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matrenka [14]
3 years ago
12

Although they're different what relationship exists between facts and theories

Physics
1 answer:
AfilCa [17]3 years ago
4 0

if a theory can be verified by an experiment ten it's a good theory. if an expt contradicts a theory, theory is wrong.


at CERN in europe, a new fundamental particle was said to be found.

Professor Peter  Higgs had a theory predicting it.

called the "higgs boson" he got a nobel prze for the boson.

a partcle of light = a photom = a boson ...

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A solid cylinder of uniform density of 0.85 g/cm3 floats in a glass of water tinted light blue by food coloring.
SVEN [57.7K]

Each side has to have at least 44 horses

F61160 N. This is further explained below.

<h3>What is the force?</h3>

Generally, We are only interested in the component that operates horizontally since the vertical components all cancel each other out. The pressure difference works on the hemisphere to generate a normal force all over the surface, but we are only concerned with that force's horizontal component. This may be determined by supposing the hemispheres to be two flat circular plates of the same radius as the hemispheres that have been forced together.

Therefore, force is equal to pressure multiplied by area, which is

F= (970 -15 )( * (0.45 m)2)

F=60754 N for each side.

Therefore, each side has to have at least 44 horses

44* 1390 = 61160 N

Read more about force

brainly.com/question/13191643

#SPJ1

3 0
2 years ago
You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C
jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0
4 years ago
A player is positioned 35 m[40 degrees W of S] of the net. He shoot the puck 25 m [E] to a teammate. What second displacement do
Lubov Fominskaja [6]

Answer:

x=22.57 m

Explanation:

Given that

35 m in W of S

angle = 40 degrees

25 m in east

From the diagram

The angle

\theta=90-40=50^o

From the triangle OAB

cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}

1340.57=35^2+25^2-x^2

x=22.57 m

Therefore the answer of the above problem will be 22.57 m

4 0
3 years ago
Why is electrostatic force able to act at a distance ​
mote1985 [20]
In Electrostatics the electrical force between Two charged objects is inversely Related to the distance of separation between the two objects .
7 0
3 years ago
A 0.49-kg cord is stretched between two supports, 7.8m apart. When one support is struck by a hammer, a transverse wave travels
katovenus [111]

To solve this problem we will apply the laws of Mersenne. Mersenne's laws are laws describing the frequency of oscillation of a stretched string or monochord, useful in musical tuning and musical instrument construction. This law tells us that the velocity in a string is directly proportional to the root of the applied tension, and inversely proportional to the root of the linear density, that is,

v = \sqrt{\frac{T}{\mu}}

Here,

v = Velocity

\mu= Linear density (Mass per  unit length)

T = Tension

Rearranging to find the Period we have that

T = v^2 \mu

T = v^2 (\frac{m}{L})

As we know that speed is equivalent to displacement in a unit of time, we will have to

T = (\frac{L}{t}) ^2(\frac{m}{L})

T = (\frac{7.8}{0.83})^2 (\frac{0.49}{7.8})

T = 5.54N

Therefore the tension is 5.54N

8 0
3 years ago
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