [17]

Answer:

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

Explanation:

Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration ( $a$ ), measured in meters per square second, is estimated by this kinematic formula:

$a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s }$ (1)

Where:

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ , $v$ - Initial and final speeds of the spaceship, measured in meters.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v = 10968\,\frac{m}{s}$ and $\Delta s = 212.8\,m$ , then the acceleration experimented by the spaceship is:

$a = \frac{\left(10968\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (212.8\,m)}$

$a = 282652.782\,\frac{m}{s^{2}}$

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

5 0

3 years ago

A baseball hit just above the ground leaves the bat 27 m/s at 45° above the horizontal. A) How far away does the ball strike the

Sedbober [7]

Answer:

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

v = u + at

0 = 19.09 - 9.81 t

t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

S = ut + 0.5 at²

H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

Maximum height of the ball = 18.57 m

6 0

3 years ago

A student has designed a circuit with one battery, two light bulbs, and two switches, pictured to the left. She would like for e

Gnoma [55]

Answer:

The bulb is not powered because the negative end of the battery is not connected to the electric current,making it dysfunctional. The electrons have to flow through the positive and negative ends of the battery in order for it to work.

Hope this helps.

5 0

3 years ago

Read 2 more answers

At this rate, how long would it take for two continents 3500 kilometers apart to collide?

meriva

Given:

3,500 kilometers

Find:

Years for two continents to collide = ?

Solution:

We know that typical motions of one plate relative to another are 1 centimeter per year.

So first, we convert 3,500 km to cm.

The solution would be like this for this specific problem:

1 km = 100,000 cm

3,500 km x 100,000 = 350,000,000 cm

Since we know that 1 cm = 1 year, then that means 350,000,000 cm is equivalent to 350,000,000 years.

Therefore, it would take 350 million years for two continents that are 3500 kilometers apart to collide.

To add, a phenomenon of the plate tectonics of Earth that occurs at convergent boundaries is called the continental collision.

5 0

3 years ago