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3 years ago

Can anyone explain how tides work

2 answers:

7 0

<span> the moon's gravitational force pulls on water in the oceans so that there are "bulges" in the ocean on both sides of the planet. The moon pulls water toward it, and this causes the bulge toward the moon. hope this helps (:</span>

Mila [183]3 years ago

6 0

The position of the sun and the moon affect how high the tide is

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An apple falls out of a tree from a height of 2.3m. what is the impact speed of the apple

Anuta_ua [19.1K]

Answer:6.71 m/s

Explanation:

Given

Apple fall from a height of $h=2.3 m$

We need to find the impact speed of apple which can be given by using

$v^2-u^2=2gh$

where v=final velocity

u=initial velocity

h=Displacement

Assuming initial velocity to be zero

substituting the value we get

$v^2-0=2\times 9.8\times 2.3$

$v=\sqrt{2\times 9.8\times 2.3}$

$v=6.71\ m/s$

5 0

3 years ago

In a game of pool, a cue ball rolls without slipping toward the stationary eight ball with a momentum of 0.23 kg. After the two

IrinaVladis [17]

This question involves the concepts of the law of conservation of momentum.

The magnitude of the final momentum of the eight ball is "0.22 N.s".

According to the law of conservation of momentum:

$P_{i1}+P_{i2}=P_{f1}+P_{f2}$

where,

$P_{i1}$ = initial momentum of the cue ball = 0.23 N.s

$P_{i2}$ = initial momentum of the eight ball = 0 N.s (since ball is initially at rest)

$P_{f1}$ = final momentum of the cue ball = 0.01 N.s

$P_{f2}$ = final momentum of the eight ball = ?

Therefore,

$0.23\ N.s + 0\N.s = 0.01\ N.s+P_{f2}\\\\P_{f2} = 0.22\ N.s$

Learn more about the law of conservation of momentum here:

brainly.com/question/1113396?referrer=searchResults

3 0

2 years ago

Which option lists a form of kinetic energy followed by a form of potential

Artemon [7]

Answer:

D. Sound Energy, Magnetic energy

Explanation:

Sound energy is in motion, and Magnetic energy is about to be in motion.

6 0

2 years ago

Explain why, under some circumstances, it is not advisable to weld a structure that is fabricated with a 3003 aluminum alloy. Hi

Scorpion4ik [409]

The 3003 aluminum alloy is made up of 1.25% Magnesium and 0.1% Copper. This combination is designed to increase the strength of the material over other types of alloys such as those of the 1000 series. This alloy provides a medium strength and can be educated by cold work.

The alloy is not heat treatable and generally has good formability, corrosion resistance and weldability.

However, being a material that hardens by cold work, welding a 3003 Aluminum structure will cause the body to undergo recrystallization which will generate a loss in the 'resistance' of the material and the force capable of withstanding. If this aluminum will be used for structural purposes, it should not be welded. It would be better to perform the structure with a 6061 aluminum, which has similar characteristics and is not so affected by welding.

7 0

2 years ago

A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym

user100 [1]

Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

- The mass of the solid cylinder M = 45 kg

- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

Find:

- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

L_(p,i) = m*V_pi*R

L_(p,i) = 3.6*3.3*0.44

L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

L_(c,i) = I*w_i

L_(c,i) = 0.5*M*R^2*0

L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

L_i = L_(p,i) + L_(c,i)

L_i = 5.2272 + 0

L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

L_(p,f) = I_p*w_f

- Similarly, for the cylinder:

L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

L_f = L_(p,f) + L_(c,f)

L_f = I_p*w_f + I_c*w_f

L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

L_i = L_f

5.2272 = w_f*(I_p + I_c)

w_f = 5.2272/ R^2*(m + 0.5M)

Plug in values:

w_f = 5.2272/ 0.44^2*(3.6 + 0.5*45)

w_f = 5.2272/ 5.05296

w_f = 1.0345 rad/s

5 0

2 years ago