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Lemur [1.5K]
2 years ago
9

Jay bought a toy car. To make the car move, he must turn a key attached to a spring. Turning the key winds the spring tight. The

toy car moves when the key is released and the spring unwinds. What’s the main transformation of energy in this example?
Physics
2 answers:
kobusy [5.1K]2 years ago
7 0
The main transformation of energy is from the chemical energy in Jay's body to kinetic energy in the car. The break up of the conversions is:

Chemical energy from Jay's body to elastic potential energy within the spring. Next, the spring releases and the elastic potential energy is converted into kinetic energy as the toy car moves forward.
faust18 [17]2 years ago
3 0

Answer:

Elastic potential energy into kinetic energy

Explanation:

Initially the energy is stored inside the spring, which is compressed. This form of energy is called elastic potential energy, and its formula is

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

where k is the spring constant, which gives the 'strength' of the spring, while x is the compression/stretching of the spring with respect to its equilibrium position.

When the spring unwinds, it returns to its equilibrium position, so x becomes zero and the potential energy converts into another form of energy, which is related to the motion of the car (in fact, the car starts moving). This form of energy is called kinetic energy, and its formula is

K=\frac{1}{2}mv^2

where m is the mass of the car and v is its speed.

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Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o
IRINA_888 [86]

Answer:

a) W=-0.0103125\ J

b) W=0.0059375\ J

c) Compressing is easier

Explanation:

Given:

Expression of force:

F=kx-bx^2+cx^3

where:

k=100\ N.m^{-1}

b=700\ N.m^{-2}

c=12000\ N.m^{-3}

x when the spring is stretched

x when the spring is compressed

hence,

F=100x-700x^2+12000x^3

a)

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, x=-0.05\ m

Now,

The spring constant at this instant:

j=\frac{F}{x}

j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}

j=-8.25\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times -8.25\times (-0.05)^2

W=-0.0103125\ J

b)

When compressing the spring by 0.05 m

we have, x=0.05\ m

<u>The spring constant at this instant:</u>

j=\frac{F}{x}

j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}

j=4.75\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times 4.75\times (0.05)^2

W=0.0059375\ J

c)

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0
1 year ago
Yasmin determines the velocity of a car to be 15 m/s. The accepted value for the velocity of the car is 15.6 m/s. What is Yasmin
snow_lady [41]

Answer:

3.8%

Explanation:

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6 0
2 years ago
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one horsepower is a unit of power equal to 746w. how much energy can a 150-horsepower engine transform in 10.0s?
Dafna1 [17]

1 watt = 1 joule/second

1 horsepower = 746 watts = 746 joule/second

   (150 horsepower) x (746 watt/HP) x (1 joule/sec  /  watt) x (10 sec)

=  (150 x 746 x 1 x 10)  joule  =  1,119,000 joules .   
if correct plz mark brainly
8 0
1 year ago
Recent findings in astrophysics suggest that the observable universe can be modeled as a sphere of radius R = 13.7 × 109 light-y
-BARSIC- [3]

Answer:

3.7\times 10^{51}) kg

Explanation:

R = radius of the sphere modeled as universe = 13\times 10^{25} m

Volume of sphere is given as

V = \frac{4\pi R^{3}}{3}

V = \frac{4(3.14) (13\times 10^{25})^{3}}{3}

V = 9.2\times 10^{78} m³

\rho = average total mass density of universe = 1\times 10^{-26} kg/m³

m = Total mass of the universe = ?

We know that mass is the product of volume and density, hence

m = \rho V

m = (1\times 10^{-26}) (9.2\times 10^{78})

m = 9.2\times 10^{52} kg

M = mass of "ordinary" matter  = ?

mass of "ordinary" matter is only about 4% of total mass, hence

M = (0.04) m

M = (0.04)(9.2\times 10^{52})

M = 3.7\times 10^{51} kg

6 0
2 years ago
Greg walks on a straight road from his home to a convenience store 3.0 km away with a speed of 6.0 km/h. On reaching the store h
VladimirAG [237]

This question was apprently selected from the "Sneaky Questions" category.

The store is 3 km from his home, and he walks there with a speed of 6 km/hr.  So it takes him (3 km) / (6 km/hr)  =  1/2 hour to get to the store.

That's 30 minutes.  So the whole part-(a.) of the question refers to only that part of the trip, and we don't care what happens once he reaches the store.  

a). Over the first 30 minutes of his travel, Greg walks 3.0 km on a straight road, and he ends up 3.0 km away from where he started.

Average speed = (distance/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>

Average velocity = (displacement/time) = (3.0 km) / (1/2 hour) = <em>6.0 km/hr</em>

There's probably some more questions in part-(b.) where you'd need to use Greg's return trip to find the answers, but johnaddy210 is only asking us for part-(a.).

8 0
2 years ago
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