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meriva
3 years ago
13

6) If nuclei A has a probability P of decaying in time t and nucleus B has a probability 2p of decaying in time t, which stateme

nts is true about the half-lives of nucleus A and nucleus B? a. The half-life of nucleus A is twice the half-life of nucleus B. b. The half-life of nucleus B is twice the half-life of nucleus A. c. The half-life of nucleus A is the same as the half-life of nucleus B. d. The half-life of nucleus A is e’ times greater than the half-life of nucleus B. ​
Physics
1 answer:
vredina [299]3 years ago
6 0

Answer:

sorry dont know so so sorry

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Ryan is looking at investment opportunities as a cloud service provider. he wants to invest in a deployment-based cloud model th
Dominik [7]

Answer : a. Community

Allows a system to be accessible by a group of organizations. It can be shared between several organizations. It may be managed by organizations or by the third party.


This should be chosen by Ryan, since this computing model is cost effective and best to share among companies and organizations.



Other options explained:

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5 0
3 years ago
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Why doesn't the skateboarder rolls as high up the ramp each time you go down the ramp ​
Alexandra [31]
Because of the law of inertia and it’s effect on the skater
5 0
3 years ago
A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 6.0 oz. Aluminum has a density of 2.70 g/cm3. Wh
SIZIF [17.4K]

Answer:

1.36 x 10^-3 cm

Explanation:

Area = 50 ft^2 = 46451.5 cm^2

mass = 6 oz = 170.097 g

density = 2.70 g/cm^3

Let t be the thickness of foil in cm.

mass = volume x density

mass = area x thickness x density

170.097 = 46451.5 x t x 2.70

t = 1.36 x 10^-3 cm

Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.

3 0
2 years ago
At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There
katen-ka-za [31]

a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

\tau_{net} = I \alpha (1)

where

\tau_{net} is the net torque acting on the object in rotation

I is the moment of inertia of the object

\alpha is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

\alpha = \frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time interval

So we can rewrite eq.(1) as

\tau_{net}=I\frac{\Delta \omega}{\Delta t}

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

\tau_{net}=0

From the previous equation, this implies that

\Delta \omega =0

Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

\tau_{net}=0

Therefore, this means that

\Delta \omega =0

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

8 0
3 years ago
Please help! Will mark best answer!!
MA_775_DIABLO [31]
The answer is (A) hope it helps 
7 0
2 years ago
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