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2 years ago

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you have one 100 ohm resistor what size resistor must you hook up in parallel in order to have a resistance of 99.0

1 answer:

romanna [79]2 years ago

7 0

Answer:

The resistor to be hooked up must be 9900 $\Omega$

Explanation:

Recall how resistors in parallel combine to render an equivalent resistor:

$\frac{1}{R_e} =\frac{1}{R_1} +\frac{1}{R_2}$

So we need to find which resistor (R2) we need to combine in parallel with resistor R1 (100 $\Omega$ ) in order to obtain an equivalent resistor Re (99 $\Omega$ ) :

$\frac{1}{99} =\frac{1}{100} +\frac{1}{R_2}\\\frac{9900\,R_2}{99} =\frac{9900\,R_2}{100} +\frac{9900\,R_2}{R_2}\\100\,R_2= 99\,R_2+9900\\R_2=9900\,\,\Omega$

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2 years ago

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A river flows down towards a lake along an incline. initially the river is 90 m above the lake flowing at a rate of 3 m/s. at th

lubasha [3.4K]

Total energy =kinetic energy +potential energy

Change in energy =change in (kinetic energy +potential energy)

potential energy, $P.E=mgh$

where m is the mass, g is acceleration due to gravity and h is the height.

potential energy per unit mass =gh

change in potential energy per unit mass = $\Delta P.E.=g\Delta h$

where, h is the height.

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change in kinetic energy per unit mass, $\Delta K.E. =\frac{1}{2}\Delta v^2$

In the given question:

Height varies from 90 m to zero as river flows from 90 m height to lake at 0 m

Velocity varies from 3m/s at top to o m/s at bottom.

Therefore,

$\Delta E =\Delta K.E.+\Delta P.E.\\ \Rightarrow \Delta E/m=\frac{1}{2}\Delta v^2+g\Delta h=\frac{1}{2}(0^2-3^2) m^2s^{-2}+9.8 (0-9)m^2s^{-2}=(-\frac{9}{2}-88.2)m^2s^{-2}=-92.7 m^2s^{-2}$

Here, it was mentioned in the question internal energy of the water is constant and there is no change in the pressure at the inlet and outlet.

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2 years ago

At time t=0 a grinding wheel has an angular velocity of 26.0 rad/s. It has a constant angular acceleration of 25.0 rad/s^2 until

Zolol [24]

Angle turned by the wheel is given by the kinematics as

$\theta = \omega t + \frac{1}{2}\alpha t^2$

now here we know that

$\omega = 26 rad/s$

$\alpha = 25 rad/s^2$

t = 2.50 seconds

$\theta = 26(2.50) + \frac{1}{2}(25)(2.50)^2$

$\theta = 143.1 rad$

So here total angle that the wheel turn is given as

$\theta = 143.1 + 440 = 583.1 rad$

Speed of the wheel after 2.5 s

$\omega = \omega_0 + \alpha t$

$\omega = 26 + (25)(2.5)$

$\omega = 88.5 rad/s$

now angular deceleration is given as

$\alpha_1 = \frac{\omega^2 - \omega_0^2}{2\theta}$

$\alpha_1 = \frac{0 - 88.5^2}{2(440)}$

$\alpha_1 = -8.9 rad/s^2$

time taken to stop

$t = \frac{\omega - \omega_0}{\alpha}$

$t = \frac{0 - 88.5}{-8.9} = 9.94 s$

now total time is given as

$T = 2.50 + 9.94 = 12.44 s$

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2 years ago

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TiliK225 [7]

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2 years ago

A driver of a car going 75 km/h suddenly sees the lights of a barrier 35 m ahead. It takes the driver 0.60 seconds before he app

frutty [35]

Answer:

A) The car will hit the barrier

b) 19.82 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

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⇒Distance = (75/3.6)×0.6

⇒Distance = 12.5 m

Distance traveled by the car during the reaction time is 12.5 m

Equation of motion

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-\frac{75}{3.6}^2}{2\times -8.5}\\\Rightarrow s=25.53\ m$

Total distance traveled is 12.5+25.53 = 38.03 m

So, the car will hit the barrier

Distance = Speed × Time

⇒d = u0.6

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-u^2}{2\times -8.5}\\\Rightarrow s=\frac{u^2}{17}$

d + s = 35

$\\\Rightarrow u0.6+\frac{u^2}{17}=35\\\Rightarrow \frac{u^2}{17}+0.6u-35=0$

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Hence, the maximum velocity by which the car could be moving and not hit the barrier 35 meters ahead is 19.82 m/s.

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2 years ago