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3 years ago

Is there a correlation between moonrise and high tide?

Physics

1 answer:

Grace [21]3 years ago

5 0

Yes when it's night time the tides get higher and when it morning there lower.

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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

3 years ago

A 65.0-kg runner has a speed of 5.20 m/s at one instant dur- ing a long-distance event. (a) What is the runner’s kinetic energy

vladimir2022 [97]

Answer:

a)KE=878.8 J

b)W=2636.4 J

Explanation:

Given that

mass ,m = 65 kg

Initial speed ,u = 5.2 m/s

We know that kinetic energy KE is given as follows

$KE=\dfrac{1}{2}mu^2$

m=mass

u=velocity

Now by putting the values in the above equation we get

$KE=\dfrac{1}{2}\times 65\times 5.2^2\ J$

KE=878.8 J

We know that

Work done by all forces = Change in the kinetic energy

The final velocity , v= 2 u = 2 x 5.2 m/s

v= 10.4 m/s

$W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

Now by putting the values in the above equation we get

$W=\dfrac{1}{2}\times 65\times 10.4^2-\dfrac{1}{2}\times 65\times 5.2^2\ J$

W=2636.4 J

a)KE=878.8 J

b)W=2636.4 J

8 0

3 years ago

An investigator collects a sample of a radioactive isotope with an activity of 490,000 Bq.48 hours later, the activity is 110,00

Daniel [21]

Answer:

The correct answer is "22.27 hours".

Explanation:

Given that:

Radioactive isotope activity,

= 490,000 Bq

Activity,

= 110,000 Bq

Time,

= 48 hours

As we know,

⇒ $A = A_0 e^{- \lambda t}$

or,

⇒ $\frac{A}{A_0}=e^{-\lambda t}$

By taking "ln", we get

⇒ $ln \frac{A}{A_0}=- \lambda t$

By substituting the values, we get

⇒ $-ln \frac{110000}{490000} = -48 \lambda$

⇒ $-1.4939=-48 \lambda$

$\lambda = 0.031122$

As,

⇒ $\lambda = \frac{ln_2}{\frac{T}{2} }$

then,

⇒ $\frac{ln_2}{T_ \frac{1}{2} } =0.031122$

⇒ $T_\frac{1}{2}=\frac{ln_2}{0.031122}$

$=22.27 \ hours$

3 0

3 years ago

ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

mixer [17]

Answer:

fffffffffffffffffffffffffffff

5 0

3 years ago

Read 2 more answers

An ideal gas in a cubical box having sides of length L exerts a pressure p on the walls of the box. If all of this gas is put in

Soloha48 [4]

Answer:

Explanation:

See attached file

5 0

4 years ago