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Potential energy can be converted into kinetic energy, a good example of this is when a pole-vaulter bends the pole during a lea

lesantik [10]

Answer:

The pole stores elastic potential energy when the pole is bent because its shape is change from its natural shape and it will want to go back to its original form, just like a spring or stretched elastic material.

Question:

Why did you lie about being in college?

6 0

3 years ago

Read 2 more answers

Do balanced forces change an objects motion

meriva

If the group of all forces acting on an object is balanced,
then the effect of all of them is the same as if there were
no forces at all on the object. In that case, the object
continues moving in a straight line at a constant speed.

8 0

3 years ago

What effect does an unbalanced force have on an object?

PSYCHO15rus [73]

Answer:

An unbalanced force can change an object's motion. An unbalanced force acting on a still object could make the object start moving. An unbalanced force acting on a moving object could make the object change direction, change speed, or stop moving.

6 0

3 years ago

The height of an object dropped from the top of a 64-foot building is given by h(t)=-16t^2+64. How long will it take the object

mojhsa [17]

Answer:

1.86 s

Explanation:

Given the expression

h(t) = -16t²+ 64...................... Equation 1

Where h = height of the object, t = time it will take the object to hit the ground.

Given: h = 64 foot.

We have to concert from foot to meters

If 1 foot = 0.3048 meters

Then, 64 foot = 0.3048×64 = 19.51 meters.

We substitute the value of h into equation

119.51 = -16t²+64

-16t² = 199.51-64

-16t² = 55.51

t² = 55.51/-16

t² = 3.469

t = √3.469

t = 1.86 s.

Hence it will take the object 1.86 s to hit the ground.

7 0

3 years ago

Problem 4 A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true ano

Nikitich [7]

Answer:

Part a: The eccentricity is 1.086.

Part b: The altitude at closest approach is 5088 km

Part c: The velocity at perigee is 8.516 km/s

Part d: The turn angle is 134.08 while the aiming radius is 5641.28 km

Explanation:

Specific energy is given by

$\epsilon=\frac{v^2}{2}-\frac{\mu}{r}$

Here

ε is the specific energy
v is the velocity which is given as 2.23 km/s
μ is the gravitational constant whose value is 398600
r is the distance between earth and the meteorite which is 402,000 km

$\epsilon=\frac{v^2}{2}-\frac{\mu}{r}\\\epsilon=\frac{2.2^2}{2}-\frac{398600}{402,000}\\\epsilon=1.495 km^2/s^2$

Value of specific energy is also given as

$\epsilon=\frac{\mu}{2a}\\a=\frac{\mu}{2\epsilon}\\a=\frac{398600}{2\times 1.495}\\a=13319 km$

Orbit formula is given as

$r=a(\frac{e^2-1}{1+ecos \theta})\\ae^2-recos\theta-(a+r)=0$

Putting values in this equation and solving for e via the quadratic formula gives

$ae^2-recos\theta-(a+r)=0\\(133319)e^2-(402000)(cos 150) e-(133319+402000)=0\\133319e^2+348142.21 e-535319=0\\\\e=\frac{-348142.21 \pm \sqrt{348142.21^2-4(133319)(535319)}}{2 (133319)}\\\\e=1.086 \, or \, -3.69$