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alexdok [17]
3 years ago
14

Calculate the KE of a 3.1 kg toy cart that moves at 4.8 m/s. Calculate the KE of the same cart at twice the speed.

Physics
1 answer:
lisov135 [29]3 years ago
4 0
When v= 4.8 m/s:

E.K=mv²/2
      =3,1(4.8)²/2
      =35.712J

when v= 2(4.8):

E.K=mv²/2
      =3.1(9.6)²/2
      =142.848J


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You charge an initially uncharged 89.9-mf capacitor through a 30.5-ω resistor by means of a 9.00-v battery having negligible int
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<span>1) The differential equation that models the RC circuit is :

(d/dt)V_capacitor </span>+ (V_capacitor/RC)​ = (V_source/<span>RC)​​</span>

<span>Where the time constant of the circuit is defined by the product of R*C

Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s


2)
C<span>harge of the capacitor 1.57 time constants

1.57*(2.742) = 4.3048 s

The solution of the differential equation is

</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T)  +  </span>V_capac(∞)

Since the capacitor is initially uncharged V_capac(0) = 0

And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery  V_capac(∞) = 9V 

This means,

V_capac (t) = (-9V)e ^(-t /T)  +  9V

The charge in a capacitor is defined as Q = C*V

Where C is the capacitance and V is the Voltage across

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V

V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s)  +  9V = -1.87V +9V

V_capac (4.3048 s) = 7.1275 V

Q (4.3048 s)  = 89.9mF*(7.1275V) = 0.6407 C

3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V

Q (∞)  = 89.9mF*(9V) = 0.8091 C
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To find the specific heat capacity of a solid of mass 600 g whose temperature was 40oC, it was placed in a calorimeter that cont
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Explanation:

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Read 2 more answers
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