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Maurinko [17]
3 years ago
6

When there are too many neutrons compared to the number of protons, the atomic nucleus is

Physics
1 answer:
Darina [25.2K]3 years ago
5 0

Answer:

99.9% sure it B. Unstable but if that’s wrong try A.stable

Explanation:

Too many protons (or too few neutrons) in the nucleus result in an imbalance between forces, which leads to nuclear instability.

if a nucleus is unstable due to too many neutrons, it will undergo Beta decay - this means they become stable by emitting a beta particle.

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Which equation could not be used to determine straight line acceration?
uranmaximum [27]

Answer:

the answer for the question is the last option

5 0
3 years ago
Allen Aby sets up an Atwood Machine and wants to find the acceleration and the tension in the string. Please help him. Two block
Vitek1552 [10]

<u>Answers:</u>

In order to solve this problem we will use Newton’s second Law, which is mathematically expressed after some simplifications as:

<h2>F=ma   (1) </h2>

This can be read as: The Net Force F of an object is equal to its mass m multiplied by its acceleration a.

We will also need to <u>draw the Free Body Diagram of each block</u> in order to know the direction of the acceleration in this system and find the Tension T of the string (<u>See figure attached).  </u>

We already know<u> m_{2} is greater than m_{1}</u>, this means the weight of the block 2 P_{2} is greater than the weight of the block 1 P_{1}; therefore <u>the acceleration of the system will be in the direction of P_{2}</u>, as shown in the figure attached.

We also know by the information given in the problem that <u>the pulley does not have friction and has negligible mass</u>, and <u>the string is massless</u>.

This means that the tension will be the same along the string regardless of the difference of mass of the blocks.

Now that we have the conditions clear, let’s begin with the calculations:

1) Firstly, we have to find the weight of each block, in order to verify that block 2 is heavier than block 1.

This is done using equation (1), where the force of the weight P is calculated using the <u>acceleration of gravity</u> g=9.8\frac{m}{s^{2}}  acting on the blocks:


<h2>P=mg   (2) </h2>

<u>For block 1: </u>

P_{1}=m_{1}g   (3)

P_{1}=1.5kg(9.8\frac{m}{s^{2}})    

<h2>P_{1}=14.7N   (4) </h2>

<u>For block 2: </u>

P_{2}=m_{2}g   (5)

P_{2}=2.4kg(9.8\frac{m}{s^{2}})    

<h2>P_{2}=23.52N      (6) </h2>

Then, we are going to <u>find the acceleration a of the whole system: </u>

F_{r}=P_{1}+P_{2}   (7)

<h2>P_{1}+P_{2}=(m_{1}+m_{2})a   (8) </h2>

Where the Resulting Force F_{r}  is equal to the sum of the weights P_{1} and P_{2}.  

In the figure attached, note that P_{1} is in opposite direction to the acceleration a, this means it must <u>have a negative sing</u>; while P_{2} is in the same direction of a.

Here we only have to isolate a from equation (8) and substitute the values according to the conditions of the system:

-14.7N+23.52N=(1.5kg+2.4kg)a  

8.82N=(3.9kg)a  

Then:

a=\frac{8.82N }{3.9kg}  

<h2>a=2.26\frac{m}{ s^{2}}  </h2><h2>This is the acceleration of the system. </h2>

2) For the second part of the problem, we have to find the tension T of the string.

We can choose either the Free Body Diagram of block A or block B to make the calculations, <u>the result will be the same</u>.  

Let’s prove it:

For m_{1}

we see in the free body diagram that the <u>acceleration is in the same direction of the tension of the string</u>, so:

F_{r}=T-P_{1}   (9)

T-P_{1}=m_{1}a   (10)

T-14.7N=(1.5kg)( 2.26\frac{m}{ s^{2}})    

Then;

<h2>T=18.09N   This is the tension of the string </h2><h2> </h2>

For m_{2}

we see in the free body diagram that the acceleration is in opposite direction of the tension of the string and must <u>have a negative sign,</u> so:

F_{r}=T-P_{2}   (9)

T-P_{2}=m_{2}a   (10)

T-23.52N=(2.4kg)(-2.26\frac{m}{ s^{2}})    

Then;

<h2>T=18.09N    This is the same tension of the string </h2>

6 0
3 years ago
What can iron filings sprinkled on a piece of paper demonstrate? A. the clockwise motion of the magnetic field B. the counter-cl
d1i1m1o1n [39]

The answer is C. because the lines will look the same as the earths gravity field (you can look up a graph)

8 0
3 years ago
Read 2 more answers
An object at rest does not _____ and an object in motion does not _____, unless an _____ force acts upon it
mel-nik [20]

Answer:

An object at rest does not move and an object in motion does not change its velocity, unless an external force acts upon it

Explanation:

This statement is also known as Newton's first law, or law of  inertia.

It states that the state of motion of an object can be changed only if there is an external force (different from zero) acting on it: therefore

- If an object is at rest, it will remain at rest if there is no force acting on it

- If an object is moving, it will continue moving at constant velocity if there is no force acting on it

This phenomenon can be also understood by looking at Newton's second law:

F = ma

where

F is the net force on an object

m is the mass

a is the acceleration

If the net force is zero, F = 0, the acceleration of the object is also zero, a = 0: therefore, the velocity of the object does not change, and it will continue moving at the same velocity (which can be zero, if the object was at rest).

5 0
3 years ago
A 4.3-kg block slides down an inclined plane that makes an angle of 30° with the horizontal. Starting from rest, the block slide
-BARSIC- [3]

Answer:

0.56

Explanation:

Let the coefficient of friction is μ.

m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s

By the free body diagram,

Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N

Friction force, f = μ N = 36.49 μ

Net force acting on the block,

Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ

Fnet = 21.07 - 36.49μ

Net acceleartion, a = Fnet / m

a = (21.07 - 36.49μ) / 4.3

use second equation of motion

s = ut + 1/2 a t^2

2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3

By solving we get

μ = 0.56

8 0
4 years ago
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