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2 years ago

14

If you have 80g of a radioactive substance whose half life is 2 days, how long will it take for the substance to decay to the po

int where there is only 10g left?

1 answer:

Ber [7]2 years ago

5 0

Answer:

6 days.

Explanation:

From radioactivity, The expression for half life is given as,

R/R' = 2⁽ᵃ/ᵇ)................... Equation 1

Where R = original mass of the radioactive substance, R' = Remaining mass of the radioactive substance after decay, a = Total time taken to decay, b = half life.

Given: R = 80 g, R' = 10 g, b = 2 days.

Substitute into equation 1

80/10 = 2⁽ᵃ/²⁾

8 = 2⁽ᵃ/²⁾

2³ = 2⁽ᵃ/²)

Equating the base and solving for a

3 = a/2

a = 2×3

a = 6 days.

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X-rays with frequency 3 ⨉ 10 18 Hz shine on a crystal, producing an interference pattern. If the first bright spot is observed a

FrozenT [24]

Answer:

$5.1645\times 10^{-11}\ m$

Explanation:

n = Order = 1

c = Speed of light = $3\times 10^8\ m/s$

f = Frequency = $3\times 10^{18}\ Hz$

$\theta$ = Angle = $75.5^{\circ}$

Lattice spacing is given by

$d=\dfrac{n\lambda}{2\sin\theta}\\\Rightarrow d=\dfrac{n\times c}{f2\sin\theta}\\\Rightarrow d=\dfrac{1\times 3\times 10^{8}}{3\times 10^{18}\times 2\times \sin75.5^{\circ}}\\\Rightarrow d=5.1645\times 10^{-11}\ m$

The lattice spacing of the crystal is $5.1645\times 10^{-11}\ m$

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2 years ago

How would you solve for x I cant remember right now 4x+6x=9x-10

-Dominant- [34]

Combine all of the x's on one side of the equation and then finish the problem!

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2 years ago

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What is the smallest sub atomic particle that orbit around the nucleus?

JulijaS [17]

Electron,a negative charged particles revolving around the nucleus.

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3 years ago

Interactive Solution 9.37 presents a method for modeling this problem. Multiple-Concept Example 10 offers useful background for

viva [34]

Answer:

21.67 rad/s²

208.36538 N

Explanation:

$\omega_f$ = Final angular velocity = $\dfrac{1}{6}78=13\ rad/s$

$\omega_i$ = Initial angular velocity = 78 rad/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

t = Time taken

r = Radius = 0.13

I = Moment of inertia = 1.25 kgm²

From equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{13-78}{3}\\\Rightarrow \alpha=-21.67\ rad/s^2$

The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=1.25\times -21.67\\\Rightarrow \tau=-27.0875$

Frictional force is given by

$F=\dfrac{\tau}{r}\\\Rightarrow F=\dfrac{-27.0875}{0.13}\\\Rightarrow F=-208.36538\ N$

The magnitude of the force of friction applied by the brake shoe is 208.36538 N

5 0

3 years ago

A skateboarder traveling at 4.45 m/s can be stopped by a strong force in 1.82 s and by a weak force in 5.34 s.

ANTONII [103]

1) Impulse: $-238.5 kg m/s$

2) Strong force: -131.1 N, weak force: -44.7 N

Explanation:

1)

The impulse exerted on an object is equal to the change in momentum of the object itself.

Mathematically:

$I=\Delta p=m(v-u)$

where

m is the mass of the object

u is the initial velocity

v is the final velocity

For the skateboarder in this problem, we have:

m = 53.6 kg

u = 4.45 m/s

v = 0 (it comes to a stop)

Therefore, the impulse is

$I=(53.6)(0-4.45)=-238.5 kg m/s$

Where the negative sign indicates that the direction is opposite to the motion of the object.

2)

The impulse is also equal to the product between the force applied and the duration of the collision:

$I=F\Delta t$

where

I is the impulse

F is the average force

$\Delta t$ is the time during which the force is applied

The strong force is applied in a time of

$\Delta t = 1.82 s$

Therefore this force is

$F=\frac{I}{\Delta t}=\frac{-238.5}{1.82}=-131.1 N$

The weak force is applied in a time of

$\Delta t = 5.34 s$

So this force is

$F=\frac{I}{\Delta t}=\frac{-238.5}{5.34}=-44.7 N$

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

6 0

2 years ago