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2 years ago

If you had started with a larger mass, how would the half-life change?

Physics

1 answer:

iogann1982 [59]2 years ago

3 0

Answer:

There is no change, unless your mass is somehow at the quantum level, at which the concept of half-life breaks down.

Half life is a property of the specific radioactive isotope...NOT of the initial sample's mass.

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A capacitor has a capacitance of 0. 40 µF at a voltage of 9. 0 V. What is the charge on each plate of the capacitor? µC.

ASHA 777 [7]

The capacitor is a device that can store electrical energy. It is a two-conductor configuration. The charge on each plate of the capacitor will be 3.6 µC.

<h3>What is a capacitor?</h3>

A capacitor is a device that can store electrical energy. It is a two-conductor configuration separated by an insulating medium that carries charges of equal size and opposite sign.

An electric insulator or vacuum, such as glass, paper, air, or a semi-conductor termed a dielectric, can be used as the non-conductive zone.

The given data in the problem is;

C is the capicitence of capicitor= 0. 40 µF

V is the voltage = 9. 0 V

Q is a charge on each plate of the capacitor=?µC.

The formula for the capacitor is given as;

$\rm Q=CV \\\\ \rm Q=0. 40 \times 9. 0 \\\\ \rm Q=3.6 \ \mu C.$

Hence the charge on each plate of the capacitor will be 3.6 µC.

To learn more about the capacitor refer to the link;

brainly.com/question/14048432

4 0

2 years ago

A high speed train is traveling at a speed of 44.7 m/s when the engineer sounds the 415 Hz warning horn. The speed of sound is 3

il63 [147K]

Answer:

Explanation:

Speed of the source of sound = v = 44.7 m/s

Speed of sound = V = 343 m/s

a) Apparent frequency as the train approaches = f = [V /(V -v) ] × f

= [343 / (343 - 44.7) ] × 415 = 477.18 Hz

Wave length = λ = v / f = 343 / 477.18 = 0.719 m

b) Frequency heard as the train leaves = f ' = [V / ( V + v) ] f

= [343 / { 343 + 44.7 ) ] x 415

= 367.2 Hz

Wavelength when leaving = v / f = 343 / 367.2 = 0.934 m

8 0

3 years ago

A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot

denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

$P_r = 345 Watt$

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

$P_r = 345 Watt$

Explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

$\Delta V = E + i r$

here we have

$E = 12 V$

$i = 56 A$

$r = 0.11$

now we have

$\Delta V = 12 + (0.11)(56) = 18.16 V$

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

Part c)

Rate of energy conversion into EMF is given as

$P_{emf} = i E$

$P_{emf} = (56)(12)$

$P_{emf} = 672 Watt$

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

$V = E - i r$

$V = 12 - (56)(0.11)$

$V = 12 - 6.16 = 5.84 V$

Part e)

now the rate of energy dissipation is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

7 0

3 years ago

Drag each label to the correct location. Sort the sentences based on whether they describe the properties of a heterogeneous or

Lisa [10]

Answer:

ijpferiukjlwbl aojh oljn,

Explanation:

5 0

3 years ago

A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220

disa [49]

First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, $\mu m g$ . For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
$ma=-\mu m g$
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
$a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2$

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
$2aS=v_f^2-v_i^2$
where $v_f=0$ is the final speed of the skier and $v_i=4.75 m/s$ is the initial speed. Substituting numbers, we find:
$S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m$

5 0

3 years ago