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MrMuchimi
4 years ago
15

A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put

on the launcher, the launcher spring compresses by 10.0 cm. The watermelon is then pushed down by an additional 30.0 cm and it’s ready to go. Just before the launch, how much energy is stored in the spring?
Physics
1 answer:
olga_2 [115]4 years ago
6 0

To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.

Since the forces are balanced the Spring force is equal to the force of the weight that is

F_s = F_g

kx = mg

Where,

k = Spring constant

x = Displacement

m = Mass

g = Gravitational Acceleration

Re-arrange to find the spring constant

k = \frac{mg}{x}

k = \frac{8*9.8}{0.1}

k = 784N/m

Just before launch the compression is 40cm, then from Potential Elastic Energy definition

PE = \frac{1}{2} kx^2

PE =\frac{1}{2} 784*0.4^2

PE = 63.72J

Therefore the energy stored in the spring is 63.72J

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A simple generator has a square armature 9.0 cm on a side. The armature has 95 turns of 0.59-mm-diameter copper wire and rotates
Marrrta [24]

Answer:

f = 3.102 Hz

Explanation:

In this case you have that the required voltage is the maximum induced emf produced by the rotating generator.

In order to calculate the frequency of rotation of the generator that allows one to obtain 12.0V you use the following formula:

emf_{max}=NBA\omega    (1)

N: turns of the armature = 95

B: magnitude of the magnetic field = 0.800T

A: area of the square armature = (9.0cm)^2 = (0.09m)^2 = 8.1*10^-3 m^2

emf_max = 12.0V

w: angular frequency

you solve the equation (1) for w:

\omega=\frac{emf_{max}}{NBA}=\frac{12.0V}{(95)(0.800T)(8.1*10^{-3}m^2)}\\\\\omega=19.49\frac{rad}{s}

Then, the frequency is:

f=\frac{\omega}{2\pi}=\frac{19.49rad/s}{2\pi}=3.102Hz

7 0
3 years ago
You take a Frisbee to the top of the Washington Monument and send it sailing along horizontally at a speed of 5 m/s from the ver
Dominik [7]

This question can be solved with the help of the equations of motion.

A) The Frisbee will remain in the air for "5.87 s".

B) The frisbee will go "29.4 m" down the range.

A)

To calculate the time, the frisbee will remain in the air, we will use the second <em><u>equation of motion</u></em>, for the vertical motion.

h = v_it+\frac{1}{2}gt^2\\\\

where,

h = height = 169.2 m

vi = initial velocity's vertical component = 0 m/s

g = acceleration due to gravity = 9.81 m/s²

t = time = ?

Therefore,

169.2\ m = (0\ m/s)(t)+\frac{1}{2}(9.81\ m/s^2)t^2\\\\t = \sqrt{\frac{(169.2\ m)(2)}{9.81\ m/s^2}}

<u>t = 5.87 s</u>

<u />

B)

Now, we will calculate the horizontal range by applying the equation for constant motion. Because the velocity in the horizontal direction will remain constant due to no air resistance

s = vt

where,

s = horizontal range = ?

v= initial velocity's horizontal component = 5 m/s

t = time = 5.87 s

Therefore,

s = (5 m/s)(5.87 s)

<u>s = 29.4 m</u>

<u />

Learn more about <em><u>equations of motion</u></em> here:

brainly.com/question/9772550?referrer=searchResults

8 0
3 years ago
In a double-slit experiment, it is observed that the distance between adjacent maxima on a remote screen is 1.0 cm. What happens
kobusy [5.1K]

Answer:

C) It increases to 2.0 cm

Explanation:

In a double-slit diffraction experiment, the distance on the screen between two adjacent maxima is given by

\Delta y = \frac{\lambda D}{d}

where

\lambda is the wavelength of the wave

D is the distance of the screen from the slits

d is the separation between the slits

In this problem, the initial distance between adjacent maxima is 1.0 cm. Later, the slit separation is cut in a half, which means that the new slit separation is

d'=\frac{d}{2}

Substituting into the equation, we find that the new separation between the maxima is

\Delta y' = \frac{\lambda D}{d/2}=2(\frac{\lambda D}{d})=2\Delta y

So, the distance increases by a factor 2: therefore, the new separation between the maxima will be 2.0 cm.

5 0
3 years ago
How many times did thomas edison mess up the light bulb
Umnica [9.8K]
<h2>Answer: 10,000 times </h2>

Thomas Alva Edison was a North American scientist and inventor who is credited with the design of the electric bulb and many other patented inventions.

However, the bulb did not come out on the first attempt, and was sometimes questioned by his repeated "failures". But his patience and perseverance paid off, after approximately 10,000 attempts, he managed to obtain an incandescent bulb that would last for a long time.

It should be noted that Edison was once asked if he considered that he had had many failures before making the bulb work, hence his famous phrase:

<h2><em>“I have not failed 10,000 times—I’ve successfully found 10,000 ways that will not work.” </em></h2>

That is why Edison is considered an example of perseverance.

4 0
3 years ago
When a baseball is hit, it travels around 65 mps (meters per second). the mass of the baseball is 0.145 kg. what is the kinetic
nika2105 [10]

Answer:

306J

Explanation:

The formula for kinetic energy is given as ½mv².

Using this, we get our answer by substituting the values into the formula.

½ × 0.145kg × 65m/s

= 306.31J

can be rounded off to 306J

4 0
2 years ago
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