How do I solve for this?

Delivery trucks that operate by making use of energy stored in a rotating flywheel have been used in Europe. The trucks are char

Marta_Voda [28]

Answer:

KE = 2.535 x 10⁷ Joules

Explanation:

given,

angular speed of the fly wheel = 940 rad/s

mass of the cylinder = 630 Kg

radius = 1.35 m

KE of flywheel = ?

moment of inertia of the cylinder

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

= $\dfrac{1}{2}\times 630\times 1.35^2$

= 574 kg m²

kinetic energy of the fly wheel

$KE = \dfrac{1}{2}I\omega^2$

$KE =\dfrac{1}{2}\times 574\times 940^2$

KE = 2.535 x 10⁷ Joules

the kinetic energy of the flywheel is equal to KE = 2.535 x 10⁷ Joules

5 0

3 years ago

Please help <br> Convert the following distance v time graph velocity graph

Reil [10]

Explanation:

for v vs t graph for t=9 to 11,v=15kmph

similarly v2=0,v3=60kmph&v4= -40kmph

5 0

3 years ago

Consider a small frictionless puck perched at the top of a xed sphere of radius R. If the puck is given a tiny nudge so that it

MakcuM [25]

Answer:

Explanation:

Let the vertical height by which it descends be h . Let it acquire velocity of v .

1/2 mv² = mgh

v² = 2gh

As it leaves the surface of sphere , reaction force of surface R = 0 , so

centripetal force = mg cosθ where θ is the angular displacement from the vertex .

mv² / r = mg cosθ

(m/r )x 2gh = mg cosθ

2h / r = cosθ

cosθ = (r-h) / r

2h / r = r-h / r

2h = r-h

3h = r

h = r / 3

5 0

3 years ago

A transformer has input voltage and current of 12 V and 4 A

Advocard [28]

Explanation:

It is given that,

Input voltage, $V_i=12\ V$

Input current, $I_i=4\ A$

Output current, $I_o=0.8\ A$

Number of turns in the secondary side of transformer, $N_s=1177$

We need to find the number of turns in the primary side of the transformer. The current to the number of turns in the input and output is given by :

$\dfrac{N_s}{N_p}=\dfrac{I_p}{I_s}$

Substituting all the above values

So,

$N_p=\dfrac{N_sI_s}{I_p}\\\\N_p=\dfrac{1177\times 4}{0.8}\\\\N_p=5885$

So, the number of turns in primary side of the transformer is 5885.

4 0

3 years ago

Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f

stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2$

When s = Db, t = 2t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2$

Dividing the two equations

$\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db$

Hence, Da=(1/4)Db

3 0

3 years ago