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NISA [10]
3 years ago
10

A light train made up of two cars is traveling at 100 km/h when the brakes are applied to both cars. Knowing that car A has a ma

ss of 35 Mg and car B a mass of 20 Mg, and that the braking force is 35 kN on each car, determine (a) the distance traveled by the train before it comes to a stop, (b) the force in the coupling between the cars while the train is showing down.

Physics
1 answer:
insens350 [35]3 years ago
8 0

Answer:

9545N

Explanation:

See attached file

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a hot piece of copper is placed in an insulated cup. what is the final temperature of the water and copper?
slavikrds [6]

Answer:

Option C. 30°C.

Explanation:

The following data were obtained from the question:

Mass of water (Mw) = 0.5 Kg

Specific heat capacity of water (Cw) = 4.18 KJ/Kg°C

Initial temperature of water (Tw1) =

22°C

Change in temperature (ΔT) = T2 – Tw1 = T2 – 22°C

Mass of copper (Mc) = 0.5 Kg

Specific heat capacity of copper (Cc) = 0.386 KJ/kg°C

Initial temperature of copper (Tc1) = 115°C

Change in temperature (ΔT) = T2 – Tc1 = T2 – 115°C

Final temperature (T2) =..?

Note: Both the water and the piece copper will have the same final temperature and the heat will be zero since the water will cool the piece of copper.

Thus, we can determine the final temperature of the water and copper as follow:

Q = MwCwΔT + McCcΔT

0 = 0.5 x 4.18 x (T2 – 22) + 0.5 x 0.386 x (T2 – 115)

0 = 2.09 (T2 – 22) + 0.193 (T2 – 115)

0 = 2.09T2 – 45.98 + 0.193T2 – 22.195

Collect like terms

2.09T2 + 0.193T2 = 45.98 + 22.195

2.283T2 = 68.175

Divide both side by the coefficient of T2 i.e 2.283

T2 = 68.175/2.283

T2 = 29.8 ≈ 30°C

Therefore, the final temperature of water and copper is 30°C.

8 0
2 years ago
An automobile starter motor has an equivalent resistance of 0.0510 Ω and is supplied by a 12.0 V battery with a 0.0090 Ω interna
Alisiya [41]

Answer:

a) 200A

b) 10.2V

c) 2.04kW

d)

I=80A

V=4.08V

P=0.326kW

Explanation:

Here we have a circuit of one power source and two resistors in series, the first question is asking for the current, so according to Ohm's Law:

I=\frac{V}{R}

Where R is the equivalent resistance of the resistors in series

R=0.0510+0.0090=0.0600[ohm]

I=\frac{12.0}{0.0600}=200A

To calculate the voltage dropped by the motor we have to apply the voltage divider rule:

V_m=V*\frac{R_m}{R_m+R_s}\\V_m=12.0*\frac{0.0510}{0.0600}\\V_m=10.2V

The power dissipated supplied to the motor is given by:

P=I^2*R_m\\P=(200)^2*0.0510=2.04kW

now solving adding a 0.0900 ohm resistor:

I=\frac{12.0}{0.15}=80A

V_m=12.0*\frac{0.0510}{0.15}\\V_m=4.08V

P=I^2*R_m\\P=(200)^2*0.0510=0.326kW

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