Give three examples of properties of elements

A 20-newton weight is attached to a spring.

Makovka662 [10]

Answer:

40 N/m

Explanation:

The diagram attached is used to answer the question

We know from Hooke's law that extension is directly proportional to the applied force hence

F=kx where x is extension, F is applied force and k is the spring constant. Making k the subject of the formula then

$k=\frac {F}{k}$

From the attached diagram extension is given by subtracting unstretched spring from stretched spring hence extension, x=1-0.5=0.5m

Substituting 20 N for F and 0.5 m for x then

$k=\frac {20}{0.5}=40 N/m$

8 0

3 years ago

Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.00 ✕ 105 kg on springs that ha

jasenka [17]

Explanation:

It is given that,

Mass of an object, $m=4\times 10^5\ kg$

(a) Time period of oscillation, T = 2.4 s

The formula for the time period of spring is given by :

$T=2\pi \sqrt{\dfrac{m}{k}}$

Where

k is the force constant

$k=\dfrac{4\pi ^2 m}{T^2}$

$k=\dfrac{4\pi ^2 \times 4\times 10^5}{(2.4)^2}$

$k=2.74\times 10^6\ N/m$

(b) Displacement in the spring, x = 2.2 m

Energy stored in the spring is given by :

$U=\dfrac{1}{2}kx^2$

$U=\dfrac{1}{2}\times 2.74\times 10^6\ N/m\times (2.2\ m)^2$

$U=6.63\times 10^6\ J$

Hence, this is the required solution.

7 0

3 years ago

Whenever two apollo astronauts were on the surface of the moon, a third astronaut orbited the moon. assume the orbit to be circu

almond37 [142]

Missing question:
"Determine (a) the astronaut’s orbital speed v and (b) the period of the orbit"

Solution

part a) The center of the orbit of the third astronaut is located at the center of the moon. This means that the radius of the orbit is the sum of the Moon's radius r0 and the altitude ( $h=430 km=4.3 \cdot 10^5 m$ ) of the orbit:
$r= r_0 + h=1.7 \cdot 10^6 m + 4.3 \cdot 10^5 m=2.13 \cdot 10^6 m$
This is a circular motion, where the centripetal acceleration is equal to the gravitational acceleration g at this altitude. The problem says that at this altitude, $g=1.08 m/s^2$ . So we can write
$g=a_c= \frac{v^2}{r}$
where $a_c$ is the centripetal acceleration and v is the speed of the astronaut. Re-arranging it we can find v:
$v= \sqrt{g r}= \sqrt{(1.08 m/s^2)(2.13 \cdot 10^6 m)}=1517 m/s = 1.52 km/s$

part b) The orbit has a circumference of $2 \pi r$ , and the astronaut is covering it at a speed equal to v. Therefore, the period of the orbit is
$T= \frac{2 \pi r}{v} = \frac{2\pi (2.13 \cdot 10^6 m)}{1517 m/s} =8818 s = 2.45 h$
So, the period of the orbit is 2.45 hours.

6 0

3 years ago

In which of the following would you find the highest population density?

marysya [2.9K]

I would think <em>B,</em> along the Amazon River. This is because of the warm climate ad lots of rainfall.

7 0

2 years ago

Read 2 more answers

A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the

makkiz [27]

Answer:

(a) She has traveled a total distance of 28958.04 m during the entire trip.

(b) Her average velocity for the trip is 6.12 m/s

Explanation:

Here is the complete question:

A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the first part, she rides for 26.8 minutes at an average speed of 7.57 m/s. During the second part, she rides for 42.4 minutes at an average speed of 3.17 m/s. Finally, during the third part, she rides for 9.69 minutes at an average speed of 15.0 m/s. (a) How far has the bicyclist traveled during the entire trip? (b) What is her average velocity for the trip?

Explanation:

(a) To determine how far the bicyclist has traveled during the entire trip, we will calculate the distance she covered in each part of the trip, and then sum up the distances to determine the total distance covered.

First, The distance covered in the first part of the trip

During the first part, she rides for 26.8 minutes at an average speed of 7.57 m/s

That is,

Average speed, $v$ = 7.57 m/s

and time, $t$ = 26.8 minutes

Convert the time to seconds

∴ $t$ = 26.8 minutes = (26.8 × 60) secs = 1608 secs

$Average speed = \frac{Distance }{ Time}$

Then, $Distance = Average speed$ × $Time$

Hence, for the first part

$Distance = 7.57$ × $1608$

$Distance = 12172.56$ m

This is the distance covered in the first part of the trip.

For the distance covered in the second part of the trip

During the second part, she rides for 42.4 minutes at an average speed of 3.17 m/s

That is, Average speed, $v$ = 3.17 m/s

and time, $t$ = 42.4 minutes

Convert the time to seconds

∴ $t$ = 42.4 minutes = (42.4 × 60) secs = 2544 secs

From,

$Distance = Average speed$ × $Time$

$Distance = 3.17$ × $2544$

$Distance = 8064.48$ m

This is the distance covered in the second part of the trip.

For the distance covered in the third part of the trip

During the third part, she rides for 9.69 minutes at an average speed of 15.0 m/s

That is, Average speed, $v$ = 15.0 m/s

and time, $t$ = 9.69 minutes

Convert the time to seconds

∴ $t$ = 9.69 minutes = (9.69 × 60) secs = 581.4 secs

From,

$Distance = Average speed$ × $Time$

$Distance = 15.0$ × $581.4$

$Distance = 8721$ m

This is the distance covered in the third part of the trip.

Now for the distance covered during the entire trip,

Total distance = distance covered in the first part of the trip + distance covered in the second part of the trip + distance covered in the third part of the trip

Hence,

Total distance = 12172 m + 8064.48 m + 8721 m

Total distance = 28958.04 m

Hence, she has traveled a total distance of 28958.04 m during the entire trip.

(b) For her average velocity for the trip

Average velocity is given by

$Average velocity = \frac{Total distance traveled}{Total time}$

Total distance traveled = 28958.04 m

Total time = 1608 secs + 2544 secs + 581.4 secs

Total time = 4733.4 secs

Hence,

$Average velocity = \frac{28958.04}{4733.4}$

Average velocity = 6.1178 m/s

Average velocity ≅ 6.12 m/s

7 0

3 years ago