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10 days ago

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The chart shows rate of decay. A 3 column table with 7 rows. The first column is Half-lives elapsed, with entries 0, 1, 2, 3, 4,

5, 6. Second column is Fraction remaining, with entries StartFraction 1 over 1 EndFraction, StartFraction 1 over 2 EndFraction, StartFraction 1 over 4 EndFraction, StartFraction 1 over 8 EndFraction, StartFraction 1 over 16 EndFraction, StartFraction 1 over 32 EndFraction, StartFraction 1 over 64 EndFraction. Third column is Perentage remaining, with entries 100, 50, 25, 12.5, 6.25, 3.125, 1.563. Which value is being measured in the columns labeled "Fraction remaining” and "Percentage remaining”? years of decay quantity of energy number of stable atoms amount of material that has not decayed

2 answers:

Sav [3.1K]10 days ago

8 0

A.

Keith_Richards [3.2K]10 days ago

6 0

D.

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It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

Maru [3345]

Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

$F_X = F_{cos \ \theta}$

Substituting 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

$F_X =64.65 \ N$

Meanwhile, the vertical component is:

$F_Y = Fsin \ \theta$

Again substituting 42° for θ and 87.0° for $F$

$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

$F_Y =58.21 \ N$

In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

$f = \mu \ N$

Where;

$\mu$ is the coefficient of friction

N = the normal force

Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting $F_Y \ for \ F_{sin \ \theta}$ . Normal reaction becomes:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Rearranging the equation above to isolate $\mu$ leads to:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

Substituting in the following values:

$F_X \ = \ 64.65 \ N$

m = 73 kg

g = 9.8 m/s²

$F_Y = \ 58.21 N$

Thus:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

Therefore, the coefficient of friction is = 0.0984

5 0

1 month ago

Point charge A with a charge of +4.00 μC is located at the origin. Point charge B with a charge of +7.00 μC is located on the x

ValentinkaMS [3465]

Response:

210.3 degrees

Justification:

The total force acting on charge A is 59.5 N

Apply the x and y components of the net force to determine the direction

atan (y/x)

8 0

28 days ago

Read 2 more answers

A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac

Keith_Richards [3271]

Answer:

The flux across the cube's surface is $2.314\ Nm^{2}/C$ .

Solution:

According to the details provided:

Cube edge length, a = 8.0 cm = $8.0\times 10^{- 2}\ m$ .

Volume charge density, $\rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}$ .

Now,

To find the electric flux:

$\phi = \frac{q}{\epsilon_{o}}$

where

$\phi$ = electric flux

$\epsilon_{o} = 8.85\times 10^{- 12}\ F/m$ = permittivity of vacuum.

The volume charge density for this scenario is described by:

$\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}$

Cube volume, $V = a^{3}$ .

Thus,

$V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}$ .

The total charge can be derived from equation (2):

$q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}$ .

$q = 2.048\times 10^{-11}\ F = 20.48\ pF$ .

Now, insert the value of 'q' into equation (1):

$\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C$ .

5 0

1 month ago

Uzupełnij zdania właściwymi sformułowaniami. Wyobraź sobie, że między linę a siodełko karuzeli łańcuchowej wmontowany jest siłom

ValentinkaMS [3465]

Explanation:

Here’s a revised version of the requirements;

Fill in the blanks with the appropriate terms. Picture a force gauge fixed between the rope and the saddle of the chain carousel. If you keep your feet off the ground while the vehicle is not in motion, the dynamometer shows A / B. When the carousel is spinning, you’ll see C / D displayed on the dynamometer.

A. Your weight including the saddle

C. Value of the rope's strength

B. Your weight

D. Value of the centripetal force

3 0

2 months ago

While dangling a hairdryer by its cord, you observe that the cord is vertical when the hairdryer isoff and, once it is turned on

Yuliya22 [3333]

Answer:

The air exiting from the hairdryer is moving at a speed of 10 m/s.

Explanation:

The thrust generated by the hairdryer enables it to maintain an elevation angled at 5° from vertical; thus, we derive from the force diagram

$(1).\: tan (5^o) = \dfrac{F_t}{Mg}$

by substituting $M =0.420kg$ , $g = 9.8m/s^2$ into the equation and resolving for $F_t$ we find:

$F_t = Mg\:tan(5^o)$

$F_t = (0.420kg)(9.8m/s^2)\:tan(5^o)$

$\boxed{F_t = 0.3601N.}$

This thrust is linked to the speed of air ejection $v$ through the equation

$(2).\: F_t = v\dfrac{dM}{dt}$

where $dM/dt$ signifies the rate of air ejection, which is known to be

$0.06m^3/2s = 0.03m^3/s$

and since $1m^3 = 1.2kg$ ,

$0.03m^3/s \rightarrow 0.036kg/s$

$\dfrac{dM}{dt} = 0.036kg/s,$

by inserting these values into equation (2), we obtain the value of $F_t$ as:

$0.3601N = 0.036v$

resulting in

$v= \dfrac{0.3601}{0.036}$

$\boxed{v =10m/s.}$

which indicates the air velocity discharged from the hairdryer.

6 0

2 months ago