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Anon25 [30]
3 years ago
9

By how much does the gravitational potential energy of a 54-kg pole vaulter change if her center of mass rises about 4.0 m durin

g the jump
Physics
2 answers:
kodGreya [7K]3 years ago
8 0
<span>the gravational potential energy of anything on the ground is zero. When calculating potential energy you take height in meters and multiply it by the mass of the object in kilograms and the acceleration of gravity to get a new unit called Joules. Any object at ground level has a potential energy of zero newtons becuase anything multiplied by zero is zero. An object with mass of 54 kg, 4 meters above the ground has a gravitatinal potential energy of 2116.8 Joules.</span>
dexar [7]3 years ago
7 0

Answer:

The gravitational potential energy of pole voulter is 2160\ J.

Explanation:

Given that,

Mass of the pole vaulter, m = 54 kg

Height it rises, h = 4 m

To find,

The gravitational potential energy.

Solution,

We know , gravitational potential energy of an object of mass m and height h is given by m\times g\times h.Here g is acceleration due to gravity=10 m/s^2.

E=m\times g\times h\\=54\times 10 \times 4\ J=2160\ J.

Therefore, the gravitational potential energy of pole voulter is 2160\ J.

Hence, this is the required solution.

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The half-life of a certain element is 100 days. How many half-lives will it be before only one-eighth of this element remains?
Natalka [10]

Answer:

3

Explanation:

The half-life is the time it takes for the amount of radioactive isotope to halve. Therefore, we have:

- After 1 half-life, only 1/2 of the element will be left

- After 2 half-lives, only 1/4 of the element will be left

- After 3 half-lives, only 1/8 of the element will be left

So, it will take 3 half-lives for the element to become 1/8 of its original amount.

Mathematically, this can be also verified by using the equation

\frac{N(t)}{N_0}=(\frac{1}{2})^\frac{t}{\tau_{1/2}}

where

N(t) is the amount of the element left at time t

N0 is the initial amount of the element

\tau_{1/2} is the half-life

Substituting t=3\tau_{1/2} (3 half-lives), we find

\frac{N(t)}{N_0}=(\frac{1}{2})^3=\frac{1}{8}

7 0
3 years ago
To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 1 T
Ronch [10]

Correct question is;

To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 mm in dim light. Find the angular resolution of the eye for 550 nm wavelength light at those extremes. In which light can you see more sharply, dim or bright?

Answer:

We'll see more sharply in dim light

Explanation:

If we consider diffraction through a circular aperture, then angular resolution is given by;

θ = 1.22λ/D

where:

θ is the angular resolution (radians) λ is the wavelength of light

D is the diameter of the lens' aperture.

Thus,

at diameter = 2mm = 2 x 10^(-3) m = 2 x 10^(6) nm

θ = (1.22 * 550)/(2 x 10^(6))

θ = 335.5 x 10^(-6) radians

Now, we need to convert this to arc seconds.

Thus;

1 arc second = 4.85 x 10^(-6) radians

So,θ = 335.5 x 10^(-6) radians = [335.5 x 10^(-6)]/[4.85 x 10^(-6)]

= 69.18 arc seconds

at diameter = 8mm = 8 x 10^(-3) m = 8 x 10^(6) nm

θ = (1.22 * 550)/(8 x 10^(6))

θ = 83.875 x 10^(-6) radians

Now, we need to convert this to arc seconds.

Thus;

1 arc second = 4.85 x 10^(-6) radians

So,θ = 83.875 x 10^(-6) radians = [83.875 x 10^(-6)]/[4.85 x 10^(-6)]

= 17.3 arc seconds

From the values of angular resolution gotten, we see that sharpness of image increases with increasing angular resolution. Thus, objects are sharper in dim light.

4 0
3 years ago
Longitudinal waves have _____.
Yakvenalex [24]
I think its either C or D. I tried, couldn't figure the last part out. Hope this helped though!! Have a great day! :D
6 0
3 years ago
Read 2 more answers
A 3.0kg mass tied to a string
dem82 [27]

Answer:

\boxed{\sf Tension \ in \ the \ string \ (T) = 3 \ kN}

Given:

Mass (m) = 3.0 kg

Uniform speed (v) = 20 m/s

Length of string (r) = 40 cm = 0.4 m

To Find:

Tension in the string (T)

Explanation:

Tension (T) is the string will be equal to centripetal force (\sf F_c).

\boxed{ \bold{ T = F_c  =  \frac{m {v}^{2} }{r} }}

Substituting value of m, v & r in the equation:

\sf \implies T =  \frac{3 \times  {20}^{2} }{0.4}  \\  \\  \sf \implies T = \frac{3 \times 400}{0.4}  \\  \\  \sf \implies T =3 \times 1000 \\  \\  \sf \implies T =3000 \: N \\  \\ \sf \implies T =3 \: kN

\therefore

Tension in the string (T) = 3 kN

5 0
3 years ago
What happens to the atomic number of an atom when the number of neutrons in the nucleus of that atom increases? a It decreases b
kupik [55]

Answer:

It remains the same

Explanation:

It remains the same. This is because the number of protons doesn't change and the number of protons determines the atomic number.

8 0
2 years ago
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