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3 years ago

To collect quantitative data, scientists use all of the following except

1 answer:

8 0

"Quantitative" means "expressed in numbers". Quantitative data is information you collect by using measuring tools, and measuring things.

You can get it using rulers, scales, and stopwatches, but not by using descriptive categories.

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What is the formula for conservation of momentum

olga55 [171]

Answer:

The general equation for conservation of momentum during a collision between n number of objects is given as: [m i ×v i a ] = [m i ×v i b ] Where m i is the mass of object i , v i a is the velocity of object i before the collision, and v i b is the velocity of object i after the collision.

Explanation:

6 0

2 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 wave front has the form of a

podryga [215]

Your answer is "surface of a sphere"

Hope this helps.

3 0

3 years ago

Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32. 4 cm. If he measures the fr

sineoko [7]

The true statement about the wave is that, the wave has traveled 97. 2 cm in 1 second.

In Physics, we define a wave as a disturbance along a medium that transfers energy. The wavelength of a wave is the distance covered by the wave while the frequency of the wave is the number of cycles of the wave completed per second.

The period of the wave is the inverse of the frequency of the wave. It is defined as the time taken for the wave to complete a cycle and it is measured in seconds.

The wave formula is given as;

v = λf

v = velocity of the wave (distance traveled by the wave in one second)

λ = wavelength of the wave

f = frequency of the wave

So;

λ = 32.4 cm

f = 3 hertz

v = 32.4 cm × 3 hertz

v = 97. 2 cms-1

Hence, the true statement about the wave is that, the wave has traveled 97. 2 cm in 1 second.

Learn more: brainly.com/question/14588679

7 0

2 years ago

What is one standard kilogramun si system

Phoenix [80]

Answer:

The kilogram (kg) is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015 ×10−34 when expressed in the unit J s, which is equal to kg m2 s−1, where the meter and the second are defined in terms of c and ∆νCs.

3 0

3 years ago

Read 2 more answers