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3 years ago

True or False: The rate of nuclear decay depends on the temperature, pressure, and surface area of the isotope.

Physics

2 answers:

Liono4ka [1.6K]3 years ago

6 0

Answer:

False.

Hope this helps!

Natali5045456 [20]3 years ago

3 0

Answer:

The correct option is False

Explanation:

Temperature, pressure and surface area are factors that affect the rate of a chemical reaction and not a nuclear reaction. It should be noted that chemical reaction involves the electrons in the shell of an atom while nuclear reaction involves the unstable nucleus of a radioactive isotope. Thus, temperature, pressure and surface area do not affect the unstable nucleus during radioactive decay/emission.

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Un pendul este suspendat de un ax cu o tijă subțire foarte ușoară.

FromTheMoon [43]

Answer:

Explanation:

Given that a pendulum is suspended by a shaft with a very light thin rod.

Followed by the given information: m = 100 g, I = 0.5 m, g = 9.8 m / s²

We can determine the answer to these questions using angular kinematics.

Angular kinematics is just derived from linear kinematics but in different symbols, and expressions.

Here are the formulas for angular kinematics:

θ = ωt
∆w =
L [Angular momentum] = mvr [mass × velocity × radius]

A) What is the minimum speed required for the pendulum to traverse the complete circle?

We can use the formula v = √gL derived from

B) The same question if the pendulum is suspended with a wire?

C) What is the ratio of the two calculated speeds?

4 0

2 years ago

Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re

leva [86]

Answer:

The correct answer is C. 45.5 lbs.

Explanation:

In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.

The formula for any problem involving a lever is:

$F_ed_e=F_ld_l$

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.

The parameter of the formula that you need is F_l:

$F_l=\frac{F_ed_e}{d_l}$

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

$F_l=\frac{25*60}{33}$

F_l=45.5 lbs

7 0

3 years ago

Two birds have the same mass but one has bigger feet. How would pressure that they produce on the ground be different ?

Mama L [17]

Answer:

the pressure per square inch is greater from the smaller feet.

Explanation:

different weight distribution

3 0

2 years ago

A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an

Nat2105 [25]

Answer:

a) $T=0.0031416s$

b) $v_{max}=6.283\frac{m}{s}$

c) $E=0.1974J$

d) $F=80N$

e) $F=40N$

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by $X=Acos(\omega t +\phi)$ (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

$\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)$ (2)

For the acceleration

$\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)$ (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

$A\omega^2=8x10^{3}\frac{m}{s^2}$

Since we know the amplitude A=0.002m we can solve for $\omega$ like this:

$\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}$

And we with this value we can find the period with the following formula

$T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s$

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

$v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}$

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

$E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J$

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens $cos(\omega t +\phi)=1$ , and we also know that the maximum acceleration is given by::

$|\frac{d^2X}{dt^2}|=A\omega^2$

So then we have that:

$F=ma=mA\omega^2$

And since we have everything we can find the force

$F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N$

6) Part e

When the mass it's at the half of it's maximum displacement the term $cos(\omega t +\phi)=1/2$ and on this case the acceleration would be given by;

$|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}$

And the force would be given by:

$F=ma=\frac{1}{2}mA\omega^2$

And replacing we have:

$F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N$

8 0

2 years ago

Stronger acids have more _______

mario62 [17]

Examples of strong acids are hydrochloric acid (HCl), perchloric acid (HClO4), nitric acid (HNO3) and sulfuric acid (H2SO4). ... For example, hydrogen chloride is a strong acid in aqueous solution, but is a weak acid when dissolved in glacial acetic acid.

7 0

3 years ago