How long must you wait (in half-lives) for a radioactive sample to drop to 2.10 % of its original activity?

Answer:100 miles

Explanation:

Answer:

10 Km

Explanation:

He is going 5km per hour and he arrived at his friend's house in 2 hours. You multiply 5 by 2 and you get 10.

Answer:

The answer is "[tex]\bold{7.56 \ Me\ V}[/tex]".

Explanation:

calculating the binding energy on per nucleon:

calculating number of proton and neutrons:

proton [tex]P_u=94[/tex]

neutron[tex]= 239-94=145[/tex]

calculating mass:

proton mass [tex]\ m_P=1.007825 \ amu\\\\[/tex]

neutron mass [tex]\ m_n=1.008665 \ amu\\\\[/tex]

neutral atom mass [tex]m = 239.05216 \ amu\\\\[/tex]

mass of prtons[tex]= 94 \times 1.007825 = 94.73555 \ amu\\\\[/tex]  

mass of neutrons[tex]= 145 \times 1.008665= 146.256425 \ amu\\\\[/tex]

Total nucleons mass formula:

[tex]\to m_n = (P+n)[/tex]

          [tex]= 94.73555+ 146.256425\\\\= 240.991975 \ amu[/tex]

calculating the mass of defect:

[tex]\to \Delta m= m_n-m\\\\[/tex]

           [tex]= 240.991975 - 239.05216\\\\= 1.939815 \ amu\\\\[/tex]

calculating the total of the binding energy:

[tex]\to BE=\Delta m\times 931.5 \ mev[/tex]

          [tex]= 1.939815 \times 931.5\\\\=1806.938 \ Me \ V\\\\[/tex]

BE in per nucleon [tex]=\frac{BE}{239}= 7.56 \ Me\ V[/tex]

The ball needs to accelerate with a magnitude a (pointed towards the center of the circle) of

a = (5.0 m/s)² / (2.0 m) = 12.5 m/s²

Then the required tension F in the string would need to be

F = (2 kg) (12.5 m/s²) = 25 N

Answer:

Object's weight = 6,839.42 N

Explanation:

Given

Above waterline = 30%

Volume of object = 1m^3

Required

Determine the weight of the object

First, we need to calculate its Mass

Mass = Density of Water * Volume of object in water

Density of water = 997kg/m³

If 30% is above waterline, then 70% is in water.

So:

Mass = Density of Water * Volume of object in water

Mass = 997kg/m³ * 70%m³

Mass = 997kg * 70%

Mass = 697.9 kg

The object weight sis then calculated as thus:

Weight = Mass * Acceleration of gravity

Weight = 697.9 kg * 9.8m/s²

Weight = 6 839.42 N

Answer:

effects of gravity on the Moon and Earth

Explanation:       Newton found the Moon's inward acceleration in its orbit to be 0.0027 metre per second per second, the same as (1/60)2 of the acceleration of a falling object at the surface of Earth.

Answer:

x=-3/4

Explanation:

Answer:

RMS velocity, [tex]v_{rms}=5748.75\ m/s[/tex]

Explanation:

We need to find the RMS speed of helium atoms near the surface of the Sun at a temperature of about 5300 K.

The formula for RMS speed of a gas is given by :

[tex]v_{rms}=\sqrt{\dfrac{3RT}{m}}[/tex]

Where

R is ideal gas constant, R = 8.314 J /mol K

T = 5300 K

m is molar mass of Helium, [tex]m = 4\times 10^{-3}\ Kg/mol[/tex]

Substituting all the values in above formula :

[tex]v_{rms}=\sqrt{\dfrac{3\times 8.314\times 5300}{4\times 10^{-3}}}\\\\=5748.75\ m/s[/tex]

So, the RMS speed Helium atoms 5748.75 m/s.

Answer: 3.

Explanation:

The correct answer is a higher amplitude and lower frequency. Since an opera singer is lowering his pitch it means that he is creating higher amplitude and because he is raising his voice for a song with that higher amplitude he is creating lower frequency.

Answer:

The half-life is [tex] t_{1/2} = 1.005 h[/tex]

Explanation:

Using the decay equation we have:

[tex]A=A_{0}e^{-\lambda t}[/tex]

Where:

We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that [tex]A = \frac{A_{0}}{2}[/tex]

[tex]\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}[/tex]

[tex]0.5=e^{-\lambda*1 h}[/tex]

Taking the natural logarithm on each side we have:

[tex]ln(0.5)=-\lambda[/tex]

[tex]\lambda=0.69 h^{-1}[/tex]

Now, the relationship between the decay constant λ and the half-life t(1/2) is:

[tex]\lambda = \frac{ln(2)}{t_{1/2}}[/tex]

[tex] t_{1/2} = \frac{ln(2)}{\lambda}[/tex]

[tex] t_{1/2} = \frac{ln(2)}{0.69}[/tex]

[tex] t_{1/2} = 1.005 h[/tex]

I hope it helps you!

Answer:

3.1

Explanation:

use formula f = v/lambda

Answer:

a) h(g) = 358,53 m

b) t = 8,16 s

c) t(t) = 16,71 s

Explanation:

Equations for vertical shooting are:

Vf = V₀ -  g * t   ;     h  =  V₀*t -  (1/2)*g*t²  ;    Vf² = V₀² - 2*g*h

And at maximum heigt Vf = 0 then

0 = V₀ - g * t

t = V₀/g                     V₀  = 80 m/s      and   g = 9,8 m/s²

t  =  80 / 9,8   (s)

t = 8,16 s

Then 8,16 s is the time to get maximum height

If we plug t = 8,16 (s) in equation  h  =  V₀*t -  (1/2)*g*t²

we get:     h (max)  =  (80)*8,16 - 0,5*9,8*(8,16)² (m)

h (max) = 652,8  -  326,27 m

h (max) = 326,53 m

Then relative to ground that height becomes

h(g) = 326,53 + 32

h(g) = 358,53 m

In order to get the time the arrow is in the air we proceed as follows:

a) for the arrow to be at the launched point will take the same time that from the launched point to the maximum height, and after that we have to find out the time the arrow takes from 32 m down to the ground level

Then  

t(t) = 8,16 + 8,16 + tₓ     (2)

Where tₓ  is the time from 32 m height to ground

h  =  V₀*tₓ -  (1/2)*g*tₓ²  but since the arrow now is going down then we change the sign of the second term on the right side of the equation

32 = (80)*tₓ  +  0,5 * 9,8 * tₓ²     Note that when the arrow is at 32 m height the speed is again V₀ = 80 m/s

32 = 80*tₓ  + 4,9*tₓ²

A second-degree equation for tₓ, solving it

4,9*tₓ² + 80*tₓ - 32 = 0

t₁,₂ = -80 ± √ 6400 + 627,2 / 9,8

t₁,₂ =( - 80 ± 83,8 ) / 9,8

there is not a negative time therefore we dismiss such solution and

t₁ = 3,8 / 9,8

t₁ = 0,39 s

And

t(t) = 8,16 + 8,16 + 0,39  s

t(t) = 16,71 s

Answer:

v=59[m/s]

Explanation:

To solve this problem we must use the principle of conservation of energy, which tells us that energy is transformed from Kinetic to potential or vice versa. At the moment when the car is at the top before falling down the cliff, we have the car moving at speed 50 [m/s] (kinetic energy) also it is 50 [m] above ground level (potential energy).

[tex]E_{k1}+E_{p1}=E_{k2}\\[/tex]

where:

Ek1 = kinetic energy before falling [J]

Ep1 = potential energy before falling [J]

Ek2 = kinetic energy in the ground [J]

The potential energy can be calculated by means of the following equation.

[tex]E_{p}=m*g*h[/tex]

where:

m = mass = 500 [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 50 [m]

Whereas the kinetic energy can be calculated by means of the following equation.

[tex]E_{k}=\frac{1}{2}*m*v^{2}[/tex]

where:

v = velocity = 50 [m/s]

Now replacing in the general equation:

[tex]\frac{1}{2} *500*(50)^{2} +500*9.81*50=\frac{1}{2} *500*v^{2}\\625000+245250=250*v^{2} \\250*v^{2} =870250\\v=\sqrt{870250/250} \\v=59[m/s][/tex]

Answer: Plates shifting

Explanation: After years and years of plates colliding into solid rock, they slowly become closer together. As recent studies have shown, Africa is currently moving closer to Europe one centimeter every year (one inch every 2.5 years).

Answer:

nvudbwasivnjlscv bwbfvsz

Explanation:

Answer:

A) 2.650 km

Explanation:

The relationship between acceleration of gravity and gravitational constant is:

[tex]g = \frac{Gm}{R^2}[/tex] ---- (1)

Where

[tex]R = 6,400 km[/tex] -- Radius of the earth.

From the question, we understand that the gravitational field of the rocket is 50% of its original value.

This means that:

[tex]g_{rocket} = 50\% * g[/tex]

[tex]g_{rocket} = 0.50 * g[/tex]

[tex]g_{rocket} = 0.5g[/tex]

For the rocket, we have:

[tex]g_{rocket} = \frac{Gm}{r^2}[/tex]

Where r represent the distance between the rocket and the center of the earth.

Substitute 0.5g for g rocket

[tex]0.5g = \frac{Gm}{r^2}[/tex] --- (2)

Divide (1) by (2)

[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}/\frac{Gm}{r^2}[/tex]

[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}*\frac{r^2}{Gm}[/tex]

[tex]\frac{1}{0.5} = \frac{1}{R^2}*\frac{r^2}{1}[/tex]

[tex]2 = \frac{r^2}{R^2}[/tex]

Take square root of both sides

[tex]\sqrt 2 = \frac{r}{R}[/tex]

Make r the subject

[tex]r = R * \sqrt 2[/tex]

Substitute [tex]R = 6,400 km[/tex]

[tex]r = 6400km * \sqrt 2[/tex]

[tex]r = 6400km * 1.414[/tex]

[tex]r = 9 049.6\ km[/tex]

The distance (d) from the earth surface is calculated as thus;

[tex]d = r - R[/tex]

[tex]d = 9049.6\ km - 6400\ km[/tex]

[tex]d = 2649.6\ km[/tex]

[tex]d = 2650\ km[/tex] --- approximated

Answer:

W = 6642 J

Explanation:

Given that,

Mass of a crate, m = 67 kg

Force with which the crate is pulled, F = 738 N

It is moved 9 m across a frictionless floor

We need to find the work done in moving the crate. Let the work done is W. It is given by :

W = F d

W = 738 N × 9 m

= 6642 J

So, the work done is 6642 J.

Answer:

The displacement is 6.4m

Explanation:

Step one:

given

we are told that Lucy runs 4 meters to the east,

then 5 meters south.

let the distance east be the displacement in the x-direction, and south be the y-direction

Step two:

The resultant of the x and y displacement is the magnitude of the total displacement z

applying Pythagoras theorem we have

z=√x^2+y^2

z=√4^2+5^2

z=√16+25

z=√41

z=6.4m

Answer:

Explanation:

Step one:

given data

period T= 4seconds

velocity v= 7m/s

wave lenght λ=?

Step two:

we know that f=1/T

the expression relating period and wave lenght is

v=λ/T

λ=v*T

λ=7*4

λ=28m

The wavelength of the wave is 28m

Answer:

in a spontaneous process, the entropy of the universe increases.

Explanation:

Entropy is a measure of of the degree of  randomness or disorderliness in a system.

The second law of thermodynamics can be stated as follows; "in any spontaneous process, the entropy of the universe increases."  

The universe here refers to the system's disorder and the disorder of the surroundings.  Therefore, a spontaneous process can occur, in which the entropy of the system decreases, only if the entropy increases in the surroundings.

For instance, when ice freezes, the entropy of liquid water decreases, that is, the entropy of the system decreases. However, heat is given off to the surroundings and the entropy of the surroundings increases. This is an obvious expression of this law.

Answer:

2008

Explanation:

2000+3+5======2008

Answer:

[tex]8\hat i-2\hat j-2\hat k[/tex]

Explanation:

Vectors in 3D

Given a vector

[tex]\vec P = P_x\hat i+P_y\hat j+P_z\hat k[/tex]

A vector [tex]\vec Q[/tex] parallel to [tex]\vec P[/tex] is:

[tex]\vec Q = k.\vec P[/tex]

Where k is any constant different from zero.

We are given the vectors:

[tex]\vec A = \hat i+4\hat j-2\hat k[/tex]

[tex]\vec B = 3\hat i-5\hat j+\hat k[/tex]

It's not specified what the 'resultant' is about, we'll assume it's the result of the sum of both vectors, thus:

[tex]\vec A +\vec B = \hat i+4\hat j-2\hat k + 3\hat i-5\hat j+\hat k[/tex]

Adding each component separately:

[tex]\vec A +\vec B = 4\hat i-\hat j-\hat k[/tex]

To find a vector parallel to the sum, we select k=2:

[tex]2(\vec A +\vec B )= 8\hat i-2\hat j-2\hat k[/tex]

Thus one vector parallel to the resultant of both vectors is:

[tex]\mathbf{8\hat i-2\hat j-2\hat k}[/tex]

Answer:

Decreasing

Hope this helps! :)

Answer:

A

Explanation:

i think

Answer:

Buoyant Force = 200N

Explanation:

Given

[tex]Object = 200N[/tex]

Required

Determine the buoyant force?

Using Archimedes principle:

When an object is immersed in a fluid, the object is acted upon on by an upward force (Buoyant force) which equals the weight of the object.

In other words;

Buoyant Force = Weight of object

Hence:

Buoyant Force = 200N

The buoyant force on the object which floats with three-fourths of its volume beneath the surface of the water is 200 N.

The buoyant force is the force which is applied by the fluid in the upward direction when a object is placed over it. This buoyant force can be calculated with the following formula.

[tex]F_b=-\rho gV[/tex]

Here, (ρ) is the density of the fluid, (g) is the gravitation force and (V) is the fluid volume.

The buoyant force is equal to the weight of the liquid displace by the object which is placed on it.

It is given that the 200-N object floats with three-fourths of its volume beneath the surface of the water.

This force applied on the object balance the 200-N object and floats it. For this case, the value of buoyant force will be equal to the weight of the object.

[tex]F_b=W\\F_b=200\rm\; N[/tex]

Thus, the buoyant force on the object which floats with three-fourths of its volume beneath the surface of the water is 200 N.

Learn more about the buoyant force here;

https://brainly.com/question/3228409

Answer:

STOP CHETING

Answer:

C is the answer to your question

Explanation:

Have a great day!!! :)

Answer:

It's C. Moving Waterrrr