What does cumulative and quantitative mean?

A car travels straight for 20 miles on a road that is 30Â° north of east. What is the east component of the carâ€™s displacement

Over [174]

The east component of the cars displacement is 17.3 miles.

Trigonometric ratio is used to show the relationship between the sides of a right angled triangle and its angles.

Let x represent the east component of the cars displacement.

Using trigonometric ratio:

cos(30) = x / 20

x = 20 * cos(30)

x = 17.3 miles

The east component of the cars displacement is 17.3 miles.

Find out more on Trigonometric ratio at: brainly.com/question/1201366

4 0

2 years ago

3. A cat walks 0.220km North, then 0. 120 km South in a time of 400 seconds. whats the displacement and average velocity?

Andru [333]

Answer:

The rate at which velocity changes with respect to a change in time is called. acceleration.

Explanation:

4 0

3 years ago

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

a) 6738.27 J

b) 61.908 J

c) $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = 6738.27 J

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

$(I_{1} +I_{2} )w$

where the subscripts 1 and 2 indicates the values first and second flywheels

$(I_{1} +I_{2} )w$ = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = $Iw^{2}$ = 6.655 x $3.05^{2}$ = 61.908 J

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = $\frac{1}{2}mv_{car} ^{2}$

where m is the mass of the car

$v_{car}$ is the velocity of the car

Equating the energy

2246.09 = $\frac{1}{2}mv_{car} ^{2}$

making m the subject of the formula

mass of the car m = $\frac{4492.18}{v_{car} ^{2} }$

3 0

3 years ago

A nickel (5 cent coin) has a mass of 5.0 g. How many nickels are there in a stack of nickels with a mass of 10.0 kg?

Marrrta [24]

Answer:

2000 nickels

Explanation:

One way to solve proportionality problems, direct and inverse: the simple 3 rule.

If the relationship between the magnitudes is direct (when one magnitude increases so does the other), the simple direct rule of three must be applied.

On the contrary, if the relationship between the magnitudes is inverse (when one magnitude increases the other decreases) the rule of three simple inverse applies.

The simple 3 rule is an operation that helps us quickly solve proportionality problems, both direct and inverse.

To make a simple rule of three we need 3 data: two magnitudes proportional to each other, and a third magnitude. From these, we will find out the fourth term of proportionality.

In the simple three rule, therefore, the proportionality relationship between two known values A and B is established, and knowing a third value C, a fourth value D is calculated.

A -> B

C -> D

Calculation

1 nickel --> 5 g

X? nickel --> 10000g

X = (10000 g * 1 nickel) / 5 g

X = 2000 nickels

7 0

3 years ago

After which action would the concentration of a solution remain constant?(1 point)

Alika [10]

Answer:

C. Evaporating water from the container.

Explanation:

The concentration of solution changes when solvent or solute are added/removed from a solution.

5 0

3 years ago