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lorasvet [3.4K]
4 years ago
10

Which of these statements would best explain the problem encountered with nuclear waste disposal? A) The isotopes have a long ha

lf-life and only remain radioactive for a long time period. B) The isotopes have a short half-life and only remain radioactive for a long time period. C) The isotopes have a long half-life and only remain radioactive for a short time period. D) The isotopes have a short half-life and only remain radioactive for a short time period. 
ANSWER IS A)
Physics
1 answer:
algol134 years ago
4 0
That's right, the correct answer is
<span>A) The isotopes have a long half-life and only remain radioactive for a long time period

The half life of an isotope is the time it takes for the amount of the sample to reduce to half of its initial value. If an isotope has a long half-life, it means it takes a long time to reduce down to a significant level, so it will remain radioactive for a long time period.</span>
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A 0.145 kg baseball is thrown with a velocity of 25.0 m/s. How much work was done on the baseball to bring it from rest to 25.0
sergey [27]

Answer:

45.31 J

Explanation:

We are given that

Mass of baseball , m=0.145 kg

Initial velocity, u=0

Final velocity, v=25 m/s

We have to find the work done on the baseball to bring it from rest to 25 m/s

We know that

Work done = Change in kinetic energy

Work done, W=\frac{1}{2}m(v^2-u^2)

Using the formula

Work done, W=\frac{1}{2}(0.145)((25)^2-0)

Work done=\frac{1}{2}(0.145)(625)

Work done, W=45.31 J

Hence, the work done on the baseball to bring it from rest to 25 m/s

=45.31 J

3 0
3 years ago
The cooking time for a roast scales like the 2/3rds power of the mass. Based on scaling laws, how much longer does a 20-lb roast
fgiga [73]

Answer:

2.726472 s more or 1.5874 times more time is taken than 10-lb roast.

Explanation:

Given:

- The cooking time t is related the mass of food m by:

                                    t = m^(2/3)

- Mass of roast 1 m_1 = 20 lb

- Mass of roast 2 m_2 = 10 lb

Find:

how much longer does a 20-lb roast take than a 10-lb roast?

Solution:

- Compute the times for individual roasts using the given relation:

                                  t_1 = (20)^(2/3) = 7.36806 s

                                  t_2 = (10)^(2/3) = 4.641588 s

- Now take a ration of t_1 to t_2, to see how many times more time is taken by massive roast:

                                  t_1 / t_2 = (20 / 10)^(2/3)

- Compute:                t_1 / t_2 = 2^(2/3) = 1.5874 s

- Hence, a 20-lb roast takes 1.5874 times more seconds than 10- lb roast.

                                  t_2 - t_1 = 2.726472 s more

3 0
3 years ago
Five-gram samples of copper and aluminum are at room temperature. Both receive equal amounts of energy due to heat flow. The spe
laila [671]

m₁ = mass of sample of copper = m₂ = mass of sample of aluminum = 5 g

T = initial temperature of copper = initial temperature of aluminum

T₁ = final temperature of copper

T₂ = final temperature of aluminum

c₁ = specific heat of copper = 0.09 cal/g°C

c₂ = specific heat of aluminum = 0.22 cal/g°C

Since both receive same amount of heat, hence

Q₁ = Q₂

m₁ c₁ (T₁ - T) = m₂ c₂ (T₂ - T)

(5) (0.09) (T₁ - T) = (5) (0.22) (T₂ - T)

T₁ - T = (2.44)  (T₂ - T)

Change in temperature of copper = (2.44) change in temperature of aluminum

hence the correct choice is

c. The copper will get hotter than the aluminum.

4 0
3 years ago
Read 2 more answers
Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of 1012 W) pu
Tcecarenko [31]

1. 7.95\cdot 10^6 J

The total energy given to the cells during one pulse is given by:

E=Pt

where

P is the average power of the pulse

t is the duration of the pulse

In this problem,

P=1.59\cdot 10^{12}W

t=5.0 ns = 5.0\cdot 10^{-9} s

Substituting,

E=(1.59\cdot 10^{12}W)(5.0 \cdot 10^{-6}s)=7.95\cdot 10^6 J

2. 1.26\cdot 10^{21}W/m^2

The energy found at point (1) is the energy delivered to 100 cells. The radius of each cell is

r=\frac{4.0\mu m}{2}=2.0 \mu m = 2.0\cdot 10^{-6}m

So the area of each cell is

A=\pi r^2 = \pi (2.0 \cdot 10^{-6}m)^2=1.26\cdot 10^{-11} m^2

The energy is spread over 100 cells, so the total area of the cells is

A=100 (1.26\cdot 10^{-11} m^2)=1.26\cdot 10^{-9} m^2

And so the intensity delivered is

I=\frac{P}{A}=\frac{1.59\cdot 10^{12}W}{1.26\cdot 10^{-9} m^2}=1.26\cdot 10^{21}W/m^2

3. 9.74\cdot 10^{11} V/m

The average intensity of an electromagnetic wave is related to the maximum value of the electric field by

I=\frac{1}{2}c\epsilon_0 E^2

where

c is the speed of light

\epsilon_0 is the vacuum permittivity

E is the amplitude of the electric field

Solving the formula for E, we find:

E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(1.26\cdot 10^{21} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12}F/m)}}=9.74\cdot 10^{11} V/m

4. 3247 T

The magnetic field amplitude is related to the electric field amplitude by

E=cB

where

E is the electric field amplitude

c is the speed of light

B is the magnetic field

Solving the equation for B and substituting the value of E that we found at point 3, we find

B=\frac{E}{c}=\frac{9.74\cdot 10^{11} V/m}{3\cdot 10^8 m/s}=3247 T

3 0
3 years ago
Use the formula h = −16t2 + v0t. (if an answer does not exist, enter dne.) a ball is thrown straight upward at an initial speed
makkiz [27]
Using the given formula with v0=56 ft/s and h=40 ft 
h = -16t2 + v0t  
40 = -16t2 + 56t 
16t2 - 56t + 40 = 0  
Solving the quadratic equation:  
t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32 
 We have two possible solutions  
t1 = (56+24)/32 = 2.5 
t2 = (56-24)/32 = 1  
So initially the ball reach a height of 40 ft in 1 second.
3 0
4 years ago
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