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2 years ago

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Every morning Ann walks her dog through the park, shown as a green square on the diagram below. They start at point 1, walk one

block up the street, take a turn at the corner labeled 2, and walk diagonally through the park to point 3. To return home, they walk two blocks down the street and turn right at the corner labeled 4. Draw the path 1→2→3→4→1 taken by Ann as she walks her dog. Represent each segment of Anna's walk with a vector.

1 answer:

bonufazy [111]2 years ago

7 0

What diagram. Edit answer and add diagram

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In 2014, the Rosetta space probe reached the comet Churyumov Gerasimenko. Although the comet's core is actually far from spheric

Viktor [21]

To solve this problem we will apply the concepts related to gravity according to the Newtonian definitions. From finding this value we will use the linear motion kinematic equations to find the time. Our values are

Comet mass $M = 1.0*10^{13} kg$

Radius $r = 1.6km = 1600 m$

Rock was dropped from a height 'h' from surface = 1m

The relation for acceleration due to gravity of a body of mass 'm' with radius 'r' is

$g = \frac{GM}{R^2}$

Where G means gravitational universal constant and M the mass of the planet

$g = \frac{(6.67408*10^{-11})(1*10^{13})}{1600^2}$

$g = 2.607*10^{-4} m/s^2$

Now calculate the value of the time

$h = \frac{1}{2} gt^2$

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2(1)}{2.607*10^{-4}}}$

$t = 87.58s$

The time taken for the rock to reach the surface is t = 87.58s

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2 years ago

A blue puck has a velocity of 0i – 3j m/s and a mass of 4 kg. A gold puck has a velocity of 12i – 5j m/s and a mass of 6 kg. Wha

Mnenie [13.5K]

By definition, the kinetic energy is given by:
K = (1/2) * m * v ^ 2
where
m = mass
v = speed
We must then find the speed of both objects:
blue puck
v = root ((0) ^ 2 + (- 3) ^ 2) = 3
gold puck
v = root ((12) ^ 2 + (- 5) ^ 2) = 13
Then, the kinetic energy of the system will be:
K = (1/2) * m1 * v1 ^ 2 + (1/2) * m2 * v2 ^ 2
K = (1/2) * (4) * (3 ^ 2) + (1/2) * (6) * (13 ^ 2)
K = <span> 525</span> J
answer
The kinetic energy of the system is<span> <span>525 </span></span>J

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2 years ago

Which of the following statements cannot be supported by Kepler's laws of planetary motion?

gladu [14]

Answer:

The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.

Explanation:

Kepler's three laws are:

1) The orbits of the planets around the Sun are ellipses, with the Sun at one of the focii

2) A line connecting the Sun with each planet sweeps out equal areas in equal time intervals

3) The cube of the semi-major axis of the orbit of one planet is proportional to the square of its orbital period

There 3 laws help explaining the following statements:

- <em>A planet's distance from the sun will not be the same in six months. --> </em>using the 1st law. In fact, since the orbit is an ellipse (and not a circle), and the Sun is at one of the focii, the distance of the planet from the Sun keeps changing during the year.

-<em> A planet's speed as it moves around the sun will not be the same in six months. -</em>-> using the 2nd law. In fact, since the line connecting the Sun to the planet must cover equal areas in the same time interval, it follows that the speed of the planet cannot be constant during the year (it will be faster when closer to the sun and slower when far from the sun).

- <em>The average distance of Saturn can be calculated using the average distance of Neptune and the orbital period of both planets. </em>--> using the 3rd law. In fact, the ratio $\frac{a^3}{T^2}$ (where a is the semi-major axis of the orbit and T the orbital period) is constant and it is the same for every planet orbiting the sun, so by knowing the data of Neptune and the orbital period of Saturn, it is possible to calculate Saturn's average distance.

Instead, the following statement:

<em>The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.</em>

Is not supported by any Kepler's law.

8 0

2 years ago

During chemistry class, Carl performed several lab tests on two white solids. The results of three tests are seen in the data ta

Nastasia [14]

Answer: it’s c) ionic

Explanation:

6 0

2 years ago

Read 2 more answers

1. Each year at a college, there is a tradition of having a hoop rolling competition. Alex rolls his 0.350 kg hoop down the cour

grigory [225]

Question 1:

Answer:

The moment of inertia of Alex's rolling hoop is 0.197 $kg \cdot cm^2$

Explanation:

<u>Given</u>:

Mass of the hoop = 0.350 g

Radius of the hoop = 75.0 cm

<u>To Find:</u>

The moment of inertia of Alex's rolling hoop = ?

<u>Solution</u><u>:</u>

The moment of inertia = $mr^2$

where

m is the mass

r is the radius

Converting cm to m, we get

75.0 cm = 0.75 m

Now substituting the values,

=> moment of inertia = $(0.350)(0.75)^2$

=> moment of inertia = $(0.350)(0.5625)$

=> moment of inertia = $(0.197)$

Question 2:

Answer:

The combined angular momentum of the masses is 1.76 $kg m^2 s^{-1}$

If she pulls her arms in to 0.12 m, her new linear speed is $18.33 m/s^2$

Explanation:

Given:

Mass = 2.0 kg

Radius = 0.8 m

Velocity = 1.2 m/s

a.The combined angular momentum of the masses:

$L = r \cdot m \cdot v_1$

Substituting the values,

$L = 0.8 \cdot 2.0 \cdot 1.1$

L= 1.76 $kg m^2 s^{-1}$

b. If she pulls her arms in to 0.12 m, what is her new linear speed

$0.12 \cdot 0.8 \cdot v_2 = 1.76$

$0.096 cdot v_2 = 1.76$

$v_2 = \frac{1.76}{0.096}$

$v_2 = 18.33 m/s^2$

6 0

2 years ago