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iren2701 [21]
3 years ago
8

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid

e a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 445-mile trip in a typical midsize car produces about 4.66 x 109 J of energy. How fast would a 19.7-kg flywheel with a radius of 0.351 m have to rotate to store this much energy
Physics
1 answer:
Sergeeva-Olga [200]3 years ago
3 0

Answer:

8.37*10^5 rpm

Explanation:

Given that rotational kinetic energy = 4.66*10^9J

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = \frac{1}{2} mr ^2

Rotational K.E is illustrated as (K.E)_{rt} = \frac{1}{2} I \omega^2

\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }

\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }

\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }

\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }

\omega = 87636.04

\omega = 8.76*10^4 rad/s

Since 1 rpm = \frac{2 \pi}{60}  rad/s

\omega = 8.76*10^4(\frac{60}{2 \pi})

\omega = 836518.38

\omega = 8.37 *10^5 rpm

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A proton accelerates from rest in a uniform electric field of 610 N/C. At some later time, its speed is 1.4 106 m/s.(a) Find the
nata0808 [166]

Answer:

a) 5.851× 10¹⁰m/s²

b) 2.411×10⁻¹¹s

c)  1.70×10⁻¹¹m

d) 1.661×10⁻²⁷KJ

Explanation:

A proton in the field experience a downward force of magnitude,

F = eE. The force of gravity on the proton will be negligible compared to the electric force

F = eE

a= eE/m

= 1.602×10⁻¹⁹ × 610/1.67×10⁻²⁷

= 5.851× 10¹⁰m/s²

b)

V = u + at

u= 0

v= 1.4106m/s

v= (0)t + at

t= v/a

= 1.4106m/s/5.851 ×10¹⁰

= 2.411×10⁻¹¹s

c)

S = ut + at²

= (o)t + 5.851×10¹⁰×(2.411×10⁻¹¹)²

= 1.70×10⁻¹¹m

d)

Ke = 1/2mv²

    = (1.67×10⁻²⁷×)(1.4106)²/2

 =  1.661×10⁻²⁷KJ

5 0
3 years ago
A test car starts from rest on a horizontal circular track of 115-m radius and increases its speed at a uniform rate to reach 90
Wewaii [24]

Answer:

a= 3.49 m/s^2

Explanation:

magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.

we know that tangential acceleration a_t= change in velocity /time taken

now 90 km/h = 25 m/s

a_t = 25/17 = 1.47 m/s^2.

radial acceleration a_r = v^2/r

v= a_t×t = 1.47×13 = 19.11 m/s

a_r = 19.11^2/115=  3.175

now,

a= \sqrt{a_t^2+a_r^2}

a= \sqrt{1.47^2+3.175^2}

a= 3.49 m/s^2

3 0
3 years ago
A roller coaster travels around a vertical 8-m radius loop. Determine the speed at the top of the loop if the normal force exert
strojnjashka [21]

Answer:

v=10m/sec

Explanation:

From the question we are told that

Radius of vertical r= 8m

Force exerted by passengers is 1/4 of weight

Generally the net force acting on top of the roller coaster is give to be

F_N+Fg

where

F_N =forceof the normal

Fg= force due to gravity

Generally the net force is given to be FC(force towards center)

F_C =F_N + Fg

F_N =F_C -Fg

F_N=F_C-F_g

F_N=\frac{mv^2}{R} -mg

Mathematical we can now derive V

m_g + \frac{8m}{4}= \frac{mv^2}{8}

\frac{5mg}{4} =\frac{mv^2}{8}

v^2 =\frac{40*10}{4}

v=10m/sec

Therefore the speed of the roller coaster is given ton be v=10m/sec

5 0
3 years ago
Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive
Burka [1]

Answer:

1.91773\times 10^{37}\ kg

Explanation:

v = Orbital speed = 130 km/s

d = Diameter = 16 ly

r = Radius = \dfrac{d}{2}=\dfrac{16}{2}=8\ ly

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

1\ ly=9.461\times 10^{15}\ m

As the centripetal force balances the gravitational energy we have the following relation

\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow M=\dfrac{v^2r}{G}\\\Rightarrow M=\dfrac{130000^2\times 8\times 9.461\times 10^{15}}{6.67\times 10^{-11}}\\\Rightarrow M=1.91773\times 10^{37}\ kg

Mass of the the massive object at the center of the Milky Way galaxy is 1.91773\times 10^{37}\ kg

4 0
3 years ago
Concrete is pumped from a cement mixer to the place it is being laid, instead of being carried in wheelbarrows. The flow rate is
max2010maxim [7]

Answer:

a. 2.4 ×109  N ⋅ m2/s

b. 48.3 N⋅s /m2  

c. 8.00×104W

Explanation:

See Attached file for explanation

3 0
3 years ago
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