Answer:
They are known as isotopes
Answer:
10250 N/C leftwards
Explanation:
QA = 4 micro Coulomb
QB = - 5 micro Coulomb
AP = 6 m
BP = 2 m
A is origin, B is at 4 m and P is at 6 m .
The electric field due to charge QA at P is EA rightwards

The electric field due to charge QB at P is EB leftwards

The resultant electric field at P due the charges is given by
E = EB - EA
E = 11250 - 1000 = 10250 N/C leftwards
There are different options here but all of them work by approximating and assuming.
i) that the boulder is above ground.
ii) that the bottom surface of the boulder is known.
iii) the shape of the boulder is taken into account.
The most accurate way is measuring it by displacement method but the boulder is immovable hence the volume can be calculated by measuring the boulder or a waterproof box to be built around the boulder and calculate the volume occupied by boulder.
All the above methods are estimating methods.
*Another way to find the density is through specific gravity.
S.G = <u>Density</u><u> </u><u>of</u><u> </u><u>object</u>
Density of water
If the material that makes the boulder is known that is if it's stone or a mineral then the specific gravity can be found.
If the boulder is purely rock then S.G lies between 3 - 3.5 and the density of water is known thus the density of the boulder can be found without moving the boulder.
This is what I think after correction and allthe best!
Answer:6 joules
Explanation:
Mass(m)=3kg
Velocity(v)=2m/s
Kinetic energy=0.5 x m x v^2
Kinetic energy=0.5 x 3 x 2^2
Kinetic energy=0.5 x 3 x 2 x 2
Kinetic energy=6
Answer:
a) v = 19,149.6 m/s
b) f = 95%
c) t = 346.5min
Explanation:
First put all values in metric units:

The equation of motion you need is:
where
is the final velocity, a is acceleration and t is time in hours.
Since the spaceship starts from 0 velocity:

Next, you need to calculate the distances traveled on each interval, considering that both starting and final intervals travel the same distance because the acceleration and time are equal. For this part you need the next motion equation:

solving for first and last interval:
Since the spaceship starts and finish with 0 velocity:

Then the ship traveled
at constant speed, which means that it traveled:

Which in percentage is 95% of the trip.
to calculate total time you need to calculate the time used during constant speed:

That added to the other interval times:
