The distance covered on the floor after leaving the ramp is the dependent variable.
- As a result of the marble's size, the substance it is constructed of, and the angle at which it is placed onto the ground, the distance it rolls varies.
- Therefore, the angle at which the marble is released onto the ground, the type of material used to make the stone, or its size can all be considered independent variables.
<h3>What is Independent variable?</h3>
- There are independent and dependent variables in every experiment.
- A variable is considered independent if its change is not influenced by the change in another variable or factor.
<h3>What is Dependent variable?</h3>
In any experiment, the dependent variable must be measured or determined, and it must change as the independent variable does.
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Answer:
True
Explanation:
This is a representation of Gauss law.
Gauss’s law does hold for moving charges, and in this respect Gauss’s law is more general than Coulomb’s law. In words, Gauss’s law states that: The net outward normal electric flux through any closed surface is proportional to the total electric charge enclosed within that closed surface. The law can be expressed mathematically using vector calculus in integral form and differential form, both are equivalent since they are related by the divergence theorem, also called Gauss’s theorem.
The answer is false because there are single celled organisms.
Here observer is moving along with an electrical charge
now magnetic field on a moving charge is given by equation
electrical force on a moving charge is given by formula
so here we can say when observer is moving with the charge it will have two forces on it
1. magnetic force due to existing magnetic field
2. electric force due to electric field
so here he must have to experience both type of field presence
So here correct answer must be
<em>C- electric and magnetic fields </em>
Answer:
Explanation:
The distance travelled in the free fall is H - h
Since the apple originally started from rest we can use v^2 = u^2 + 2 x g x s where v is the final velocity, g the accln due to gravity and s the distance travelled and u is the initial velocity = 0
So the velocity just before it enters the grass is sq rt [2 x g x (H - h)]
Once in the grass, it slows down at a constant rate which means that the acceleration (a) during this period is constant.
So once again using the same formula we have v = O and u = sq rt[2 x g x (H-h)]
so since v^2 = u^2 + 2 x a x s then
O^2 = 2 x g x (H-h) + 2 x a x h
{O^2 - 2 x g x (H - h)}/(2 x h) = a