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Snezhnost [94]
3 years ago
13

A hollow spherical shell has mass 8.05kg and radius 0.215m . It is initially at rest and then rotates about a stationary axis th

at lies along a diameter with a constant acceleration of 0.895rad/s2 .
What is the kinetic energy of the shell after it has turned through 5.25rev ?

Express your answer with the appropriate units.

Physics
2 answers:
hram777 [196]3 years ago
4 0

Answer: The kinetic energy of the shell after it has turned through 5.25rev is 7.32J

Explanation: Please see the attachments below

Neporo4naja [7]3 years ago
3 0

Answer:

930.21J

Explanation:

The kinetic energy of a solid body that is rotating, is given  by:

E_k=\frac{1}{2}I\omega^2

where I is the moment of inertia and w is the angular velocity.

The moment of inertia for a spherical shell is:

I=\frac{2}{3}MR^2

where M is the mass and R is the radius of the sphere. By replacing we have

I=\frac{2}{3}(8.05kg)(0.215m)^2=0.55\ kgm^2

To calculate w we have to use the equation

\omega^2=\omega_0^2+2\alpha \theta\\\\\omega=\sqrt{2(0.895rad/s^2)(5.25rev*\frac{360\°}{1\ rev})}=58.16\frac{rad}{s}

where we have taken w0=0 rad/s.

Finally, by replacing I and w we obtain:

E_k=\frac{1}{2}I\omega^2 =\frac{1}{2}(0.55\ kgm^2)(58.16\frac{rad}{s})^2=930.21J

HOPE THIS HELPS!

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An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electric
ELEN [110]

Answer:

a. 0 W b. ε²/R c. at R = r maximum power = ε²/4r d. For R = 2.00 Ω, P = 227.56 W. For R = 4.00 Ω, P = 256 W. For R = 6.00 Ω, P = 245.76 W

Explanation:

Here is the complete question

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electrical power output of the source. By conservation of energy, P is equal to the power consumed by R. What is the value of P in the limit that R is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when R = r . What is this maximum P in terms of ε and r? (d) A battery has ε= 64.0 V and r=4.00Ω. What is the power output of this battery when it is connected to a resistor R, for R=2.00Ω, R=4.00Ω, and R=6.00Ω? Are your results consistent with the general result that you derived in part (b)?

Solution

The power P consumed by external resistor R is P = I²R since current, I = ε/(R + r), and ε = e.m.f and r = internal resistance

P = ε²R/(R + r)²

a. when R is very small , R = 0 and P = ε²R/(R + r)² = ε² × 0/(0 + r)² = 0/r² = 0

b. When R is large, R >> r and R + r ⇒ R.

So, P = ε²R/(R + r)² = ε²R/R² = ε²/R

c. For maximum output, we differentiate P with respect to R

So dP/dR = d[ε²R/(R + r)²]/dr = -2ε²R/(R + r)³ + ε²/(R + r)². We then equate the expression to zero

dP/dR = 0

-2ε²R/(R + r)³ + ε²/(R + r)² = 0

-2ε²R/(R + r)³ =  -ε²/(R + r)²

cancelling out the common variables

2R =  R + r

2R - R = R = r

So for maximum power, R = r

So when R = r, P = ε²R/(R + r)² = ε²r/(r + r)² = ε²r/(2r)² = ε²/4r

d. ε = 64.0 V, r = 4.00 Ω

when R = 2.00 Ω, P = ε²R/(R + r)² = 64² × 2/(2 + 4)² = 227.56 W

when R = 4.00 Ω, P = ε²R/(R + r)² = 64² × 4/(4 + 4)² = 256 W

when R = 6.00 Ω, P = ε²R/(R + r)² = 64² × 6/(6 + 4)² = 245.76 W

The results are consistent with the results in part b

8 0
3 years ago
A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg
Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

p_i=mv+0=mv

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

p_f=mv'+MV'

Equating initial and the final momenta we get

mv=mv'+MV'\\\\m(v-v')=MV'.....i

Now since the surface is frictionless thus the energy is also conserved thus

E_i=\frac{1}{2}mv^2

Similarly the final energy becomes

E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\

Equating initial and final energies we get

\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)

Solving i and ii we get

v+v'=V'

Using this in equation i we get

v'=\frac{v(m-M)}{(M-m)}=-v

Thus putting v = -v' in equation i  we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

4 0
3 years ago
anice is watching her granddaughter drive a Barbie Jeep with a 6 V battery and an electric motor with 5 ohm of resistance. How m
Scrat [10]

Answer:

Power required will be 7.2 watt

Explanation:

We have given battery of Barble jeep is 6 volt

So potential difference V = 6 volt

Resistance of electric motor R = 5 ohm '

We have to find the power motor using to drive the jeep

Power is given by P=\frac{V^2}{R}, here V is voltage and R is resistance

So P=\frac{V^2}{R}=\frac{6^2}{5}=7.2watt

5 0
3 years ago
Read 2 more answers
What is the force on a 15.5 kg ball that is falling freely due to the pull of gravity
velikii [3]

Answer:

151.9 N

Explanation:

Force = mass x acceleration

Acceleration due to gravity is 9.8 m/s^2 (you should memorize this number).

F = ma

F = (15.5)(9.8)

F = 151.9

3 0
3 years ago
A current-carrying wire is bent into a circular loop of radius R and lies in an xy plane. A uniform external magnetic field B in
Ilya [14]

Answer:

 F = 2π I R B

Explanation:

The magnetic force is described by the equation.

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Where i is the current, L is a vector that points in the direction of the current (length) and B is the magnetic field.

This equation can be used in scalar form and the direction of the force found by the right hand ruler, the thumb goes in the direction of L, the fingers extended in the direction of B and the palm of the hand indicates the direction of the force if the load is positive

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In this case the wire is in the xy plane and the z-axis field whereby they are perpendicular, θ = 90º and sin 90 = 1

     F = i L B

The loop length is

    L = 2π R

    F = i 2π R B

    F = 2π I R B

The force is in the loop

8 0
3 years ago
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