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3 years ago

How would the elastic potential energy change if the displacement is multiplied by five?

1 answer:

3 0

Answer: Work is done when a spring is extended or compressed . Elastic potential energy is stored in the spring. The elastic potential energy stored can be calculated using the equation: elastic potential energy = 0.5 × spring constant × (extension) 2.

Explanation: can you please mark me as brainliest

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Light does not move infinitely fast but has a finite speed. We normally use ""c"" to indicate the speed of light in science. Wri

Ede4ka [16]

Answer:

6.71 × 10^8 mi/hr

Explanation:

Light is usually defined as an electromagnetic wave that is comprised of a definite wavelength. It is of both types, visible and invisible. The light emitted from a source usually travels at a speed of about 3 × 10^8 meter/sec. This speed of light is commonly represented by the letter 'C'.

To write it in the metric system, it has to be converted into miles/hour.

We know that,

1 minute = 60 seconds

60 minutes = 1 hour

1 kilometer = 1000 meter

1 miles = 1.6 kilometer

Now,

= $\frac{3 \times\ 10^8 meter \times\ 60 sec \times\ 60 min}{1 sec \times\ 1 min \times\ 1 hr}$

= 1.08 × 10^12 m/ hr (meter/hour)

= $\frac{1.08 \times\ 10^{12} meter \times\ 1 km \times\ 1 miles}{1 hr \times\ 1000 meter \times\ 1.6 km}$

= 6.71 × 10^8 mi/hr (miles/hour)

Thus, the value for speed of light (C) in metric unit is 6.71 × 10^8 mi/hr.

5 0

3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago

In a classroom demonstration, the pressure inside a soft drink can is suddenly reduced to essentially zero. Assuming the can to

umka2103 [35]

Answer:

3141N or 3.1 ×10³N to 2 significant figures. The can experiences this inward force on its outer surface.

Explanation:

The atmospheric pressure acts on the outer surface of the can. In order to calculate this inward force we need to know the total surface area of the can available to the air outside the can. Since the can is a cylinder with a total surface area given by 2πrh + 2πr² =

A = 2πr(r + h)

Where h = height of the can = 12cm

r = radius of the can = 6.5cm/2 = 3.25cm

r = diameter /2

A = 2π×3.25 ×(3.25 + 12) = 311.4cm² = 311.4 ×10-⁴ = 0.031m²

Atmospheric pressure, P = 101325Pa = 101325 N/m²

F = P × A

F = 101325 ×0.031.

F = 3141N. Or 3.1 ×10³ N.

5 0

3 years ago

Which describes an object in projectile motion? Select three options.

Karo-lina-s [1.5K]

Description of an object in projectile motion is;

Gravity acts to pull the object down.
The object’s inertia carries it forward.
The path of the object is curved.

Explanation:

The path of the projectile is usually curved, and NOT straight, due to the influence of gravity on it which is teh only force acting on it-, causing it motion path to fall towards the earth. Most projectiles follow a parabolic path. The projectile, even though it was launched, its motion is then only due to its own inertia – tendency to stay in motion in a straight line, or rest, unless an external force is acting on it - such as drag or friction. An example of such projectile motion is of ballistic missiles.

8 0

3 years ago

Read 2 more answers

Rx: hydrochloric acid 2 percent solution 500 ml. your stock solution of hydrochloric acid is 10 percent. how much of the stock s

Allisa [31]

100 ml

100 ml of the stock solution is required to prepare the order.

We know that C1V1 = C2V2

where C1= 2%

V1 = 500ml

C2= 10%

V2 = ?

V2 = C1V1 / C2

= 500 * 2% / 10%

=100

V2 = 100 ml

<h3>What is meant by stock solution?</h3>

A stock solution is a sizable amount of a typical reagent in a standardized concentration, like sodium hydroxide or hydrochloric acid.
This phrase is frequently used in analytical chemistry while doing operations like titrations where it's crucial to employ precise solution concentrations.

<h3>What distinguishes a standard solution from a stock solution?</h3>

The main distinction between stock solution and standard solution is that the former is a highly concentrated solution while the later is a solution whose concentration is precisely known.
Because standard solutions frequently arrive as stock solutions, the phrases "stock solution" and "standard solution" are connected.

To learn more about stock solution preparation visit:

brainly.com/question/14667249

#SPJ4

8 0

2 years ago