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zhenek [66]
3 years ago
9

Johnny stretches a spring 40 cm from its resting position. The spring’s constant of proportionality is 20 N/m. What is the sprin

g potential energy?
J=
Physics
1 answer:
sergiy2304 [10]3 years ago
6 0

The spring potential energy is 1.6 J

Explanation:

The elastic potential energy of a spring is given by the equation:

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

where

U is the potential energy

k is the spring constant

x is the compression/stretching of the spring

For the spring in this problem, we have:

x = 40 cm = 0.40 m is the stretching

k = 20 N/m is the spring constant

Substituting into the equation, we can find the spring potential energy:

U=\frac{1}{2}(20)(0.40)^2=1.6 J

#LearnwithBrainly

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7nadin3 [17]
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5 0
3 years ago
A car drives over a hilltop that has a radius of curvature 0.120 km at the top of the hill. At what speed would the car be trave
Mashutka [201]

Answer:

34.3 m/s

Explanation:

Newton's Second Law states that the resultant of the forces acting on the car is equal to the product between the mass of the car, m, and the centripetal acceleration a_c (because the car is moving of circular motion). So at the top of the hill the equation of the forces is:

mg-R = m a_c = m\frac{v^2}{r}

where

(mg) is the weight of the car (downward), with m being the car's mass and g=9.8 m/s^2 is the acceleration due to gravity

R is the normal reaction exerted by the road on the car (upward, so with negative sign)

v is the speed of the car

r = 0.120 km = 120 m is the radius of the curve

The problem is asking for the speed that the car would have when it tires just barely lose contact with the road: this means requiring that the normal reaction is zero, R=0. Substituting into the equation and solving for v, we find:

v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(120 m)}=34.3 m/s

6 0
3 years ago
Solve the problem using dimensional analysis. Every number must have a unit. Work must be shown. Conversion factors are in the t
anygoal [31]

The number of grams of gold you can purchase for $15 if Gold cost $8.00 per ounce is 53.26g

Dimensional analysis is a means of representing values using units.

From the question, we are told that Gold cost $8.00 per ounce, this means that:

$8.00 = 1 ounce

In order to calculate the amount in grams you can purchase for $15, we need to know the price in ounces first.

$15.00 = x

Divide both expressions

\dfrac{8}{15} = \dfrac{1}{x}\\x = \dfrac{15}{8}\\x=  1.875 ounces

Next is to convert the ounces to pounds.

1 pound = 16 ounces

y = 1.875 ounces

y = 1.875/16

y = 0.1171875 pounds

Conver 0.1171875 pounds to kg

Since 1kg = 2.2 pounds

z = 0.1171875 pounds

2.2z = 0.1171875

z = 0.1171875/2.2

z = 0.053267kg

Finally, convert 0.053267kg to grams using the conversion rate

1kg = 1000g

0.053267kg = y

y =  0.05326 * 1000

y = 53.26g

This shows that the amount of grams you can purchase for $15 is 53.26g

Learn more here: brainly.com/question/11819069

5 0
3 years ago
Find the circumference of a circle whose area is 452.16 square meters. (Use 3.1416 as the value of .)
kakasveta [241]

1) The area of a circle is given by

A=\pi r^2

where r is the radius of the circle. In this problem, we are told that the area is

A=452.16 m^2

so we can rearrange the previous equation to calculate the radius:

r=\sqrt{\frac{A}{\pi}}=\sqrt{\frac{452.16 m^2}{3.1416}}=12.0 m


And now we can find the circumference, by using the formula:

c=2 \pi r=2 \pi (12.0 m)=75.3984 m

Therefore, the correct answer is D. 75.3984 meters.



2) The volume of a cube is given by:

V=L^3

where L is the length of the edge. In this problem, L=6 cm, therefore the volume of the cube is

V=L^3=(6 cm)^3=216 cm^3


and so, the correct answer is A. 216 cu. cm.

5 0
4 years ago
The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de
spin [16.1K]

Answer:

the spring constant k = 5.409*10^4 \ N/m

the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

Explanation:

From Hooke's Law

F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

Thus; the spring constant k = 5.409*10^4 \ N/m

The amplitude is decreasing 37% during one period of the motion

e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}

b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) }    }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s

Therefore; the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

5 0
4 years ago
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