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2 years ago

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A 1500-kg car traveling due north with a speed of 25 m/s collides head-on with a 4500-kg truck traveling due south with a speed

of 15 m/s. The two vehicles stick together after the collision. What is the total momentum of the system prior to the collision

1 answer:

Goryan [66]2 years ago

8 0

Answer:

OMG IM ON THE SAME QUESTION

Explanation:

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What is the theory of light?

Lostsunrise [7]

Answer:

Corpuscular theory of light

Explanation:

In optics, the corpuscular theory of light, arguably set forward by Descartes in 1637, states that light is made up of small discrete particles called "corpuscles" which travel in a straight line with a finite velocity and possess impetus. This was based on an alternate description of atomism of the time period.

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3 years ago

1. Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless

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Since momentum is a vector quantity, take any direction as positive and other as negative. Answer won't change.

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Current that moves in one direction from negative to positive. May be created by a battery. Is generally NOT found in U.S. elect

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That's "<em>DC</em>" . . . Direct Current .

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You are on a school bus driving north at 20 m/s. You walk toward the back of the bus with a velocity of 6 m/s. What is your velo

Ganezh [65]

I answered the question but it got deleted?? why?

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2 years ago

A circuit consists of a series combination of 6.50 −kΩ and 4.50 −kΩ resistors connected across a 50.0-V battery having negligibl

Vlada [557]

Answer:

Part A: 16.1 V

Part B: 20.5 V

Part C: 21.5%

Explanation:

The voltmeter is in parallel with the 4.5-kΩ resistor and the combination is in series with the 6.5-kΩ resistor. The equivalent resistance of the parallel combination is given as

$\dfrac{1}{R_E}=\dfrac{1}{4.50}+\dfrac{1}{10.0}$

$R_E=\dfrac{4.50\times10.0}{4.50+10.0} = 3.10$

Part A

The voltmeter reading is the potential difference across the parallel combination. This is found by using the voltage-divider rule.

$V_1 = \dfrac{3.10}{3.10+6.50}\times50.0 = \dfrac{3.10}{9.60}\times50.0 = 16.1 \text{ V}$

Part B

Without the voltmeter, the potential difference across the 4.5-kΩ resistor is found using the same rule as above:

$V_2 = \dfrac{4.50}{4.50+6.50}\times50.0 = \dfrac{4.50}{11.0}\times50.0 = 20.5 \text{ V}$

Part C

The error in % is given by

$\dfrac{20.5-16.1}{20.5}\times100\% = \dfrac{4.4}{20.5}\times100\% = 21.5\%$

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2 years ago

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