The word focus means ______

Water has a mass per mole of 18.0 g/mol, and each water molecule (H20) has 10 electrons. (a) How many electrons are there in one

Allushta [10]

Answer:

total number of electron in 1 litter is 3.34 × $10^{26}$ 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 × $10^{23}$ 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 × $10^{23}$ × 1

total number of electron in water is = 6.02× $10^{24}$ electron

and

we know

mass = density × volume ..........1

here we know density of water is 1000 kg/m

and volume = 1 litter = 1 × $10^{-3}$ m³

mass of 1 litter = 1000 × 1 × $10^{-3}$

mass = 1000 g

so

total number of electron in 1 litter = mass of 1 litter × $\frac{molecules per gram mole}{mass per mole}$

total number of electron in 1 litter = 1000 × $\frac{6.02*10{24}}{18}$

total number of electron in 1 litter is 3.34 × $10^{26}$ electron

8 0

3 years ago

A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just

bagirrra123 [75]

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.

5 0

3 years ago

a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien

Law Incorporation [45]

Answer:

The magnitude of the friction force is 8197.60 N

Explanation:

Using the definition of the centripetal force we have:

$\Sigma F=ma_{c}=m\frac{v^{2}}{R}$

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:

$F_{f}=m\frac{v^{2}}{R}$

$F_{f}=2100\frac{18^{2}}{83}$

$F_{f}=8197.60 \: N$

Therefore the magnitude of the friction force is 8197.60 N

I hope it helps you!

7 0

3 years ago

Observe the given figure and find the the gravitational force between m1 and m2.

Leno4ka [110]

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;

$F =G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}$

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;

$F = 6.67430 \times 10^{-11} \ N \cdot m^2/kg \times \dfrac{20 \ kg\times 50 \ kg}{(0.1 \ m+ 0.15 \ m)^{2}} \approx 1.06789 \times 10^{-6} \ N$

The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N

8 0

2 years ago

A 2.00-m rod of negligible mass connects two very small objects at its ends. The mass of one object is 1.00 kg and the mass of t

8090 [49]

Answer:

Explanation:

Step one:

given

length of rod=2m

mass of object 1 m1=1kg

let the unknown mass be x

center of mass c.m= 1.6m

hence 1kg is 1.6m from the c.m

and x is 0.4m from the c.m

Taking moment about the c.m

clockwise moment equals anticlockwise moments

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

3 0

2 years ago

The word focus means _______.