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mestny [16]
2 years ago
15

2. A 4.0 kg magnetic toy car traveling at 3.0 m/s east collides and sticks to a 5.0 kg toy magnetic car also traveling at 2.0 m/

s east. Calculate the final speed and direction of the magnetic car (coupled) system?
Physics
1 answer:
Afina-wow [57]2 years ago
4 0

Answer:

2.44 m/s due East

Explanation:

From the question given above, the following data were obtained:

Mass of 1st car (m₁) = 4 Kg

Velocity of 1st car (u₁) = 3 m/s

Mass of 2nd car (m₂) = 5 Kg

Velocity of 2nd car (u₂) = 2 m/s

Final velocity (v) =?

The final velocity can be obtained as follow:

v(m₁ + m₂) = m₁u₁ + m₂u₂

v(4 + 5) = (4×3) + (5×2)

9v = 12 + 10

9v = 22

Divide both side by 9

v = 22/9

v = 2.44 m/s

Thus, the final velocity is 2.44 m/s.

Since both cars was moving due East before collision, and after collision, they stick together, then their direction will be due East.

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(a) What is the cost of heating a hot tub containing 1440 kg of water from 10.0°C to 40.0°C, assuming 75.0% efficiency to take h
BaLLatris [955]

Answer:

a) E = 6.024\,USD, For m kilograms, it is 4184m J., 3600000 joules, b) i = 88.200\,A

Explanation:

a) The amount of heat needed to warm water is given by the following expression:

Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})

Where:

m_{w} - Mass of water, measured in kilograms.

c_{w} - Specific heat of water, measured in \frac{J}{kg\cdot ^{\circ}C}.

T_{f}, T_{i} - Initial and final temperatures, measured in ^{\circ}C.

Then,

Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)

Q_{needed} = 180748800\,J

The energy needed in kilowatt-hours is:

Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)

Q_{needed} = 50.208\,kWh

The electric energy required to heat up the water is:

E = \frac{50.208\,kWh}{0.75}

E = 66.944\,kWh

Lastly, the cost of heating a hot tub is: (USD - US dollars)

E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)

E = 6.024\,USD

The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.

b) The current required for the electric heater is:

i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}

i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}

i = 88.200\,A

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