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Vikentia [17]
3 years ago
9

A 0.02kg dart is loaded into a toy spring gun by compressing the spring 6cm. If the spring has a constant of 20 N/m, find… a. Th

e work needed to compress the spring. (0.036 J) b. The total energy stored in the compressed spring. (0.036 J) c. The kinetic energy of the dart as it leaves the gun. (0.036 J) d. The speed of the dart as it leaves the gun. (1.90 m/s) e. The height the dart would go if shot straight up. (0.18 m)
Physics
1 answer:
katrin2010 [14]3 years ago
6 0

Answer:

a) 0.036 J b) 0.036J c) 0.036 d) 1.9m/s e) 0.18 m

Explanation:

Mass of the dart = 0.02kg, the spring was compressed to 6cm

Work needed to compress the spring = 1/2*k*x ^2 where k is the force constant of the spring in N/m, x is the distance it was compressed in m

Work needed to compress the spring = 0.5 * 20* 0.06^2 since 6cm = 6 / 100 = 0.06 m

Work needed to compress the spring = 0.036J

b) the total energy stored in the spring = the work done to compress the spring = 0.036J

c) kinetic energy of the dart as it leaves the the spring = elastic potential energy stored in the spring = the work done in compressing the = 0.036J using the law of conservation of energy; energy is neither created nor destroyed but transformed from one form to another.

d) 1/2mv^2 = 0.036

mv^2 = 0.036*2

v^2 = 0.036*2 / 0.02 = 3.6

v = √3.6 = 1.897 approx 1.9m/s

e) kinetic energy of the dart = work done against gravity to get the body to height h

Work done against gravity = potential energy conserved at height = -mgh g is negative because the motion is upward while gravity acts downward

0.036 = 0.02 * 9.81 * h

0.036 / ( 0.02*9.81) = h

h = 0.18 m

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3 years ago
2. A stone is thrown horizontally at a speed of 6.0 m/s from the top of a cliff 78.4 m high.
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3 years ago
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mrs_skeptik [129]
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F= 4x1.2
F= 4.8 N

F= 4gsin30 - Friction
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5 0
3 years ago
In the year 2000, a company made $4.7 million in profit. For each consecutive year after that, their profit increased by 15%. Ho
olya-2409 [2.1K]

Answer: $8.2 million.

Explanation:

If we have a quantity A, and we have an increase of x%, this can be written as:

A + (x%/100%)*A

Now, for this particular case we have:

In year 2000 (we can define this year as the year zero, y = 0) the initial value is $4.7 million.

The next year, y = 1, there is an increase of 15%, then we will have a profit of:

P = $4.7 million + (15%/100%)*$4.7 million = $4.7 million +  0.15*$4.7 million

P = $4.7 million*(1 + 0.15) = $4.7 million*(1.15)

in the next year, y = 2, the profit will be:

P =  $4.7 million*(1.15) + (15%/100%)* $4.7 million*(1.15)

  = $4.7 million*(1.15) + 0.15* $4.7 million*(1.15)

   = $4.7 million*(1.15)^2

We already can see the pattern, the profit in the year y will be:

P(y) = $4.7 million*(1.15)^y

In particular, in the year 2004 we have y = 4, then the profit that year will be:

P(4) = $4.7 million*(1.15)^4 = $8.2 million.

8 0
3 years ago
A person walks 25 m east 35 m north 25 me west and 5 m south. What's the distance traveled?
Zolol [24]
With these questions, drawing it out would always help, the answer for this would be 90m if you add them all up. If it’s displacement, it would be 30m. But since it’s asking for the distance TRAVELED then it’s 90m

ANSWER: 90m

8 0
3 years ago
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