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3 years ago

15

Sr-90 has a half-life of T1/2 = 2.85 a (years). How much Sr-90 will remain in a 5.00 g sample after 5.00 a? Show all of your wor

k. (2 marks)

1 answer:

Molodets [167]3 years ago

6 0

Answer:

1.48 g

Explanation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

A = (5.00 g) (½)^(5.00 a / 2.85 a)

A = 1.48 g

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Murljashka [212]

Answer:

I love that video it’s so funny!!!

Explanation:

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3 years ago

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What is your speed (in m/s) if you walk 1000m in 20 minutes? (Hint: how many seconds are in a minute)

labwork [276]

Answer:

20min = 20 × 60 = 1200sec.

Speed in m per sec.

V = 1000/1200

V = 0.833m per sec.

Explanation:

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3 years ago

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed

kogti [31]

Answer:

a) $y_{first}=5.3mm$

b) $y_{second}=10.6-5.3 =5.3 mm$

Explanation:

a)

The width of the central bright in this diffraction pattern is given by:

$y=\frac{m\lambda D}{a}$ when m is a natural number.

here:

m is 1 (to find the central bright fringe)
D is the distance from the slit to the screen
a is the slit wide
λ is the wavelength

So we have:

$y_{first}=\frac{633*10^{9}*3.35}{0.0004}$

$y_{first}=5.3mm$

b)

Now, if we do m=2 we can find the distance to the second minima.

$y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}$

$y_{2}=10.6 mm$

Now we need to subtract these distance, to get the width of the first bright fringe :

$y_{second}=10.6-5.3 =5.3 mm$

I hope it heps you!

4 0

3 years ago

A guitar string with a linear density of 2.0 g/m is stretched between supports that are 60 cm apart. The string is observed to f

Butoxors [25]

<h2>The frequency of driver is 700 Hz</h2>

Explanation:

The frequency of wave in a string is given by the relation

n = $\frac{p}{2l} \sqrt{\frac{T}{m} }$

here n is the frequency

p is the number of antinodes and l is the length of string .

T is the tension in string and m is the mass per unit length

Thus 420 = $\frac{3}{120} \sqrt{\frac{T}{2} }$ I

Now if there is 5 antinodes , the value of p = 5

Thus n = $\frac{5}{120} \sqrt{\frac{T}{2} }$ II

Dividing II by I , we have

n/420 = 5/3

or n = 5/3 x 420 = 700 Hz

3 0

3 years ago

A wooden bucket filled with water has a mass of 68 kg and is attached to a rope that is wound around a cylinder with a radius of

Aleonysh [2.5K]

Answer: 210.2N

Explanation:

Assume a bucket of water with a total mass of 68kg is attached to a rope, which in turn is tied around a 0.078m radius cylinder at the top of a well. A crank with a turning radius of 0.250 m is attached to the end of the cylinder.

the minimum force directed perpendicular to the crank handle required to raise the bucket is

(Assume the rope's mass is negligible, that the cylinder turns on friction-less bearings, and that g = 9.8 m/s2

The crank handle provides a torque T=0.25F where F is the force we are looking for.

A free body diagram will show that the tension in the rope times the cylinder radius R is equal to the torque on the cylinder. But the tension in the rope is just the weight of the bucket

W=mg= 68kg

W(0.078)=T=0.25F

F=0.312W=0.312(68kg)=21.216kg= 210.2N

7 0

3 years ago