Answer:
total number of electron in 1 litter is 3.34 ×
electron
Explanation:
given data
mass per mole = 18 g/mol
no of electron = 10
to find out
how many electron in 1 liter of water
solution
we know molecules per gram mole is 6.02 ×
molecules
no of moles is 1
so
total number of electron in water is = no of electron ×molecules per gram mole × no of moles
total number of electron in water is = 10 × 6.02 ×
× 1
total number of electron in water is = 6.02×
electron
and
we know
mass = density × volume ..........1
here we know density of water is 1000 kg/m
and volume = 1 litter = 1 ×
m³
mass of 1 litter = 1000 × 1 × 
mass = 1000 g
so
total number of electron in 1 litter = mass of 1 litter × 
total number of electron in 1 litter = 1000 × 
total number of electron in 1 litter is 3.34 ×
electron
Answer: C
Frictional force
Explanation:
The description of the question above is an example of a circular motion.
For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.
Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.
Therefore, the correct answer is option C - the frictional force.
Answer:
<u>The magnitude of the friction force is 8197.60 N</u>
Explanation:
Using the definition of the centripetal force we have:

Where:
- m is the mass of the car
- v is the speed
- R is the radius of the curvature
Now, the force acting in the motion is just the friction force, so we have:
<u>Therefore the magnitude of the friction force is 8197.60 N</u>
I hope it helps you!
Answer:
The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N
Explanation:
The details of the given masses having gravitational attractive force between them are;
m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m
The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;

Where;
F = The gravitational force between m₁ and m₂
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
r₂ = 0.1 m + 0.15 m = 0.25 m
Therefore, we have;

The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N
Answer:
<h2> 4kg</h2>
Explanation:
Step one:
given
length of rod=2m
mass of object 1 m1=1kg
let the unknown mass be x
center of mass<em> c.m</em>= 1.6m
hence 1kg is 1.6m from the <em>c.m</em>
and x is 0.4m from the <em>c.m</em>
Taking moment about the <em>c.m</em>
<em>clockwise moment equals anticlockwise moments</em>
1*1.6=x*0.4
1.6=0.4x
divide both sides by 0.4 we have
x=1.6/0.4
x=4kg
The mass of the other object is 4kg