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AURORKA [14]
3 years ago
8

HELP... 3.00 amu = _____ Mev. 3.22 x 10-3 2.79 x 103 3.10 x 102

Physics
2 answers:
Nutka1998 [239]3 years ago
7 0

Answer:

2.79x103

Explanation:

DanielleElmas [232]3 years ago
5 0
The correct answer is:
2.79 \cdot 10^3 MeV
Let's see why.

1 amu corresponds to the mass of the proton, which is:
m_p = 1.66 \cdot 10^{-27} kg
if we convert this into energy, using Einstein equivalence between mass and energy, we find:
E=mc^2 = (1.66 \cdot 10^{-27} kg)(3\cdot 10^8 m/s)^2 = 1.49 \cdot 10^{-10} J
Now we can convert it into electronvolts:
E= \frac{1.49 \cdot 10^{-10}kg}{1.6 \cdot 10^{-19} J/eV} =9.34 \cdot 10^9 eV = 934 MeV

So, 1 amu = 934 MeV. Therefore, 3 amu corresponds to 3 times this value:
3 amu = 3 \cdot 934 MeV  \sim 2790 MeV = 2.79 \cdot 10^3 MeV
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A ceiling fan rotates at the rate of 45 degree every 0.75sec. what is the angular speed of the fan in rad/sec
babymother [125]

45° × π/180 = 0,7854 rad

ω = θ /t

ω = 0.7854 / 0.75 = 1.0472 rad/s

6 0
2 years ago
What type of reaction occurs when an egg cooks in a pan?
Elan Coil [88]

Answer:

exothermic

Explanation:

energy is absorbed by the surroundings

5 0
3 years ago
Read 2 more answers
The A 400 kg spacecraft has a momentum of 30,000 kg·m/s up. What is the velocity? A. 12,000,000 m/s B. 1200 m/s C. 300 m/s D. 75
tia_tia [17]
Momentum of an object is calculated by multiplying the mass by the velocity. 

p = mv 
where:
p = momentum
m = mass
v = velocity

Let's take your given into account and put it in the equation:

p = mv
30,000 kg.m/s = (400kg)v

Velocity is our unknown, so to get it all we need to do is transfer mass (m) to the other side of the equation and isolate the velocity (v). When we do this, we need to use the opposite operation (rules of transposition). 

(30,000kg.m/s)/(400kg) = v

Cancel out the kg and you are left with m/s.

75m/s = v

The answer is then D. 75 m/s.

Now for your second question, as you can see in the formula, mass and velocity is directly proportional to momentum. That means that the higher the mass or the velocity, the higher the momentum.

So if the velocity increases, the momentum increases as well. 




 


8 0
2 years ago
What is the term for trade without barriers, tariffs, or quotas?
Amanda [17]
I think its free trade
5 0
2 years ago
Starting from rest, a wheel undergoes constant angular acceleration for a period of time T. At what time after the start of rota
rjkz [21]

Answer:

<em>At t=T/2 the angular speed equals to its average angular speed </em>

Explanation:

<u>Angular Motion</u>

Let w be the angular speed of a rotating object, \alpha its angular acceleration, and T the time the acceleration is acting upon the object. The basic formula for the angular motion is

w=w_o+\alpha t

We are told the initial speed is zero, so

w=\alpha t

The average angular speed from t=0 to t=T can be found by

\displaystyle \bar w=\frac{1}{T}\int_0^T{w.dt}

\displaystyle \bar w=\frac{1}{T}\int_0^T{\alpha t.dt}

\displaystyle \bar w=\frac{1}{T}\frac{\alpha T^2}{2}

\displaystyle \bar w=\frac{\alpha T}{2}

This value is reached at a certain time we need to compute, knowing that

w=\bar w

Or equivalently

\displaystyle \alpha t=\frac{\alpha T}{2}

Simplifying we have

\displaystyle t=\frac{T}{2}

At t=T/2 the angular speed equals to its average angular speed

4 0
2 years ago
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