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3 years ago

10

What are 5 examples of balanced and unbalanced forces in your home?

1 answer:

ryzh [129]3 years ago

8 0

Balance:
a book resting on a table
a car driving at 10 miles per hour in constant velocity
a cat sitting on a chair
a bulb that attach to the ceiling
your grandma sleeping on a bed
Unbalance:
your brother sprinting across the kitchen
a ball rolling at 5 m/s^2
your mom trying to run at 2 m/s^2 to spank you
you dropping your coffee mug on a floor
a cat jumping out of your bed
a tear from your eye falling through the floor
Hope this helps

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2 years ago

An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in ser

s2008m [1.1K]

Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

$V_L=V_oe^{-\frac{Rt}{L}}$

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

$\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s$

hence, after 1.6s the inductor will have a potential difference of 24V

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3 years ago

Tarzan has foolishly gotten himself into another scrape with the animals and must be rescued once again by Jane. The 60.0 kg Jan

Brut [27]

Complete part of Question: What is Jane's (and the vine's) angular speed just before she grabs Tarzan

Answer:

Jane's (and the vine's) angular speed just before she grabs Tarzan, w = 1.267 rad/s

Explanation:

According to the law of energy conservation:

Total change in kinetic energy = Total change in potential energy

Mass of Jane = 60 kg

Mass of the vine = 32 kg

Mass of Tarzan = 72 kg

Height of Tarzan = 5.50 m

Length of the vine = 8.50 m

Jane's change in gravitational potential energy,

$U_J = 60 * 9.8 * 5.5\\U_J = 3234 J$

Vine's gravitational potential energy,

$U_v = Mgh/2\\U_v = 32*9.8*5.5/2\\U_v = 862.4J$

Vine's Kinetic energy :

$KE_V = 0.5 I w^{2} \\I_V = \frac{ML^2}{3} = \frac{32 * 8.5^2}{3} = 770.67 kg m^2\\ KE_V = 0.5 *770.67 * w^{2}\\KE_V = 385.33 w^{2}$

Jane's Kinetic energy:

$KE_J = 0.5m(wL)^2\\KE_J = 0.5*60(w * 8.5)^2\\KE_J = 2167.5 w^2$

$U_J + U_V = KE_J + KE_V$

3234 + 862.4 = 2167.5w² + 385.33w²

4096.4 = 2552.83w²

w² = 4096.4/2552.83

w² = 1.605

w = √1.605

w = 1.267 rad/s

5 0

2 years ago

Find deacceleration An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When

earnstyle [38]

Answer:

The deceleration is $a = -0.7273 \ m/s^2$

Explanation:

From the question we are told that

The distance of the car from the crossing is $d = 360 \ m$

The speed is $u = 16 \ m/s$

The reaction time of the engineer is $t = 0.53 \ s$

Generally the distance covered during the reaction time is

$d_r = u * t$

=> $d_r = 16 * 0.53$

=> $d_r = 8.48 \ m$

Generally distance of the car from the crossing after the engineer reacts is

$D = d- d_r$

=> $D = 360 - 8.48$

=> $D = 352 \ m$

Generally from kinematic equation

$v^2 = u^2 + 2as$

Here v is the final velocity of the car which is 0 m/s

So

$0^2 = 16^2 + 2 * a * 352$

=> $a = -0.7273 \ m/s^2$

3 0

2 years ago