A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:

PE = $\frac{1}{2}$ k $x^{2}$

Where PE is the energy, k is the spring constant and x is extension.

i. Given that: PE = 10 J and x = 10 cm, then;

PE = $\frac{1}{2}$ k $x^{2}$

10 = $\frac{1}{2}$ k $10^{2}$

20 = 100k

k = 0.2 J/cm

ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.

PE = $\frac{1}{2}$ k $x^{2}$

20 = $\frac{1}{2}$ (0.2) $x^{2}$

40 = 0.2 $x^{2}$

$x^{2}$ = 200

x = $\sqrt{200}$

= 14.1421

x = 14.14 cm

So that;

x is approximately 14.00 cm.

Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.

For more information, check at: brainly.com/question/1352053.

8 0

3 years ago

Let the displacement function of a particle is x(t)=(20t^2-15t+200). Find the total displacement, instantaneous velocity and ins

irga5000 [103]

Answer:

A.) 39.5 m

B.) 0

C.) 60m/s^2

Explanation:

Given that a displacement function of a particle is x(t)=(20t^2-15t+200).

To Find the total displacement,

Reduce everything by dividing them by 5

X(t) = 4t^2 - 3t + 40 ...... (1)

For instantaneous velocity, differentiate x(t). That is,

dy/dt = 60t - 15 ...... (2)

But dy/dt = velocity.

If dy/dt = 0, then

60t - 15 = 0

60t = 15

t = 15/60

t = 0.25s

Substitutes t in equation (1)

Total displacement will be

X(t) = 4(0.25)^2 - 3(0.25) + 40

X(t) = 0.25 - 0.75 + 40

Total displacement = 39.5 m

To calculate instantaneous velocity, substitute t into equation (2)

V = 60 (0.25) - 15

V = 0.

and to find instantaneous acceleration, differentiate dv/dt

dv/dt = 60

Therefore, acceleration = 60 m/s^2

4 0

3 years ago

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist

e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > $38 \frac{ft}{s^2} x \frac{1mi}{5280ft} x \frac{(3600s)^2}{(1h)^2} = 93272.27 \frac{mi}{h^2}$

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > $v_2^2=v_1^2 + 2as$

The initial velocity is 50mi/h $(v_1=50)$

When it stops the final velocity is $(v_2=0)$

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

$0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s = 2500/(185644.54)

s=0.0134
$

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > $0.0134 mi * \frac{5280ft}{1mi} = 70.8 ft$
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0

3 years ago

two clear, colorless liquids are poured together. A bright yellow solid forms. What physical properties have changed?

jek_recluse [69]

Change in state(from liquid to solid) and change in colour I believe.

3 0

3 years ago