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valentinak56 [21]
3 years ago
12

Explain two scenarios where a large truck can have the same momentum as a small car.

Physics
1 answer:
KengaRu [80]3 years ago
5 0

The momentum, p, of any object having mass m and the velocity v is

p=mv\cdots(i)

Let M_L and M_S be the masses of the large truck and the car respectively, and V_L and V_S be the velocities of the large truck and the car respectively.

So, by using equation (i),

the momentum of the large truck = M_LV_L

and the momentum of the small car = M_SV_S.

If the large truck has the same momentum as a small car, then the condition is

M_LV_L=M_SV_S\cdots(ii)

The equation (ii) can be rearranged as

\frac {M_L}{M_S}=\frac {V_S}{V_L} \; or \; \frac{M_L}{V_S}=\frac{M_S}{V_L}

So, the first scenario:

\frac {M_L}{M_S}=\frac {V_S}{V_L}

\Rhghtarrow M_L:M_S=V_S:V_L

So, to have the same momentum, the ratio of mass of truck to the mass of the car must be equal to the ratio of velocity of the car to the velocity of the truck.

The other scenario:

\frac{M_L}{V_S}=\frac{M_S}{V_L}

\Rhghtarrow M_L:V_S= M_S:V_L

So, to have the same momentum, the ratio of mass of truck to the velocity of the car must be equal to the ratio of mass of the car to the velocity of the truck.

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The burner on an electric stove has a power output of 2.0 kW. A 710 g stainless steel tea kettle is filled with 20∘C water and p
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Answer:

The volume of water that was in the kettle is  1170 cm^{3}

Explanation:

Given:

Power, P = 2.0 kW = 2000 W, Mass of stainless steel, m_{s} = 710 g = 0.71 kg at temperature of  20^{0} C

Part A:

If it takes time, t = 3.5 minutes to reach boiling point of water 100^{0} C, then from conservation of energy,

Total energy supplied by the burner = Total heat gained by the water and the stainless steel to rise from 20^{0} C to 100^{0} C

i.e. Pt = m_{s}c_{s}(100 - 20 ) + m_{w}c_{w}(100 - 20 )

m_{w} = \frac{pt - 80m_{s} C_{s} }{80c_{w} } = \frac{2000*3.5*60 - 80*0.71* 450}{80*4200}

m_{w} = 1.17 kg

where c_{w} = 4200 J/Kgk (specific heat capacity of water), c_{s} = 450 J/Kgk (specific heat capacity of steel)

But volume of water in the the kettle, v = \frac{mass}{density} = \frac{1.17}{1000}= 1.17 *10^{-3} m^{3}

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A projectile is fired at an upward angle of 45.0º from the top of a 265-m cliff with a speed of .sm 185 What will be its speed w
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The final velocity of the projectile when it strikes the ground below is 198.51 m/s.

<h3>Time of motion of the projectile</h3>

The time taken for the projectile to fall to the ground is calculated as follows;

h = vt + ¹/₂gt²

where;

  • h is height of the cliff
  • v is velocity
  • t is time of motion

265 = (185 x sin45)t + (0.5)(9.8)t²

265 = 130.8t + 4.9t²

4.9t² + 130.8t - 265 = 0

solve the quadratic equation using formula method,

t = 1.89 s

<h3>Final velocity of the projectile</h3>

vyf = vyi + gt

where;

  • vyf is the final vertical velocity
  • vyi is initial vertical velocity

vyf = (185 x sin45) + (9.8 x 1.89)

vyf = 149.322 m/s

vxf = vxi

where;

  • vxf is the final horizontal velocity
  • vxi is the initial horizontal velocity

vxf = 185 x cos(45)

vxf = 130.8 m/s

vf = √(vyf² + vxf²)

where;

  • vf is the speed of the projectile when it strikes the ground below

vf = √(149.322²  +  130.8²)

vf = 198.51 m/s

Learn more about final velocity here: brainly.com/question/6504879

#SPJ1  

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