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meriva
3 years ago
5

A train travels 250 km westward from Carthage to Johnson City. The train arrives 2.5 hours after it left. What was the average v

elocity of the train between the two stations?
A) 2.5 km/h west
B) 25 km/h west
C) 100 km/h west
D) 250km/h west
Physics
1 answer:
den301095 [7]3 years ago
5 0
The answer is C 100 km/h west
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Is it possible for a circuit to be connected both in series and in parallel?
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3 years ago
A spherical soap bubble with a surface-tension of 0.005 lbf/ft is expanded from a diameter of 0.5 in to 3.0 in. How much work, i
Sladkaya [172]

Answer:W = 1.23×10^-6BTU

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A2= final surface area

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D1= 0.0417 ft

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8 0
3 years ago
A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m
madam [21]

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR   where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

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I₂  = .8874 + .045 x (2 x .95 / 3)²

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4 0
3 years ago
A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap
viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

V_r + V_L + V_c = V_{net}

now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

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c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

5 0
3 years ago
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