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Vadim26 [7]
3 years ago
6

Scenario A: A 3 spring is compressed a distance of 1.0 m. Scenario B: A 6 spring is compressed a distance of 0.8 m. Scenario C:

A 9 spring is compressed a distance of 0.6 m. Scenario D: A 12 spring is compressed a distance of 0.4 m. Which scenario generates the most elastic potential energy?
Physics
2 answers:
lapo4ka [179]3 years ago
6 0

The CORRECT answer is B

Westkost [7]3 years ago
5 0

Answer:

Scenario B)

Explanation:

The elastic potential energy stored in a spring is given by:

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

where k is the spring constant and x is the compression/stretching of the spring.

Let's calculate the potential energy of each spring by using this formula:

A) U=\frac{1}{2}(3 N/m)(1.0 m)^2=1.5 J

B) U=\frac{1}{2}(6 N/m)(0.8 m)^2=1.92 J

C) U=\frac{1}{2}(9 N/m)(0.6 m)^2=1.62 J

D)  U=\frac{1}{2}(12 N/m)(0.4 m)^2=0.96 J

Therefore, the scenario with most elastic potential energy is B.

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3 years ago
Balanced forces can change an<br> object's direction.<br> A. True<br> B. False
makvit [3.9K]

Answer:

False

Explanation:

Balanced forces result in a net force of 0N. This means no direction or acceleration change will be applied to the object. A torque may be applied, but with no other external forces, the object will not move.

8 0
3 years ago
(TCO 5) The relationship between Celsius (º C) and Fahrenheit (º F) degree of measuring temperature is linear. Find the linear e
gtnhenbr [62]

Answer:

C=-\dfrac{10}{17}(F-32)

Explanation:

Given that

32° F corresponds to 0 °C.  ---Point 1

212° F corresponds to 100 °C.----Point 2

We know that if two point is given that equation of line can be found as

y-y_1=\dfrac{y_2-y_1}{x_2-x_2}(x-x_1)

Lets C in y- direction and F in x- direction,so we can say that

C-C_1=\dfrac{C_2-C_1}{F_2-F_2}(F-F_1)

C-0=\dfrac{100-0}{32-212}(F-32)

C=-\dfrac{10}{17}(F-32)

So the linear relationship is

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8 0
3 years ago
Which question requires the collection of data to answer it?
iren [92.7K]

Answer:

a

Explanation:

b, c, and d are all opinion based, a is the only one that you need factual evidence and observations.

3 0
2 years ago
A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic
barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, \Delta E=\dfrac{1}{2}m(v^2-u^2)

\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)

\Delta E=170418.5\ J

\Delta E=1.7\times 10^5\ J

(b) Change in momentum of the truck, \Delta p=m(v-u)

\Delta p=2000\ kg\times (16.11-9.44)

\Delta p=13340\ kg-m/s

Hence, this is the required solution.

6 0
3 years ago
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