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White raven [17]
1 year ago
11

Hydrogen-3 has a half-life of 12.35 years. What mass of hydrogen-3 will remain form a 100.0 MG initial sample after 5.0 years? A

) 7 MG B) 24 MG C) 76 MG D) 18 MG
Physics
1 answer:
bija089 [108]1 year ago
8 0
I haven't worked out the number, but it won't be necessary ... it's easy to pick the answer out of the choices that are given. ... The time (5 yrs) is less than the half-life (12.35 yrs), so more than half of the original sample (100mg) remains. Only one of the choices is more than 50 mg. (Can you find it ?)
You might be interested in
Two window washers, Bob and Joe, are on a 3.00 m long, 395 N scaffold supported by two cables attached to its ends. Bob weighs 8
WINSTONCH [101]

Answer:

- the forces on the left hand side is 1.038 kN

- the forces on the right hand side is 1.483 kN

Explanation:

Given the data in the question, as illustrated in the image below;

Length of the scaffold = 3 m

weight of the scaffold = 395 N

Weight of Bob = 805 N and stands 1 m from the left end

weight of washing equipment = 500N and on sits 2 m from the left end

Weight of Joe = 820 N and stand 0.500 m from the right end

so the force on the left cable will be;

T_{left = \frac{1}{3m}[ (805 N)( (3-1) m) + ( 395 N )( \frac{3}{2} m) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]

T_{left =  \frac{1}{3m}[ 1610 + 592.5 + 500 + 410 ]

T_{left =  \frac{1}{3m}[ 3112.5 ]

T_{left =  1037.5 N

T_{left =  1.038 kN

Therefore, the forces on the left hand side is 1.038 kN

On the right hand side;

T_{Right =  ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N

T_{Right =  2520 N - 1037.5 N

T_{Right =  1482.5 N

T_{Right =  1.483 kN

Therefore, the forces on the right hand side is 1.483 kN

5 0
1 year ago
How many significant figures do each of the following numbers have: (a) 214, (b) 81.60, (c) 7.03, (d) 0.03, (e) 0.0086, (f) 3236
Korolek [52]

In determining the number of significant figures in a given number, there are three rules to always remember / follow:

First: All integers except zero are always significant.

<span>Second: Any zeros located between  non zeroes are always significant.</span>

Third: A zero located after a non zero in a decimal is always significant whether it is before or after the decimal

 

Therefore using this rule, the number of significant digits in the given numbers are:

(a) 214 = 3

(b) 81.60 = 4

(c) 7.03 = 3

(d) 0.03 = 1

(e) 0.0086 = 2

(f) 3236 = 4

(g) 8700 = 2

4 0
2 years ago
Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p
worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

  • U1/U2=6
  • U1/U2=6

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy  is mathematically given as

U=\frac{-k*qQ}{r}

Therefore

The Equations of U1 and U2 is

For U1

U1=\frac{-k*q_1Q}{r}

For U2

U2=\frac{-k*q_1Q}{3*2r}

Since

U is a function of q and  q2=q1/3

Therefore

U1/U2=6

For Question 2

For U1

U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}

Therefore

U1/U2=6

For more information on this visit

brainly.com/question/23379286?referrer=searchResults

7 0
1 year ago
An alloy is made of a material of specific gravity 7.87 and another material of specific gravity 4.50. The alloy of mass 750g ha
julsineya [31]

Answer:

13.9

Explanation:

Apparent weight is the normal force.  Sum of the forces on the alloy when it is submerged:

∑F = ma

N + B − W = 0

N + ρVg − mg = 0

6.6 + (0.78 × 1000) V (9.8) − (0.750) (9.8) = 0

V = 9.81×10⁻⁵

If x is the volume of the first material, and y is the volume of the second material, then:

x + y = 9.81×10⁻⁵

(7.87×1000) x + (4.50×1000) y = 0.750

Two equations, two variables.  Solve with substitution:

7870 (9.81×10⁻⁵ − y) + 4500 y = 0.750

0.772 − 7870 y + 4500 y = 0.750

0.0222 = 3370 y

y = 6.58×10⁻⁶

x = 9.15×10⁻⁵

The ratio of the volumes is:

x/y = 13.9

8 0
2 years ago
How high above the earth's surface is g reduced to 8.80m/^2?
Sladkaya [172]
Gravity changes as the altitude change.<span> The gravitational force is proportional to 1/R2,  where R is the distance from the center of the Earth the radius of earth where gravity is 9.8 m/s^2 is 6400 km this will serve as the zero mark.

g1/(g2) = R2^2/(R1)^2

so we set the constant values to R1 and the unknown distance as x

(9.8)/(8.80) = (6400-x)2/(6400)^2

solving for x we will get 

x = 345.85 km above the earths surface
</span>

<span>Hope my answer would be a great help for you.    
If </span>you have more questions feel free to ask here at Brainly.

<span> </span>


6 0
2 years ago
Read 2 more answers
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