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Goshia [24]
3 years ago
6

Extend your thinking: household appliances are usually connected in a parallel circuit. why do you think it might be a problem i

f too many appliances are turned on at once? (hint: current in a wire also produces heat.)
Physics
1 answer:
alexandr1967 [171]3 years ago
6 0
The equivalent resistance of several devices connected in parallel is given by
\frac{1}{R_{eq}} =  \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n}
where R_i are the resistances of the various devices. We can see that every time we add a new device in parallel, the term \frac{1}{R_{eq}} increases, therefore the equivalent resistance of the circuit R_{eq} decreases.

But Ohm's law:
I= \frac{V}{R_{eq}}
tells us that if the equivalent resistance decreases, the total current in the circuit increases. The power dissipated through the circuit (and so, the heat produced) depends on the square of the current:
P=I^2 R
therefore if there are too many devices connected in parallel, this can be a problem because there could be too much power dissipated (and too much heat) through the circuit.
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A medicine ball has a mass of 5 kg and is thrown with a speed of 2 m/s. what is its kinetic energy?
Fantom [35]
The kinetic energy of the ball is given by:
K= \frac{1}{2}mv^2
where m=5 kg is the mass of the ball and v=2 m/s is its speed. Substituting these numbers, we find the kinetic energy:
K= \frac{1}{2} (5kg)(2m/s)^2=10 J
7 0
3 years ago
A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total accelera
AlekseyPX

Answer:

Explanation:

Given

Radius of bicycle wheel r=0.3\ m

Initial angular velocity \omega _0=0

It rotates 3 revolution in 5 s therefore

\omega =2\pi 3=\6\pi =18.85\ rad/s

using \omega =\omega _0+\alpha t

where \alpha =angular\ acceleration

\omega =Final\ angular\ velocity

t=time

\alpha =\frac{18.85}{5}=3.77 rad/s^2

Total acceleration of any point will be a vector sum of tangential acceleration and centripetal acceleration

\omega at t=1

\omega =0+3.77\times 1=3.77 rad/s

a_c=\omega ^2\cdot r

a_c=(3.77)^2\cdot 0.3=4.26 m/s^2

Tangential acceleration a_t=\alpha \times r

a_t=3.77\times 0.3=1.13 m/s^2

a_{net}=\sqrt{a_t^2+a_c^2}

a_{net}=\sqrt{(1.13)^2+(4.26)^2}

a_{net}=4.41 m/s^2

                       

7 0
3 years ago
A 0.46-kg cord is stretched between two supports, 7.2 m apart. When one support is struck by a hammer, a transverse wave travels
Kryger [21]

Answer:

      T = 6.0 N

Explanation:

given,

mass of the cord = 0.46 Kg

length of the supports = 7.2 m

time taken to travel = 0.74 s

tension in the chord = ?

using formula for tension calculation

T = \dfrac{v^2.m}{l}

v = \dfrac{l}{s}

v = \dfrac{7.2}{0.74}

v = 9.73 m/s

now, calculation of tension

T = \dfrac{9.73^2\times 0.46}{7.2}

      T = 6.0 N

The tension in the cord is equal to 6.0 N.

7 0
3 years ago
Hello please help me out
agasfer [191]
The answer is c.........
8 0
2 years ago
Read 2 more answers
A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit
lapo4ka [179]

Answer:

The  angle is  \theta  =  15.48^o

Explanation:

From the question we are told that  

     The distance of the dartboard from the dart is  d  =  3.66  \ m

     The time taken is  t =  0.455 \ s

   

The  horizontal component of the speed of the dart is mathematically represented as

      u_x =  ucos \theta

where u is the the velocity at dart is lunched

  so

      distance =  velocity \ in \ the\  x-direction  *  time

substituting values

      3.66 =   ucos  \theta *  (0.455)

 =>   ucos \theta =  8.04  \ m/s

From projectile kinematics the time taken by the dart can be mathematically represented as

         t  =  \frac{2usin \theta }{g}

=>    usin \theta =  \frac{g  * t}{2 }

       usin \theta =  \frac{9.8  * 0.455}{2 }

      usin \theta = 2.23

=>   tan \theta =  \frac{usin\theta }{ucos \theta }  =  \frac{2.23}{8.04}

       \theta  =  tan^{-1} [0.277]

      \theta  =  15.48^o

     

4 0
3 years ago
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