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stealth61 [152]
3 years ago
8

If two 100 ohms resistors are placed in series, their total resistance is what?

Physics
1 answer:
Sindrei [870]3 years ago
3 0

Answer:

add the Resistance

2(100)= 200

2) 200 ohms

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What is the motional kinetic energy of a 25 kg object moving at a speed of 10 m/s?
Soloha48 [4]

1250kgm²/s is the motional kinetic energy of a 25kg object moving at a speed of 10m/s

Kinetic energy of an object is defined as the energy which is possessed when that is  in motion. It is the energy of the kinetic mass of an object. Kinetic energy is never negative and is a scalar quantity. That is, it shows only size, not orientation.

Given to us

Mass of the object, m=25kg

Velocity of the object, v=10m/s  

K.E=1/2x25x10²

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Kinetic energy is directly proportional to the mass and velocity squared (K.E.) of an object. =1/2xMxV². If the mass is in kilograms and the velocity is in meters/second, then the kinetic energy is in kilograms - meters squared/second.

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Explanation:

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A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

Compute U if the bulb remains on for 5h

Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

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3 years ago
What is the largest four subsystems?​
AnnZ [28]

Explanation:

Everything in Earth's system can be placed into one of four major subsystems: land, water, living things, or air. These four subsystems are called "spheres." Specifically, they are the "lithosphere" (land), "hydrosphere" (water), "biosphere" (living things), and "atmosphere" (air).

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2. An athlete of average size is hanging from the end of a 20 m long rope, which has a mass of 4 kg and is attached to a hook in
a_sh-v [17]

Answer:

  t = 0.319 s

Explanation:

With the sudden movement of the athlete a pulse is formed that takes time to move along the rope, the speed of the rope is given by

             v = √T/λ

Linear density is

           λ = m / L

           λ = 4/20

           λ = 0.2 kg / m

The tension in the rope is equal to the athlete's weight, suppose it has a mass of m = 80 kg

           T = W = mg

           T = 80 9.8

           T = 784 N

The pulse rate is

          v = √(784 / 0.2)

          v = 62.6 m / s

The time it takes to reach the hook can be searched with kinematics

          v = x / t

          t = x / v

          t = 20 / 62.6

          t = 0.319 s

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3 years ago
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