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balu736 [363]
3 years ago
5

A bus travels a distance of 120 km with a speed of 40km per hour and returns with a speed of 30km per hour calculate the average

speed for the entire journey
Physics
1 answer:
Vsevolod [243]3 years ago
3 0

Answer:

35 km/hr

Explanation:

Average speed = (total of the speed)/(the sets of speeds given)

Direction does not matter in this instance since speed is only magnitude,

Average speed = (30 + 40)/2

Average speed = 70 ÷ 2

= 35 km/hr

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An air track car with a mass of 6 kg and velocity of 4 m/s to the right collides with a 3 kg car moving to the left with a veloc
sammy [17]

Remember that moment before collision is equal to the moment after collision.

(m1 \times u1) + (m2 \times u2) = (m1 \times v1) + (m2  \times u2)

Plugging in our values,

(6 \times 4) - (3 \times 2) = (6 \times 1) + (3 \times v2) \\ 24 - 6 = 6 \times 3v2 \\ 18 = 18v2 \\ v2 = 1 {ms}^{ - 1}

8 0
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An unknown substance was observed to have the following properties: colorless liquid that feels slippery, pH of 8.5, turns red l
Mashcka [7]

Answer:

It is a base

Explanation:

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3 0
3 years ago
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Marc attaches a falling 500-kg object with a rope through a pulley to a paddle wheel shaft. He places the system in a well-insul
Klio2033 [76]

Answer:

4.68227 °C

Explanation:

m_o = Mass of object = 500 kg

m_w = Mass of water = 25 kg

c = Specific heat of water at 20°C = 4186 J/kg°C

h = Height from which the object falls = 100 m

g = Acceleration due to gravity = 9.8 m/s²

The potential energy and heat will balance each other

PE=Q\\\Rightarrowmc m_ogh=m_oc\Delta T\\\Rightarrow \Delta T=\frac{m_ogh}{m_oc}\\\Rightarrow \Delta T=\frac{500\times 9.8\times 100}{25\times 4186}\\\Rightarrow \Delta=4.68227\ ^{\circ}C

The temperature change in the water is 4.68227 °C

5 0
3 years ago
g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value v_2  =  4 \sqrt{10} \  m/s

Explanation:

From the question we are told that

   The  charge on the first sphere is  q_1  =  2\mu C  =  2*10^{-6} \  C

    The charge on the second sphere is  q_2 =  8 \mu C = 8*10^{-6} \  C

     The  mass of the second charge is m  =  1.50 \  g  =  1.50 *10^{-3} \ kg

      The  distance apart is  d =  0.4 \  m

      The  speed of the second  sphere is  v_1  =  20 \  ms^{-1}

Generally the total energy possessed by when q_2 and  q_1 are separated by 0.8 \  m is mathematically represented

     Q =  KE + U

Here KE   is  the kinetic energy which is mathematically represented as

     KE  =  \frac{1 }{2}  m (v_1)^2

substituting value

     KE  =  \frac{1 }{2}  * ( 1.50 *10^{-3}) (20 )^2

     KE  =  0.3 \  J

And  U is  the  potential  energy which is mathematically represented as

        U  =  \frac{k *  q_1 *  q_2  }{d }

substituting values

       U  =  \frac{9*10^9 *  2*10^{-6} * 8*10^{-6}  }{0.8 }

      U  =  0.18 \  J

So

       Q =  0.3 +  0.18

       Q =  0.48 \  J

Generally the total energy possessed by when q_2 and  q_1 are separated by 0.4 \  m is mathematically represented

         Q_f =  KE_f + U_f

Here KE_f is  the kinetic energy which is mathematically represented as

     KE_f  =  \frac{1 }{2}  m (v_2^2

substituting value

     KE_f  =  \frac{1 }{2}  * ( 1.50 *10^{-3}) (v_2 )^2

     KE_f  =  7.50 *10^{ -4} (v_2 )^2

And  U_f is  the  potential  energy which is mathematically represented as

        U_f  =  \frac{k *  q_1 *  q_2  }{d }

substituting values

       U_f  =  \frac{9*10^9 *  2*10^{-6} * 8*10^{-6}  }{0.4 }

      U_f  =  0.36 \  J

From the law of energy conservation

     Q =  Q_f

So

    0.48 =  0.36 +(7.50 *10^{-4} v_2^2)

   v_2  =  4 \sqrt{10} \  m/s

     

   

6 0
3 years ago
My sis needs help with this and I don't wanna help her!!! Can you guys help her????
Katena32 [7]

Answer and Explanation:

Cup 1. The ice cubes are in a solid state.

Cup 2: The cup has water that is liquified.

Cup 3: The cup is empty and only contains free floating air particles.

***If you found my answer helpful, please give me the brainliest. :) ***

7 0
3 years ago
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