1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Gnom [1K]
3 years ago
7

As mentioned before, our asteroid is in the shape of a sphere and has a mass of 1000 kilograms. Determine the density (in grams

per cubic centimeter) of this asteroid if its diameter is known to be 1.2 meters. Useful information: 1 kg = 1000 g, 1 m = 100 cm, volume of sphere = 4/3 ? r3. Remember that the radius of a sphere is equal to half its diameter. Show all of your work. (20 points)
Physics
1 answer:
notsponge [240]3 years ago
4 0

Answer: 1.1052g/cm^{3}

Explanation:

Density D is a characteristic property of a material and is defined as the relationship between the mass m and volume V of a specific substance or material. So, the density of the asteroid is given by the following equation:

D=\frac{m}{V}   (1)

On the other hand, we know the asteroid has a mass m=1000kg and is spherical. This means its volume is given by the following formula:

V=\frac{4}{3}}\pi r^{3}   (2)

Where r=\frac{d}{2}=\frac{1.2m}{2}=0.6m  is the radius of the sphere and is half its diameter d.

Knowing this, we can calculate the volume:

V=\frac{4}{3}}\pi (0.6m)^{3}   (3)

V=0.904m^{3}   (4)

Substituting (4) in (1):

D=\frac{1000kg}{0.904m^{3}}=1105.242\frac{kg}{m^{3}}   (5) This is the density of the asteroid, but we were asked to find it in \frac{g}{cm^{3}}. This means we have to make the conversion:

D=1105.242\frac{kg}{m^{3}}.\frac{1000g}{1kg}.\frac{1m^{3}}{(100cm)^{3}}

Finally:

D=1.1052\frac{g}{cm^{3}}

You might be interested in
Which angle (A, B, or C) is the diffraction angle?
lisov135 [29]
C is the diffraction angle.... step by step explanation= I think it’s that I might be wrong lol
8 0
3 years ago
Read 2 more answers
If 2 objects have the same momentum which statement is true
rewona [7]

The correct answer is B

8 0
3 years ago
Read 2 more answers
Ms. Stafford wants to try racing with a push start. A student pushes her at 5 m/s before the rocket kicks in. The rocket still o
Gnom [1K]
The initial velocity of Ms. Stafford is v_0 = 5 m/s, while her acceleration is 
a=4 m/s^2
This is a uniform accelerated motion, so we can calculate the total distance travelled by Ms. Stafford in a time of t=7.0 s using the law of motion for a uniform accelerated motion:
S=v_0t +  \frac{1}{2} at^2 = (5 m/s)(7.0 s)+ \frac{1}{2}(2 m/s^2)(7.0 s)^2 = 84 m
8 0
3 years ago
In 1993, Ileana Salvador of Italy walked 3.0 km in under 12.0 min. Suppose that during 115 m of her walk Salvador is observed to
dexar [7]

Answer:

t = 25 seconds

Explanation:

Given that,

Distance, d = 115 m

Initial speed, u = 4.2 m/s

Final speed, v = 5 m/s

We need to find the time taken in increasing the speed.

We know that,

Acceleration, a=\dfrac{v-u}{t} ....(1)

The third equation of kinematics is as follows :

v^2-u^2=2ad\\\\\text{Put the value of a in above equation}\\\\v^2-u^2=2\times \dfrac{v-u}{t}\times d\\\\\because (a^2-b^2)=(a-b)(a+b)\\\\(v-u)(v+u)=\dfrac{2\times (v-u)d}{t}\\\\t=\dfrac{2d}{v+u}\\\\\text{Putting all the values}\\\\t=\dfrac{2\times 115}{4.2+5}\\\\t=25\ s

Hence, it will take 25 seconds to increase the speed.

6 0
3 years ago
An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.
CaHeK987 [17]

A) 11.1 cm

We can find the focal length of the lens by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p = 16.0 cm is the distance of the object from the lens

q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)

Solving the equation for f:

\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.09 cm^{-1}\\f=\frac{1}{0.09 cm^{-1}}=11.1 cm

B) Converging

The focal length is:

- Positive for a converging lens

- Negative for a diverging lens

In this case, the focal length is positive, so it is a converging lens.

C) 18.0 mm

The magnification equation states that:

\frac{h_i}{h_o}=-\frac{q}{p}

where

h_i is the heigth of the image

h_o is the height of the object

q=36.0 cm

p=16.0 cm

Solving the formula for h_i, we find

h_i = -h_o \frac{q}{p}=-(8.00 mm)\frac{36.0 cm}{16.0 cm}=-18.0 mm

So the image is 18 mm high.

D) Inverted

From the magnification equation we have that:

- When the sign of h_i is positive, the image is erect

- When the sign of h_i is negative, the image is inverted

In this case, h_i is negative, so the image is inverted.

4 0
3 years ago
Other questions:
  • Convert 100dyne into joule​
    11·1 answer
  • What did Issac Newton think he could describe?
    13·2 answers
  • What is the awnser to Blank occurs when a substance changes from a liquid to a gas
    7·1 answer
  • Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3>s. (a) How fast will it shoot out of a hole 4
    10·1 answer
  • A wave is incident on the surface of a mirror at an angle of 41° with the normal. What can you say about its angle of reflection
    6·2 answers
  • Who was the first president of the united states​
    12·2 answers
  • I NEED THIS ASAP I WILL MARK YOU THE BRAINLIEST NO LINKS !!!!
    8·2 answers
  • A man has been guarding a house for one hour . Why is it not considered work in science
    9·1 answer
  • The following data were collected during a short race between two friends. Velocity (m/s) 0 0.5 1 1.5 2 2 4 6 2 0 Time (s) 0 2 4
    11·1 answer
  • Astronaut X of mass 50kg floats next to Astronaut Y of mass 100kg while in space, as shown in the figure. The positive direction
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!