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Digiron [165]
2 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]2 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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The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
5 0
2 years ago
An elementary particle of mass m completely absorbs a photon, after which its mass is 1.01m. (a) what was the energy of the inco
sdas [7]
A.) We use the famous equation proposed by Albert Einstein written below:

E = Δmc²
where
E is the energy of the photon
Δm is the mass defect, or the difference of the mass before and after the reaction
c is the speed of light equal to 3×10⁸ m/s

Substituting the value:

E = (1.01m - m)*(3×10⁸ m/s) = 0.01mc² = 3×10⁶ Joules

b) The actual energy may be even greater than 3×10⁶ Joules because some of the energy may have been dissipated. Not all of the energy will be absorbed by the photon. Some energy would be dissipated to the surroundings.
8 0
2 years ago
A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m
Marrrta [24]

Answer:

-1.47 m/s^2

Explanation:

We can use the following SUVAT equation to solve the problem:

v^2 - u^2 = 2ad

where

v = 0 is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d = 196 m is the displacement of the car before coming to a stop

Solving the equation for a, we find the acceleration:

a=\frac{v^2-u^2}{2d}=\frac{0-(24)^2}{2(196)}=-1.47 m/s^2

4 0
2 years ago
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
MArishka [77]

Complete Question:

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer:

m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

6 0
2 years ago
A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the drive train, and 70% reaches the wheels.
shutvik [7]

Answer:

a = 6.53 m/s^2

v = 11.5689 m/s

Explanation:

Given data:

engine power is 217 hp

70 % power reached to wheel

total mass ( car + driver) is 1530 kg

from the data given

2/3 rd of weight is over the wheel

w = 2/3rd mg

maximum force

F = \mu W

we know that F = ma

ma =  \mu (2/3 mg)

a_{max} = 2/3(1.00) (9.8) = 6.53 m/s^2

the new power is p  = 70\% P_[max} = 0.7 P_{max}

P =f_{max} v

0.7P_{max} = ma_{max} v

solving for speed v

v =0.7 \times \frac{P_{max}}{ma_{max}}

v = 0.7 \frac{217 [\frac{746 w}{1 hp}]}{1500 \times 6.53}

v = 11.5689 m/s

7 0
2 years ago
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