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Liono4ka [1.6K]

2 years ago

11

After you cross the finish line, you slow down and come to a stop, breathing hard. It takes you 10 seconds to come to a stop. Wh

at is your acceleration?

1 answer:

Otrada [13]2 years ago

4 0

Acceleration = (change in speed) / (time for the change)

<em>Acceleration = -- (speed as you crossed the finish line) / (10 seconds)</em>

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What is the equation for potential energy?

Mrrafil [7]

You can calculate potential energy by:
U = m.g.h

Where, U = potential energy
m = mass
g = acceleration due to gravity
h = height

Hope this helps!

7 0

3 years ago

Read 2 more answers

When light is directed on a metal surface, the kinetic energies of the photoelectrons a) are random b) vary with the frequency o

jekas [21]

Answer:

b) vary with the frequency of the light

Explanation:

The phone electric effect can be expressed as

K.E=(hv -W•)

Where K.E is the Kinectic energy

W• = work function of the metal

ν =frequency of the radiation

h = Planck's constat

Then, we can see that K.E is proportional linearly to "v" in the equation above.

Therefore, When light is directed on a metal surface, the kinetic energies of the photoelectrons vary with the frequency of the light

5 0

2 years ago

what are the speeds of (a) a proton that is accelerated from rest through a potential difference of −1000 v−1000 v and (b) an el

Evgen [1.6K]

Answer:

This is the answer: The speed of a proton is about 5.0 × 10⁵ m/s

Explanation:

Because of the speeds of protons! :D

5 0

1 year ago

What is one benefit of lifelong physical activity?

kvasek [131]

The answer would be C

5 0

3 years ago

Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left

Dahasolnce [82]

Answer:

$x_c= \dfrac{5}{9}L$

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any distance x from point A mass density

$\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x$

$\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x$

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

$dI=\lambda x^2dx$

So total moment of inertia

$I=\int_{0}^{L}\lambda x^2dx$

By putting the values

$I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx$

By integrating above we can find that

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Now to find location of center mass

$x_c = \dfrac{\int xdm}{dm}$

$x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}$

Now by integrating the above

$x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}$

$x_c= \dfrac{5}{9}L$

So mass moment of inertia $I=\dfrac {7}{12}\lambda_ 0 L^3$ and location of center of mass $x_c= \dfrac{5}{9}L$

8 0

3 years ago