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KengaRu [80]
3 years ago
10

A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is c

ompressed a distance 2x. How much faster does the second dart leave the gun compared to the first? The book gives an answer of "two times as much" but I not totally clear on what equation this is coming from. Is it the equation: W(ext) = 1/2mvf^2 - 1/2mvi^2 where the initiial velocity would be 0?
Physics
1 answer:
bezimeni [28]3 years ago
7 0

Answer:

The second dart leaves the gun two times as faster than the first one.

Explanation:

Assuming no energy loss during the spring-dart energy transfer, we have by the conservation of energy principle

U_s = K_d \\ \frac{1}{2} kx^2 = \frac{1}{2}mv^2 \\ v = \sqrt{\frac{k}{m}x^2}.

Given an arbitrary x and its double, 2x, launch velocities are

v_1 = \sqrt{\frac{k}{m}x^2} \text{ and} \\ v_2 = \sqrt{\frac{k}{m}\left(2x\right)^2} = \sqrt{\frac{k}{m}4x^2} = 2\sqrt{\frac{k}{m}x^2} = \mathbf{2v_1}.

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A 5.00 kg object oscillates back and forth at the end ofa spring whose spring constant is 49.3 N/m. An obersever istraveling at
defon

Answer:

5.571 sec

Explanation:

angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s

Period To = 2π / angular frequency

Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got

T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec

t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been  ( To / (√ (1 - (v²/c²))).  where To = 2.00 sec

8 0
3 years ago
If an object were released in space far away from planted or stars and given an initial momentum, describe what would happen to
mamaluj [8]
Strange as it may seem, the object would keep moving, in a straight line and at the same speed, until it came near another object. Its momentum and kinetic energy would never change. It might continue like that for a billion years or more.

Have a look at Newton's first law of motion.
7 0
3 years ago
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A. What evidence have you discovered to explain how
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Answer:

Well if it gets warmer the ice will weaken and if it gets hot enough will melt. Of course if the weather gets colder the ice will thicken.

Explanation:

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3 years ago
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When light bulbs A and B are connected in series to a battery, A glows brightly and B glows dimly. You remove bulb B so that the
torisob [31]

Answer:

The bulb B glows brighter.

Explanation:

Given that,

A glows brightly and B glows dimly.

According to ohm's law,

Two light bulbs A and B are connected in series to a battery then the current will be same in both bulbs and the resistance is high of bulb A and low in bulb B.

If bulb A connect to a battery and bulb B connect to a same battery separately.

Then bulb B glows brighter because the resistance is high in bulb A so the current will be low.

The resistance is low in bulb B so the current will be high.

Hence, The bulb B glows brighter.

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3 years ago
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A small glass bead has been charged to 20 nC. What is the magnitude of acceleration in m/s^2 of an electron that is 1.0 cm from
MAVERICK [17]

Answer:

The acceleration is 3.16x10¹⁷ m/s².

Explanation:

First, we need to find the magnitude of the Coulombs force (F):

|F| = \frac{Kq_{1}q_{2}}{d^{2}}

<u>Where</u>:

K is the Coulomb constant = 9x10⁹ Nm²/C²

q₁ is the charge = 20x10⁻⁹ C  

q₂ is the electron's charge = -1.6x10⁻¹⁹ C

d is the distance = 1.0 cm = 1.0x10⁻² m

|F| = \frac{Kq_{1}q_{2}}{d^{2}} = \frac{9\cdot 10^{9}Nm^{2}/C^{2}*20 \cdot 10^{-9} C*(-1.6\cdot 10^{-19} C)}{(0.01 m)^{2}} = 2.88 \cdot 10^{-13} N                                      

Now, we can find the acceleration:

a = \frac{F}{m} = \frac{2.88 \cdot 10^{-13} N}{9.1 \cdot 10^{-31} kg} = 3.16 \cdot 10^{17} m/s^{2}

Therefore, the acceleration is 3.16x10¹⁷ m/s².

I hope it helps you!    

7 0
3 years ago
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