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Scorpion4ik [409]
3 years ago
11

If an electrical component with a resistance of 53 Q is connected to a 128-V source, how much current flows through the componen

t?
Physics
1 answer:
natta225 [31]3 years ago
8 0

Answer:

the current that flows through the component is 2.42 A

Explanation:

Given;

resistance of the electrical component, r = 53 Ω

the voltage of the source, V = 128 V

The current that flows through the component is calculated using Ohm's Law as demonstrated below;

V = IR\\\\I = \frac{V}{R} = \frac{128 \ V}{53 \ ohms} = 2.42 \ A

Therefore, the current that flows through the component is 2.42 A

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Jose is creating a game for class. In his game, fellow students will have to identify different types of communities. Which of t
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Answer:

Citizens and neighborhood.

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3 years ago
Which car has the most kinetic energy?
mixas84 [53]

Answer:

B. A car of mass 2000 kg with speed 7 m/s

Explanation:

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where m is the mass of the object and v is its speed.

From the formula, we see that the larger the mass and the speed of the object, the larger its kinetic energy. Among the choices given, we see that the car with largest mass and largest speed is car B, which has a mass of 2000 kg and speed of 7 m/s. Its kinetic energy is:

K=\frac{1}{2}(2000 kg)(7 m/s)^2=49,000 J

We can verify that the other cars have smaller kinetic energy. In fact:

- Car A: K=\frac{1}{2}(1000 kg)(3 m/s)^2=4,500 J

- Car C: K=\frac{1}{2}(1000 kg)(7 m/s)^2=24,500 J

- Car D: K=\frac{1}{2}(2000 kg)(3 m/s)^2=9,000 J

So, car B is the one which has most kinetic energy.

5 0
4 years ago
Assume that, when we walk, in addition to a fluctuating vertical force, we exert a periodic lateral force of amplitude 25 NN at
dexar [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

   The amplitude of the lateral  force is  F = 25 \  N

   The frequency is   f = 1 \  Hz

   The mass of the bridge per unit length is  \mu  =  2000 \  kg /m

    The length of the central span is  d =  144 m

     The oscillation amplitude of the section  considered at the time considered is  A = 75 \ mm =  0.075 \  m

      The time taken for the undriven oscillation to decay to \frac{1}{e}  of its original value is  t = 6T

Generally the mass of the section considered is mathematically represented as

            m =  \mu  *  d

=>        m =  2000 * 144

=>        m =  288000 \ kg

Generally the oscillation amplitude of the section after a  time period  t is mathematically represented as

                 A(t) = A_o e^{-\frac{bt}{2m} }

Here b is the damping constant and the A_o is the amplitude of the section when it was undriven

So from the question  

               \frac{A_o}{e}  = A_o e^{-\frac{b6T}{2m} }

=>            \frac{1}{e}  =e^{-\frac{b6T}{2m} }

=>          e^{-1} =e^{-\frac{b6T}{2m} }

=>           -\frac{3T b}{m}  =  -1

=>         b  = \frac{m}{3T}

Generally the amplitude of the section considered is mathematically represented as

           A =  \frac{n * F }{ b *  2 \pi }

=>       A =  \frac{n * F }{ \frac{m}{3T}  *  2 \pi }

=>       n =  A  *  \frac{m}{3}  *  \frac{2\pi}{25}

=>       n = 0.075 *  \frac{288000}{3}  *  \frac{2* 3.142 }{25}

=>       n = 1810 \ people

3 0
3 years ago
13 points!
skelet666 [1.2K]
I think it may be C!
4 0
3 years ago
Read 2 more answers
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Answer: Qualifying as an Olympian is an extremely difficult feat for anyone to accomplish and is regarded as a very prestigious title. To qualify as a Paralympian is an even tougher challenge. ForParalympic athletes, the sports in which Olympians compete are harder to do, harder to train for, and pose challenges that would prove nearly impossible for a fully able-bodied person to overcome. The disparity in level of difficulty between the same sports in Olympics and Paralympics was evident to me when watching the events. Whether skiing down a mountain, swimming, or playing on ice, Paralympic athletes must demonstrate a higher level of skill in order to succeed in their respective event.

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