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Annette [7]
3 years ago
13

Any girls come to talk on insta or here​

Physics
1 answer:
xxMikexx [17]3 years ago
7 0

Answer:

No

Explanation:

Sorry hun, I'm a lesbian lol

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Which technology is most efficient in mapping large areas of the ocean floor?
qaws [65]
''Sonar'' is the answer.
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What is a suspension bridge- in your own words please and ty
PIT_PIT [208]
A bridge supported by vertical cables which then leads to more support from larger cables.
4 0
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A truck travels up a hill with a 5.7° incline. The truck has a constant speed of 22 m/s. What is the horizontal component of the
tester [92]

Answer:

The horizontal component of the velocity is 21.9 m/s.

Explanation:

Please see the attached figure for a better understanding of the problem.

Notice that the vector v and its x and y-components (vx and vy) form a right triangle. Then, we can use trigonometry to find the magnitude of vx, the horizontal component of the velocity.

To find vx, let´s use the following trigonometric rule of right triangles:

cos α = adjacent / hypotenuse

cos 5.7° = vx / 22 m/s

22 m/s · cos 5.7° = vx

vx = 21.9 m/s

The horizontal component of the velocity is 21.9 m/s.

8 0
3 years ago
Two sealed 1 l containers full of gas are at room temperature. Container a has a pressure of 4 atm and container b has a pressur
loris [4]

Answer:

The number density of the gas in container A is twice the number density of the gas in container B.

Explanation:

Here we have

P·V =n·R·T

n = P·V/(RT)

Therefore since V₁ = V₂ and T₁ = T₂

n₁ = P₁V₁/(RT₁)

n₂ = P₂V₂/(RT₂)

P₁ = 4 atm

P₂ = 2 atm

n₁ = 4V₁/(RT₁)

n₂ =2·V₁/(RT₁)

∴ n₁ = 2 × n₂

Therefore, the number of moles in container A is two times that in container B and the number density of the gas in container A is two times the number density in container B.

This can be shown based on the fact that the pressure  of the container is due to the collision of the gas molecules on the walls of the container, with a kinetic energy that is dependent on temperature and mass, and since the temperature is constant, then the mass of container B is twice that of A and therefore, the number density of container A is twice that of B.

5 0
3 years ago
A 4.00-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is fired into a wooden block with mass 0.650 kg ,
Maru [420]

Answer:

a) Coefficient of kinetic friction between block and surface = 0.12

b) Decrease in kinetic energy of the bullet = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = 0.541 J

Explanation:

Given,

Mass of bullet = 4.00 g = 0.004 kg

Initial velocity of the bullet = 400 m/s

Mass of wooden block = 0.65 kg

Initial velocity of the wooden block = 0 m/s (since it was initially at rest)

Final velocity of the bullet = 190 m/s

Distance slid through by the block after the collision = d = 72.0 cm = 0.72 m

Let the velocity of the wooden block after collision be v

According to the law of conservation of momentum,

Momentum before collision = Momentum after collision

Momentum before collision = (Momentum of bullet before collision) + (Momentum of wooden block before collision)

Momentum of bullet before collision = (0.004×400) = 1.6 kgm/s

Momentum of wooden block before collision = (0.65)(0) = 0 kgm/s

Momentum after collision = (Momentum of bullet after collision) + (Momentum of wooden block after collision)

Momentum of bullet after collision = (0.004×190) = 0.76 kgm/s

Momentum of wooden block after collision = (0.65)(v) = (0.65v) kgm/s

Momentum balance gives

1.6 + 0 = 0.76 + 0.65v

0.65v = 1.6 - 0.76 = 0.84

v = (0.84/0.65)

v = 1.29 m/s

The velocity of the wooden block after collision = 1.29 m/s

To obtain the coefficient of kinetic friction between block and surface, we will apply the work-energy theorem.

The work-energy theorem states that the work done in moving the block from one point to another is equal to the change in kinetic energy of the block between these two points.

The points to consider are the point when the block starts moving (immediately after collision) and when it stops as a result of frictional force.

Mathematically,

W = ΔK.E

W = workdone by the frictional force in stopping the wooden block (since there is no other horizontal force acting on the block)

W = -F.d (minus sign because the frictional force opposes motion)

d = Distance slid through by the block after the collision = 0.72 m

F = Frictional force = μN

where N = normal reaction of the surface on the wooden block and it is equal to the weight of the block.

N = W = mg

F = μmg

W = - μmg × d = (-μ)(0.65)(9.8) × 0.72 = (-4.59μ) J

ΔK.E = (final kinetic energy of the block) - (initial kinetic energy of the block)

Final kinetic energy of the block = 0 J (since the block comes to a rest)

(Initial kinetic energy of the block) = (1/2)(0.65)(1.29²) = 0.541 J

ΔK.E = 0 - 0.541 = - 0.541 J

W = ΔK.E

-4.59μ = -0.541

μ = (0.541/4.59)

μ = 0.12

b) The decrease in kinetic energy of the bullet

(Decrease in kinetic energy of the bullet) = (Kinetic energy of the bullet before collision) - (Kinetic energy of the bullet after collision)

Kinetic energy of the bullet before collision = (1/2)(0.004)(400²) = 320 J

Kinetic energy of the bullet after collision = (1/2)(0.004)(190²) = 72.2 J

Decrease in kinetic energy of the bullet = 320 - 72.2 = 247.8 J

c) Kinetic energy of the block at the instant after the bullet passes through it = (1/2)(0.65)(1.29²) = 0.541 J

Hope this Helps!!!

4 0
3 years ago
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