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2 years ago

How is elastic energy related to mechanical energy?

1 answer:

6 0

Elastic energy is potential energy and mechanical energy is kinetic energy.

Many times, elastic energy is a consequence of mechanical energy. When energy is transferred to an object through mechanical energy, it can often displace or deform the object. This stored energy because of the deformation is called elastic energy.

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Name the fundamental units involved in pascal

gtnhenbr [62]

Explanation:

Hey there!

Here,

Pascal is a unit of pressure.

$pressure = \frac{f}{a}$

Now, As per the formula the units are:

kg, m and s^2.

Hope it helps...

5 0

3 years ago

on june 21 some earth locations have 24 hours of daylight these locations are all between the latitudes of

andrew-mc [135]

66° N and 90° N

the area of the artic circle in the northern hemisphere

4 0

2 years ago

On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const

IceJOKER [234]

Answer:

Explanation:

Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

$x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} (1)$

Since the car starts from rest, v₀ =0.
We know the value of t = 5 sec., but we need to find the value of a.
Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

$a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2 (2)$

Replacing a and t in (1):

$x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m. (3)$

Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

$a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2 (4)$

Replacing v₀, at and t in (1), we have:

$x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m (5)$

Therefore, as the truck travels twice as far as the car, the right answer is a).

7 0

2 years ago

Q11) If you were standing at the top of a building and you dropped a rock.

Dafna1 [17]

Answer:

Part A

The distance travel by the rock is approximately 132.496 m

Part B

The speed when the rock hits the ground is approximately 50.96 m/s

Explanation:

Part A

The question is focused on the kinetics equation of a free falling object

The given parameter is the time it takes the rock to hit the ground, t = 5.2 s

For an object in free fall, we have;

h = 1/2·g·t²

Where;

h = The height from which the object is dropped

g = The acceleration due to gravity ≈ 9.8 m/s²

t = The time taken to travel the distance, h = 5.2 s

∴ h = 1/2 × 9.8 m/s² × (5.2 s)² ≈ 132.496 m

The distance travel by the rock, h ≈ 132.496 m

Part B

The speed, 'v', when the rock hits the ground, is given by the following kinematic equation,

v = g·t

∴ v = 9.8 m/s² × 5.2 s = 50.96 m/s

The speed when the rock hits the ground, v ≈ 50.96 m/s.

8 0

2 years ago

Does water expand more as the temperature drops even more below 0 deg? I am worried about the coolant in my rad. as the temp.is

alex41 [277]

Water is, indeed, an exception. Normally when temperature drops, material shrinks. Water doesn't, water expands instead.

4 0

3 years ago