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lozanna [386]
3 years ago
10

How does the currently accepted model of the nucleus provide evidence of the existence of the strong nuclear force?

Physics
1 answer:
olganol [36]3 years ago
4 0

Answer:

Protons and neutrons are all attracted to each other as a result - the strong nuclear force. This is an attractive force that only has an effect over a very short range in the nucleus.

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A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 21.2 N; when it is completely i
blsea [12.9K]

Answer:

7066kg/m³

Explanation:

The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:

Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)

= 7066kg/m³

Explanation:

8 0
3 years ago
Read 2 more answers
A 52.0-kg person, running horizontally with a velocity of +3.63 m/s, jumps onto a 15.2-kg sled that is initially at rest. (a) Ig
trasher [3.6K]

Answer:

The coefficient of kinetic friction between the sled and the snow is 0.0134

Explanation:

Given that:

M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

After collision, the person and the sled would move with the same velocity V.

a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

V=\frac{MU+mu}{M+m}

Substituting values:

V=\frac{MU+mu}{M+m}=\frac{52(3.63)+15.2(0)}{52+15.2} =2.81m/s

The velocity of the sled and person as they move away is 2.81 m/s

b) acceleration due to gravity (g) = 9.8 m/s²

d = 30 m

Using the formula:

V^2=2\mu(gd)\\\mu=\frac{V^2}{2gd} \\\mu=\frac{2.81^2}{2*9.8*30} =0.0134

The coefficient of kinetic friction between the sled and the snow is 0.0134

3 0
3 years ago
A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220
disa [49]
First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, \mu m g. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
ma=-\mu m g
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
2aS=v_f^2-v_i^2
where v_f=0 is the final speed of the skier and v_i=4.75 m/s is the initial speed. Substituting numbers, we find:
S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m
5 0
3 years ago
An egg of mass 0.060 kg is dropped from the top of a building. Just before it reaches the ground, it has a total kinetic energy
3241004551 [841]

The velocity is 14 m/s

The parameters given on the question are

mass= 0.060 kg

kinetic energy= 5.9 joules

K.E= 1/2mv²

5.9= 1/2 × 0.060 × v²

5.9= 0.5 × 0.060v²

5.9= 003v²

v²= 5.9/0.03

v²= 196.66

v= √196.66

v= 14 m/s

Hence the velocity of the egg before it strikes the ground is 14 m/s

brainly.com/question/2084569?referrer=searchResults

3 0
2 years ago
Dans car has broken down. He decides to push it to a garage 800m away qlong a flat road. He pushes the car with qn average force
Ede4ka [16]

Answer:

Answer = 480000 joule.

mark me the Brainliest.. pls and follow me

3 0
2 years ago
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