Answer:
- I have fond the answer
Explanation:
but my camera doesn't work
Answer: To answer this question, we will need the following equation: SPEED = DISTANCE/TIME (A multiplication and division triangle will be shown)i) The speed of the car is calculated by doing 100 metres/ 20 seconds which gives us 5 metres per second. ii) Rearranging the equation earlier, we can make the distance the subject of the equation so that we get SPEED x TIME = DISTANCE. We worked out the speed and the time was given as 1 minute 40 seconds but we cannot plug in the numbers yet as the time has to be converted to units of seconds (because our speed is in meters per second). 1 minute 40 seconds = 60 seconds + 40 seconds = 100 secondsWe then plug in the numbers to get the distance travelled = 5 metres per second x 100 seconds = 500 metres.
Explanation:
Answer:
force is the answer because force is pushing the item
Answer:
Mass of the steel cube = 7800 kg
Volume of the steel = 1.025 cubic centimetre
Explanation:
Given:
The density of the steel = 7.8
Side of the cube = 12 cm
<u>(1)The mass of steel cube :</u>
We know that,

We are given with density and sides of the cube
then volume of the cube
=
= 
= 1000 cubic centimetre
Now


mass = 7800 kg
<u>(2)volume of steel:</u>
Given the mass = 8 kg

Substituting the values


volume = 1.025 cubic centimetre
Answer:
Explanation:
The volume of a sphere is:
V = 4/3 * π * a^3
The volume charge density would then be:
p = Q/V
p = 3*Q/(4 * π * a^3)
If the charge density depends on the radius:
p = f(r) = k * r
I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.





Since p = k*r
Q = p*π^2*r^3 / 2
Then:
p(r) = 2*Q / (π^2*r^3)