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Vadim26 [7]
2 years ago
13

The half-life of Cs-137 is 30.2 years. If the initial mass of the sample is 1.00 kg, how much will remain after 151 years?

Physics
2 answers:
nexus9112 [7]2 years ago
8 0

Answer: Its B. There will be 0.031 kg left after 151 years,

Explanation: Im doing my assessment right now.

Xelga [282]2 years ago
3 0

Answer:

i found something on a website that will help you

Explanation:

do you want me to send you it???

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If you were to be drawn into a black hole, what would happen? To the black hole, not to you.
love history [14]

Answer:

It grows

Explanation:

The blacks holes will absorb

Me hoizontally stretching me like a noodle by the spaghtification process,thus growing bigger.

7 0
3 years ago
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0
3 years ago
If the change in kinetic energy of a tennis ball hit by the racket
Marrrta [24]

Answer: .36 m

Explanation:

5 0
2 years ago
What causes the spiral structural in milky way?
Mrac [35]
The spiral structure of the milky way can be explained by long lived quasi-static density waves<em>, </em><em>according to the lin-shu hypothesis. </em>Curiously, the waves of higher density gas and stars (seen as spiral arms) appear to remain static as stars move around the galaxy. This explained by assuming that the gravitational disturbances cause by the 'clumping' material in the arms does not affect the gravitational field of the galaxy as whole and is therefore negligible.

source: Astrophysicist
6 0
3 years ago
Help it’s multiple choice 11 through 15 please!
riadik2000 [5.3K]

1. • Here, force of gravity on the block = 20 N.

• Therefore, the normal force will also be the same, i.e., 20 N [According to Newton's Third Law, on every action, there is an equal and opposite reaction]

• The coefficient

u_{k} = 0.4

• Force of friction =

u_{k} \times  \: normal \:  \:  \: force \\  = 0.4 \times 20N \\  = 8N

• Hence, the force of sliding friction between the block and the ground is 8 N.

• So, it is option c. 8 N

2. The answer is option d. continue in the same direction with no change in speed.

We know, force = mass × acceleration. When force is 0, then acceleration will also be 0 since mass cannot be 0. So, there will be no change in speed.

3. It is option b. force that is required to give a one kilogram object the acceleration of 1 m/s^2.

Newton is the SI unit of force. As mentioned earlier, force = mass × acceleration. The SI unit of mass and acceleration is Kg and m/s^2 respectively.

So, 1 N = 1 Kg × 1 m/s^2.

4. It is d. not zero.

Acceleration is the change in speed. So, if the force is zero, then acceleration will not occur.

5. Force = 2 N

Acceleration of the object A = 2 m/s^2.

Acceleration of the object B = 1 m/s^2.

Therefore, mass of the object A = 2 N ÷ 2 m/s^2 = 1 Kg

And, mass of the object B = 2 N ÷ 1 m/s^2 = 2 Kg

So, the mass of object B is greater than that of object A.

Hence, the answer is option c. Object B has more mass.

Hope you could get an idea from here.

Doubt clarification - use comment section.

3 0
2 years ago
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