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1 year ago

The conservation of momentum is most closely related to

Physics

1 answer:

BartSMP [9]1 year ago

6 0

Answer:

<h2>B) Newton's 2nd law</h2>

Explanation:

<h2>From; force= mass × acceleration </h2><h2> f= m×a </h2><h2>where a(acceleration)= velocity/time</h2><h3> force = mv/t</h3><h3>But momentum(p) = Mass × velocity </h3><h2>hence force =p/t </h2><h3>that is Momentum = force × time ( Newton's 2nd law)</h3>

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A meteor moving 468 km per minute traveling in a south-to-north direction passed near Earth in 2013. Does this statement describ

Thepotemich [5.8K]

Answer:

this statement describes meteor's velocity,

because velocity is a vector quantity which has both magnitude as well as a specific direction and here the meteor's direction is specified in the statement hence we conclude that this statement describes meteor's velocity as well as speed too.

3 0

3 years ago

Calculate the potentia energy of a car with a mass of 3800kg that is on a hill 110 meters above sea level? USE 10 instead of 9.8

Arada [10]

Potential energy (PE ) = m g h

Where:

m = mass = 3800 kg

g = acceleration due gravity = 10 m/s^2

h = heigth = 110 meters

Replacing:

PE = 3800 * 10 * 110 = 4,180,000 J

7 0

11 months ago

As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road i

Darya [45]

<span>As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road is kinetic friction.

We exert force to move the object from rest and in this case, static friction works. But, when the object comes in motion, then kinetic friction works. Here, since the car is driving without slipping means, kinetic friction acts on it. Its also called sliding or dynamic friction.</span>

5 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:

<em>c) </em> $\frac{4492.18}{v_{car} ^{2} }$

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 $31.82^{2}$ = <em>6738.27 J</em>

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 $3.05^{2}$ = <em>61.908 J</em>

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

Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 30

dybincka [34]

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

<em><u>Support Cy:</u></em>

∑ M at Ay = 0

(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

( L ab + L bc )

Cy = 1.3 x 10³ k-N

<em><u>Support Ay:</u></em>

Since ∑ F = 0, A + C - F - D = 0

A = F + D - C

Ay = 200 k-N

4 0

2 years ago