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Sav [38]
3 years ago
12

Can two similar charges attracts each other ? ​

Physics
2 answers:
frez [133]3 years ago
8 0

Answer:

No, they cannot.

Explanation:

Two like charges have a repulsive effect; they deflect each other.

Hope this helped! :)

VMariaS [17]3 years ago
5 0

Answer:

no they cannot, only opposite charges attract

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Each wheel of a 320 kg motorcycle is 52 cm in diameter and has rotational inertia 2.1 kg m2 . The cycle and its 75 kg rider are
Amiraneli [1.4K]

Answer:

The value is  h  = 32.91 \  m

Explanation:

From the question we are told that

    The diameter of each wheel is  d =  52 \ cm  =  0.52 \  m

    The mass of the motorcycle is  m  =  320 \ kg

    The rotational kinetic inertia is  I  =  2.1 \  kg \  m^2

    The  mass of the  rider is  m_r =  75 \ kg

     The  velocity is  v  =  85 \  km/hr = 23.61 \  m/s

      Generally the radius of the wheel is mathematically represented as

      r =  \frac{d}{2}

=>     r =  \frac{0.52}{2}

=>    r =  0.26 \  m

Generally from the law of energy conservation

     Potential energy  attained  by  system(motorcycle and rider )  =  Kinetic  energy of the system  +  rotational kinetic energy of  both wheels of the motorcycle

=>  Mgh  =  \frac{1}{2}  Mv^2  +   \frac{1}{2}  Iw^2  +  \frac{1}{2}  Iw^2

=>    Mgh  =  \frac{1}{2}  *  Mv^2  +  Iw^2

Here  w is the angular velocity which is mathematically represented as

     w =  \frac{v }{r }

So

    Mgh  =  \frac{1}{2}  *  Mv^2  +  I \frac{v}{r} ^2

Here M  =  m_r +  m

         M  =  320 + 75

          M  = 395 \  kg

395 *  9.8 *  h  =   0.5    *   395 *  (23.61)^2 +  2.1  *[\frac{ 23.61}{ 0.26} ] ^2

=>   h  = 32.91 \  m

   

7 0
4 years ago
You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c
Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

  • Since the block slides across the floor at constant speed, this means that it's not accelerated.
  • According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
  • This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

       F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N  (1)

  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

      F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N  (2)

⇒    169 N + Fn = Fg = 216 N  (3)

  • This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
  • Fn = 216 N - 169 N = 47  N (4)

6 0
3 years ago
Am un test de rezolvat la fizică mă puteți ajuta
OleMash [197]

Answer:

umm

Explanation:

i don't speak Spanish sorry

5 0
3 years ago
An electric pump has 2kW power . How much water will it lift every minute if the height is 10 m ?
valkas [14]

Answer:

Given that,

  • Power = 2000 W
  • time = 60 seconds
  • distance= 10m

Power = work done ÷ time

Here, since the movement is vertical, w = mgh

So,

Power = mgh÷t

2000 = (m × 9.8 ×10) ÷ 60

m = (2000 ×60) ÷98

m = 1224.5kg

3 0
3 years ago
Car A hits car B (initially at rest and of equal mass) from behind while going 15 m/s Immediately after the collision, car B mov
Mamont248 [21]

Given :

Initial speed of car A is 15 m/s and initial speed of car B is zero.

Final speed of car A is zero and final speed of car B is 10 m/s.

To Find :

What fraction of the initial kinetic energy is lost in the collision.

Solution :

Initial kinetic energy is :

K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}

Final kinetic energy is :

K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}

Now, fraction of initial kinetic energy loss is :

Loss = \dfrac{\dfrac{225m}{2}-\dfrac{100m}{2}}{\dfrac{100m}{2}}\\\\Loss = \dfrac{125}{100}\\\\Loss = 1.25

Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .

6 0
3 years ago
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