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labwork [276]
3 years ago
8

Nuclear fusion occurs when a heavy element decays to a lighter one hydrogen nuclei divide a helium atom splits atomic nuclei joi

n or come together
Physics
1 answer:
Cloud [144]3 years ago
5 0
Fusion usually means something is joining to make something else. Nuclear fusion happened, for example, in the Big Bang, where subatomic particles came together to make Hydrogen and Helium. In short, the last option is your answer. :)
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Over the years most technology has gotten?
WITCHER [35]
More expensive & probably less complicated
3 0
4 years ago
How many atoms are in 3.5 moles of sulfur
irina [24]
There are 2.1077x10^24 atoms of sulfur in 3.50 mols of sulfur.
5 0
3 years ago
Use the formula h = −16t2 + v0t. (if an answer does not exist, enter dne.) a ball is thrown straight upward at an initial speed
makkiz [27]
Using the given formula with v0=56 ft/s and h=40 ft 
h = -16t2 + v0t  
40 = -16t2 + 56t 
16t2 - 56t + 40 = 0  
Solving the quadratic equation:  
t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32 
 We have two possible solutions  
t1 = (56+24)/32 = 2.5 
t2 = (56-24)/32 = 1  
So initially the ball reach a height of 40 ft in 1 second.
3 0
4 years ago
Initilal velocity of projectile at an angle of 22.5° is 10m/s. Then what is the magnitude of range of projectile?
crimeas [40]

Answer:

0.707m

Explanation:

from formula of range i.e R=Usin2Q/g

3 0
3 years ago
A cat leaps into the air to catch a bird with an initial speed of 2.74 m/s at an angle of 60.0° above the ground. What is the hi
Volgvan

Answer: D. 0.29 m

Explanation:

We will use the following equations to describe the leap of the cat:

y=V_{o}sin\theta t-\frac{gt^{2}}{2}   (1)

V_{y}=V_{oy}-gt   (2)

Where:

y  is the height of the cat  

V_{oy}=V_{o}sin\theta is the cat's initial velocity

\theta=60\°

g=9.8m/s^{2}  is the acceleration due gravity

t is the time

V_{y} is the y-component of the velocity

Now the cat will have its maximum height y_{max} when V_{y}=0. So equation (2) is rewritten as:

0=V_{oy}-gt   (3)

Finding t:

t=\frac{V_{oy}}{g}=\frac{V_{o}sin\theta}{g}   (4)

t=\frac{2.74 m/s sin(60\°)}{9.8m/s^{2}}   (5)

t=0.24 s   (6)

Substituting (6) in (1):

y_{max}=(2.74 m/s)sin(60\°) (0.24 s)-\frac{(9.8m/s^{2})(0.24 s)^{2}}{2}   (7)

Finally:

y_{max}=0.287 m \approx 0.29 m   (8)

3 0
3 years ago
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