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Digiron [165]
3 years ago
8

Mantles for gas lanterns contain thorium, because it forms an oxide that can survive being heated to incandescence for long peri

ods of time. Natural thorium is almost 100% 232Th, with a half-life of 1.405 ✕ 1010 y. If an average lantern mantle contains 200 mg of thorium, what is its activity (in Bq)?
Physics
1 answer:
maksim [4K]3 years ago
5 0

Answer:

The activity is 811.77 Bq

Solution:

As per the question:

Half life of Thorium, t_{\frac{1}{2}} = 1.405\times 10^{10}\ yrs

Mass of Thorium, m = 200 mg = 0.2 g

M = 232 g/mol

Now,

No. of nuclei of Thorium in 200 mg of Thorium:

N = \frac{N_{o}m}{M}

where

N_{o } = Avagadro's number

Thus

N = \frac{6.02\times 10^{23}\times 0.2}{232} = 5.19\times 10^{20}

Also,

Activity is given by:

\frac{0.693}{t_{\frac{1}{2}}}\times N

A= \frac{0.693}{1.405\times 10^{10}}\times 5.19\times 10^{20} = 2.56\times 10^{10}\ \yr

A = \frac{2.56\times 10^{10}}{365\times 24\times 60\times 60} = 811.77\ Bq

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Explain the following defects of a simple electric cell:
eimsori [14]

Answer:

Explanation:

The two major defects of simple electric cells causes current supplied to be for short time. These defects are: polarization and local action.

a. Polarization: This is a defect caused by an accumulation of hydrogen bubbles at the positive electrode of the cell. It can be prevented by the use of vent, using a hydrogen absorbing material or the use of a depolarizer.

b. Local Action: This is the gradual wearing away of the electrode due to impurities in the zinc plate. It can be controlled by the amalgamation of the zinc plate before it is used.

4 0
3 years ago
A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?
dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0
3 years ago
An arrow movirg 48.3 m/s has 5.22<br> kg•m/s of momentum. What is its<br> mass?
spin [16.1K]

Answer:

0.11 kg

Explanation:

Ft = MV

Ft = momentum 5.22kg m/s

M = mass

V = velocity 48.3m/s

Therefore

5.22 = M x 48.3

Divide both sides by 48.3

5.22/48.3 = M x 48.3/48.3

0.11 = M

M = 0.11kg

6 0
4 years ago
A long ramp made of cast iron is sloped at a constant angle θ = 52.0∘ above the horizontal. Small blocks, each with mass 0.42 kg
dezoksy [38]

Answer:

For cast iron we have

h = 0.92 m

For copper

h = 1.05 m

For Lead

h = 1.23 m

For Zinc

h = 2.43 m

Explanation:

As we know that final speed of the block is calculated by work energy theorem

W_f + W_g = \frac{1}{2}mv^2

now we have

-\mu_k mg cos\theta(\frac{h}{sin\theta}) + mgh = \frac{1}{2}mv^2

now we have

v^2 = 2gh - 2\mu_k g h cot\theta

v = \sqrt{2gh(1 - \mu_k cot\theta)}

For cast iron we have

4 = \sqrt{2(9.81)(h)(1 - 0.15cot52)}

h = 0.92 m

For copper

4 = \sqrt{2(9.81)(h)(1 - 0.29cot52)}

h = 1.05 m

For Lead

4 = \sqrt{2(9.81)(h)(1 - 0.43cot52)}

h = 1.23 m

For Zinc

4 = \sqrt{2(9.81)(h)(1 - 0.85cot52)}

h = 2.43 m

4 0
3 years ago
If you want to know how energy will move between two objects, what do you need to know about the objects?
Inessa [10]
You need to know how much friction that object.
8 0
4 years ago
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