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3 years ago

What is inertia? That is all but I need 20 characters to ask lol

1 answer:

8 0

Inertia, property of a body by virtue of which it opposes any agency that attempts to put it in motion or, if it is moving, to change the magnitude or direction of its velocity. Inertia is a passive property and does not enable a body to do anything except oppose such active agents as forces and torques.

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A stone that is dropped freely from rest traveled half the total height in the last second. with what velocity will it strike th

alexira [117]

Answer:

hellooooo :) ur ans is 33.5 m/s

At time t, the displacement is h/2:

Δy = v₀ t + ½ at²

h/2 = 0 + ½ gt²

h = gt²

At time t+1, the displacement is h.

Δy = v₀ t + ½ at²

h = 0 + ½ g (t + 1)²

h = ½ g (t + 1)²

Set equal and solve for t:

gt² = ½ g (t + 1)²

2t² = (t + 1)²

2t² = t² + 2t + 1

t² − 2t = 1

t² − 2t + 1 = 2

(t − 1)² = 2

t − 1 = ±√2

t = 1 ± √2

Since t > 0, t = 1 + √2. So t+1 = 2 + √2.

At that time, the speed is:

v = at + v₀

v = g (2 + √2) + 0

v = g (2 + √2)

If g = 9.8 m/s², v = 33.5 m/s.

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

Which of the following is/are true? Electromagnetic waves are created by accelerating charges. Electromagnetic waves are transve

Vikki [24]

Answer:

All of the above are true.

Explanation:

(a). true

whenever charge particle move back and froth from its mean position then it will produce oscillating electric and magnetic fields, . so an em wave can be obtain by accelerating charge

(b). true

the electric field and the magnetic field have vibrations in the perpendicular direction along the motion of the wave so electromagnetic wave is a transverse wave. therefore, the EM wave is a Transverse wave

(c) true .

The Electromagnetic wave consists of the two mutually perpendicular electric and magnetic fields and also both fields are perpendicular to the direction of propagation of the wave.

(d) true .

An electromagnetic wave carry energy through vacuum with a speed of $3 \times 10^8 m/s$

so , all of the above are true.

4 0

3 years ago

Please help me !! Thank you

ira [324]

I think it would be Fusing of 2 or more small atoms into a larger one

4 0

3 years ago

Read 2 more answers

g A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has

Nikitich [7]

Answer:

Explanation:

gravitational acceleration of meteoroid

= GM / R²

M is mass of planet , R is radius of orbit of meteoroid from the Centre of the planet .

R = (.9 x 6370 + 600 )x 10³ m

= 6333 x 10³ m

M , mass of the planet = 5.97 x 10²⁴ kg .

gravitational acceleration of meteoroid

= GM / R²

= (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ kg / (6333 x 10³ m)²

9.92m/s²

3 0

3 years ago