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BlackZzzverrR [31]

2 years ago

12

A 432 g sample of 60/27Co has a decay constant of 4.14 x 10-9 s-1. How long will it take before only 1/3 of the original sample

remains?

1 answer:

Musya8 [376]2 years ago

7 0

Answer:

remain 1s60

Explanation:

I took away sample

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Desde lo alto de un acantilado de 140 m, se lanza verticalmente un objeto hacia abajo con velocidad de 3m/s. Entonces la magnitu

grin007 [14]

Answer:

54

Explanation:

d = vo*t + ½*g*t²

d = 3*3 + ½*10*3²

d = 9 + 45

d = 54 m

entonces el objeto tiene 54 m de desplazamiento

7 0

3 years ago

Does uranium use steam?

Neko [114]

Answer:

Yes.

Explanation:

Reactors use uranium for nuclear fuel. The uranium is processed into small ceramic pellets and stacked together onto sealed metal tubes called fuel rods. The heat created by fission turns the water into steam.

4 0

3 years ago

You can also describe position by using coordinates of _____and _______

SOVA2 [1]

Answer:

x and y

Explanation:

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3 years ago

A ball initially at rest rolls down a hill with an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move?

goldfiish [28.3K]

<u>Answer:</u>

Ball will move 92.8125 meter along the cliff in 7.5 seconds.

<u>Explanation:</u>

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case initial velocity = 0 m/s, acceleration = 3.3 $m/s^2$ , we need to calculate displacement when time = 7.5 seconds.

Substituting

$s=0*7.5+\frac{1}{2} *3.3*7.5^2\\ \\ =92.8125 meter$

So ball will move 92.8125 meter along the cliff in 7.5 seconds.

5 0

3 years ago

Two 2 kg masses is placed at either end of a rod that has a mass of .5 kg and a length of 3 m. What is the moment of inertia if

sergij07 [2.7K]

Explanation:

a) $I=\displaystyle \sum_{i}m_ir_i^2$

where $r_i$ is the distance of the mass $m_i$ from the axis of rotation. When the axis of rotation is placed at the end of the rod, the moment of inertia is due only to one mass. Therefore,

$I= mr^2 = (2\:kg)(3\:m)^2 = 18\:kg-m^2$

b) When the axis of rotation is placed on the center of the rod, the moment is due to both masses and the radius r is 1.5 m. Therefore,

$\displaystyle I= \sum_{i}m_ir_i^2 = 2(2\:kg)(1.5\:m)^2 = 9\:kg-m^2$

7 0

3 years ago