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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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The specification limits for a product are 8 cm and 10 cm. A process that produces the product has a mean of 9.5 cm and a standa
My name is Ann [436]

Answer:

The value of Cpk is 0.83.

Explanation:

Given that,

Upper specification limits = 10 cm

lower specification limits = 8 cm

Mean = 9.5

Standard deviation = 0.2 cm

We need to calculate the process capability

Using formula of Cpk

Cpk=min(\dfrac{USL-mean}{3\times SD}, \dfrac{mean-LSL}{3\times SD})

Put the value into the formula

Cpk=min(\dfrac{10-9.5}{3\times0.2}, \dfrac{9.5-8}{3\times0.2})

Cpk=min(0.83,2.5)

Cpk=0.83

Hence, The value of Cpk is 0.83.

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3 years ago
Given the vector A with components Ax = 2.00, Ay = 6.00, the vector B with components Bx = 2.00, By = 22.00, and the vector D =
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Answer:

<em>The magnitude of vector d is 16 and the angle with the x-axis is 270°</em>

Explanation:

<u>Operations With Vectors</u>

Given two vectors in rectangular components:

\vec a=(ax,ay)\ ,\  \vec b=(bx,by)

The sum of the vectors is:

\vec a+\vec b=(ax+bx,ay+by)

The difference between the vectors is:

\vec a-\vec b=(ax-bx,ay-by)

The magnitude of \vec a is:

|\vec a|=\sqrt{ax^2+ay^2}

The angle \vec a makes with the horizontal positive direction is:

\displaystyle \tan\theta=\frac{ay}{ax}\\

The question provides the vectors:

\vec a=(2,6)

\vec b=(2,22)

\vec d=\vec a-\vec b

Calculate:

\vec d=(2,6)-(2,22)=(0,-16)

The magnitude of \vec d is:

|\vec d|=\sqrt{0^2+(-16)^2}=\sqrt{0+256}=16

The angle is calculated by:

\displaystyle \tan\theta=\frac{-16}{0}

The division cannot be calculated because the denominator is zero. We need to estimate the correct angle by looking at the components of the vector. Since the x-coordinate is zero and the y-coordinate is negative, the vector points downwards (south), thus the angle must be -90° or 270° if the range goes from 0° to 360°.

The magnitude of vector d is 16 and the angle with the x-axis is 270°

4 0
3 years ago
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