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cluponka [151]
3 years ago
12

A car tire rotates with an average angular speed of 32 rad/s. In what time interval will the tire rotate 3.5 times? Answer in un

its of s.
Physics
1 answer:
stiks02 [169]3 years ago
3 0

Answer:

\Delta t = 0.687\ s

Explanation:

given,

Angular speed of the tire = 32 rad/s

Displacement of the wheel = 3.5 rev

Δ θ = 3.5 x 2 π

        = 7 π rad

now,

Time interval of the car to rotate 7π rad

using equation

\omega_{avg} =\dfrac{\Delta \theta}{\Delta t}

\Delta t=\dfrac{7\pi}{32}

\Delta t = 0.687\ s

Time taken to rotate 3.5 times is equal to 0.687 s.

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What is the meaning of the universe?
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Explanation:

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The letter "p" in the symbol 4p3 indicates the ____? The letter "p" in the symbol 4p3 indicates the ____ B orbital shape. D spee
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3 years ago
(a) You short-circuit a 20 volt battery by connecting a short wire from one end of the battery to the other end. If the current
erica [24]

(a) 1.11 \Omega

When the battery is short-cut, the only resistance in the circuit is the internal resistance of the battery. Therefore, we can apply Ohm's law:

r=\frac{V}{I}

where

V = 20 V is the voltage across the internal resistance of the battery

I = 18 A is the current flowing through it

Solving the equation,

r=\frac{20 V}{18 A}=1.11\Omega

(b) 360 W

The power generated by the battery is given by the equation

P=VI

where

V = 20 V is the voltage of the battery

I = 18 A is the current

Substituting into the formula,

P=(20 V)(18 A)=360 W

(c) 360 J

The energy dissipated by the internal resistance is given by

E=Pt

where

P = 360 W is the power generated

t = 1 s is the time

Solving the equation, we find

E=(360 W)(1 s)=360 J

(d) 1.65 A

The battery is now connected to a R=11 \Omega resistor. This means that the internal resistance of the battery is now connected in series with the other resistor R: so, the total resistance of the circuit is

R_T = r+R=1.11 \Omega +11 \Omega = 12.11 \Omega

And so, the current flowing through the circuit is

I=\frac{V}{R_T}=\frac{20 V}{12.11\Omega}=1.65 A

(e) 29.9 W

The power dissipated in the external resistor is given by

P=I^2 R

where

I = 1.65 A is the current

R=11 \Omega is the resistance

Solving the equation, we find

P=(1.65 A)^2(11 \Omega)=29.9 W

(f) 18.17 V

The terminals of the voltmeter are placed at the two end of the battery. The battery provides an emf of 20 V, however due to the internal resistance, some of this voltage is dropped across the internal resistance. Therefore, the actual potential difference that will be read by the voltmeter will be:

V=\epsilon - Ir =20 V -(1.65 A)(1.11 \Omega)=18.17 V

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