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kow [346]
3 years ago
6

A bicycle has a momentum of 23.4

Physics
1 answer:
Snowcat [4.5K]3 years ago
4 0
  • Momentum=23.4kgm/s
  • Velocity=2m/s

\\ \rm\longmapsto Momentum=Mass\times velocity

\\ \rm\longmapsto Mass=\dfrac{Momentum}{Velocity}

\\ \rm\longmapsto Mass=\dfrac{23.4}{2}

\\ \rm\longmapsto Mass=11.7kg

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A football is thrown horizontally with an initial velocity of (16.6 m/s)x^. ignoring air resistance, the average acceleration of
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Solution:

According to the equations for 1-D kinematics. The only change to them is that instead one equation that describes general motion.

So we will have to use the equations twice: once for motion in the x direction and another time for the y direction.  

v_f=v_o + at ……..(a)

 [where v_f and v_o are final velocity and initial velocity, respectively]

Now ,

Initially, there was y velocity, however gravity began to act on the football, causing it to accelerate.  

Applying this value in equation (a)  

v_yf = at = -9.81 m/s^s * 1.75 = -17.165 m/s in the y direction  

For calculating the magnitude of the equation we have to square root the given value

         (16.6i - 17.165y)  

\\ \left | V  \right |=sqrt{16.6^{2}+17.165^{2}}\\ = \sqrt{275.56+294.637225}\\= \sqrt{570.197225}\\= 23.87[/tex]

5 0
3 years ago
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Room temperature water is placed in an Erlenmyer flask and heated to the boiling point. After the flask is removed from the heat
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Explanation:

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What is mesothermal climates
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1 year ago
A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

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2gH = v^{2}

v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

65.0\times 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

Now, time taken for the fall:

h = \frac{1}{2}gt^{2}

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

7 0
3 years ago
Read 2 more answers
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