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Arlecino [84]
3 years ago
8

A resistor with resistance R is connected to a battery that has emf 13.0 V and internal resistance r = 0.350 Ω . For what two va

lues of R will the power dissipated in the resistor be 81.0 W ?
Physics
1 answer:
makvit [3.9K]3 years ago
8 0

Answer:

R = 1,746 Ω

Explanation:

The power dissipated in the circuit is

   P = V I = V² / R

Let's find the current

   R = V² / P

Let's calculate

  R = 13²/81

   R = 2,096 Ω

This is total resistance

  R_total = R + r

   R = R_total - r

   R = 2,096 -0,350

   R = 1,746 Ω

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A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball slides on the lane with initial speed <img src="https:
s344n2d4d5 [400]

Answer:

a) v_com= Rω

b) -2.254 m/s^{2}

c) 51.2 rad/s^{2}

d) t=1.08 seconds

e) x=7.865m

f) v_roll=6.07m

Explanation:

Initially, the ball is travelling with v_com=v_0

Wen not rotating, at the initial stage the ball must be sliding along the surface.

This motion therefore generates a frictional force F_r at the point of contact.

Let the velocity at the point of contact be v_bottom

v_bottom=v_com-Rω

Therefore when ω=0, v_bottom=v_com

So when the ball begins rolling

v_com= Rω

F_r=μ_rmg

〖-F〗_r=ma_com

a_com=(〖-μ〗_r mg)/m

a_com=-μ_rg

a_com=-(0.23)(9.8)

a_com=-2.254m/s^2

Te negative sow decrearse  

\alpha=(μ_r mgR)/I  =  (〖5μ〗_r mgR)/2mRR

=(〖5μ〗_r g)/2R

=(5*(0.23)*(9.8))/(2*0.11)

=51.2 rad/s^2

t=v_0/(〖-a〗_com+Rα)

=8.5/(2.255+0.11*(51.2))

=8.5/7.886

=1.08 seconds

X=v_0 t+1/2 a_com t^2

X=8.5*(2.254) -  1/2 (2.254)*〖1.08〗^2

=7.865m

v_roll=v_0+a_com t_r

=8.5-(2.254)(1.08)

        =6.07m/sec

Initially, the ball is travelling with v_com=v_0

Wen not rotating, at the initial stage the ball must be sliding along the surface.

This motion therefore generates a frictional force F_r at the point of contact.

a) Let the velocity at the point of contact be v_bottom

v_bottom=v_com-Rω

Therefore when ω=0, v_bottom=v_com

So when the ball begins rolling

v_com= Rω

b)    F_r=μ_rmg

〖-F〗_r=ma_com

a_com=(〖-μ〗_r mg)/m

a_com=-μ_rg

a_com=-(0.23)(9.8)

a_com=-2.254m/s^2

Te negative sow decrearse  

c) α=(μ_r mgR)/I  =  (〖5μ〗_r mgR)/2mRR

=(〖5μ〗_r g)/2R

=(5*(0.23)*(9.8))/(2*0.11)

=51.2 rad/s^2

d) t=v_0/(〖-a〗_com+Rα)

=8.5/(2.255+0.11*(51.2))

=8.5/7.886

=1.08 seconds

e) X=v_0 t+1/2 a_com t^2

X=8.5*(2.254) -  1/2 (2.254)*〖1.08〗^2

=7.865m

f) v_roll=v_0+a_com t_r

=8.5-(2.254)(1.08)

        =6.07m/sec

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