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3 years ago

11

Determine the amount of time for polonium-210 to decay to one fourth its original quantity. The half-life of polonium-210 is 138

days. Explain.

1 answer:

Tju [1.3M]3 years ago

5 0

Answer:

276 days

Explanation:

1/4 th of the original means <u><em>2 half lives</em></u>

1 half life = 138 days

So,

2 half lives = 276 days

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BRAINLIEST AND MAY DOUBLE POINTS!!!

raketka [301]

Given that:

Energy of bulb (Work ) = 30 J,

Time (t) = 3 sec

The power consumption = ?

We know that, Power can be defined as rate of doing work

Power (P) = Work(Energy supplied) ÷ time

= 30 ÷ 3

= 10 Watts

<em> The power consumption is 10 W.</em>

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3 years ago

If the kinetic energy of an electron is 4.1e-18 j, what is the speed of the electron? (you can use the approximate (nonrelativis

arlik [135]

The kinetic energy of the electron is
$K= \frac{1}{2}mv^2$
where $m=9.1 \cdot 10^{-31} kg$ is the mass of the electron and v its speed. Since we know the value of the kinetic energy, $K=4.1 \cdot 10^{-18} J$ , we can find the value of the speed v:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2\cdot 4.1 \cdot 10^{-18}J}{9.1 \cdot 10^{-31}kg} } = 3\cdot 10^6 m/s$

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3 years ago

a sphere of diameter 6•0cm is moulded into a thin uniform wire of diameter 0•2mm.calculate the length of the wire in metres

Scilla [17]

The length of the wire is 36 m.

<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

$\text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}$

Similarly,

$\text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}$

We know the below formulas,

$\text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}$

$\text {volume of wire}=\pi \times r^{2} \times l$

When equating both the equations, we can find length of wire as below, where $\pi=\frac{22}{7}$

$\frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l$

$\frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l$

The $\pi$ value gets cancelled as common on both sides, we get

$\frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l$

The $10^{-6}$ value gets cancelled as common on both sides, we get

$l=4 \times 9=36 m$

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3 years ago

Pam, wearing a rocket pack, stands on frictionless ice. She has a mass of 49 kg. The rocket supplies a constant force for 22.0 m

ss7ja [257]

Answer:

Magnitude of the force is 4350N

Explanation:

As the woman accelerates at a distance of 22 m to go from rest to 62.5 m / s, we can use the kinematics to find the acceleration

v² = v₀² + 2 a x

v₀ = 0

a = v² / 2x

a = 62.5²/(2 × 22)

a = 88.78m/s²

the time you need to get this speed

v = v₀ + a t

t = v / a

t = 62.5 / 88.78

t = 0.704s

Let's caculate the magnitude of the force

F = ma

= 49 × 88.78

= 4350.22

≅ 4350N

Magnitude of the force is 4350N

t = 1,025 s

a = 55.43 m / s²

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3 years ago

A sailboat is heading directly north at a speed of 20 knots (1 knot 50.514 m/s). the wind is blowing toward the east with a spee

stellarik [79]

(a) The magnitude of the wind as it is measured on the boat will be the result of the two vectors. Since they are at 90°, the resultant can be determined by the Pythagorean theorem.

R = sqrt ((20 knots)² + (17 knots)²)
R = sqrt (400 + 289)
R = 26.24 knots

The direction of the wind will have to be angle between the boat and the resultant.
cos θ = (20 knots)/(26.24 knots)
θ = 40.36°

Hence, the direction is 40.36° east of north.

(b) As stated, the wind is blowing in the direction that is to the east. This means that it only has one direction. Parallel to the motion of the boat, the magnitude of the wind velocity will have to be zero.

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4 years ago