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Kitty [74]
3 years ago
14

8. Contrast. The energy that can be released during a nuclear fission reaction with the energy that can be released during a nuc

lear fission reaction.
9. Contest. Alpha particles, beta particles, and gamma rays


10. Explain. Why mass-energy equivalence is not apparent for chemical reactions


11. Think critically. Explain why high temperatures are needed for fission reactions to occur but not for fission reactions to occur
Physics
1 answer:
Cloud [144]3 years ago
7 0
8) the energy released by fusion is generally 3 to 4 times larger than with fission.  Fission has very few by-products but fusion releases large amounts of radioactive particles because it starts with large nuclei.
9) Alpha particles are 2 protons and 2 neutrons all put together.  It's really the nucleus of a helium atom.  It is most dangerous if you ingest it but it can be stopped with a sheet of paper so outside the body it's not as dangerous as others and due to its size it can't get very far in the air before hitting air molecules
beta particles are high energy electrons or positrons.  They travel further due to their small size but can be stopped by a thin barrier of plastic or wood.
Gamma rays are high frequency photons (light)  They are stopped by metal plates and go through human tissue.  They are quite dangerous.
10) The mass that is lost in chemical reactions is very small. Solve E=mc² for mass and you get m=E/c².  This says the mass you lose is equal to the energy you gained divided by the speed of light squared.  c² is a VERY big number so you need a lot of energy produced to notice it.  Chemical reactions are simply too inefficient to get that much energy out.
11)You need high temperatures for fusion because you're trying to push two atoms together (to "fuse" them as the name suggests)  The electrons in one atom repel the other electrons in the other atoms.  When stripped down to only protons, you still have to overcome this repulsion (Coulomb repulsion).  High temperatures means high velocity of the particles in the plasma.  This gives them enough "oomph" to get close enough to fuse.  Once close enough to each other, the nuclear force takes over and overwhelms the Coulomb repulsion and the nuclei fuse and release energy in doing so.
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Witch statement about polymers is true
Igoryamba
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3 0
3 years ago
A volleyball player bumps a ball across a net with the velocity and angle shown below. What is the maximum height of the ball?
Marrrta [24]

Answer:

D. 12.4 m

Explanation:

Given that,

The initial velocity of the ball, u = 18 m/s

The angle at which the ball is projected, θ = 60°

The maximum height of the ball is given by the formula

                             h = u² sin²θ/2g  m

Where,

                           g - acceleration due to gravity. (9.8 m/s)

Substituting the values in the above equation

                            h = 18² · sin²60 / 2 x 9.8

                               = 18² x 0.75 / 2 x 9.8

                               = 12.4 m

Hence, the maximum height of the ball attained, h = 12.4 m

6 0
3 years ago
A skier of mass 60 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him 75 m up a 30° slop
postnew [5]

Answer:

Explanation:

Given

mass of skier=60 kg

distance traveled by skier=75 m

inclination(\theta )=30^{\circ}

speed (v)=2.4 m/s

as the skier is moving up with a constant velocity therefore net force is zero

F_{net}=0

Force applied by cable=mg\sin \theta

F=60\times 9.8\times \sin (30)=294 N

work done=F\cdot x

W=294\cdot 75=22.125 J

(b)Power=F\cdot v

P=294\cdot 2.4=705.6 W\approx 0.946 hp

5 0
3 years ago
(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

8 0
3 years ago
What is the weight of a 1-kilogram brick resting on a table?
MakcuM [25]

Answer:

The weight if the block is 10Newtons

Explanation:

The weight of any object is quantity of matter the object contains and it is always acting downwards on such body. This shows that the object is under the influence of gravity.

The weight of an object is calculated as mass of the object × its acceleration due to gravity

W = mg

Give the mass of the brick to be 1kg

g is the acceleration due to gravity = 10m/s²

Weight of the object = 1 × 10

= 10kgm/s² or 10Newtons

5 0
3 years ago
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