In the early twentieth century, _______ redefined psychology as "the science of observable behavior."

The following symbols show the wind speed and direction for West Palm Beach, FL, from 8:00 a.m. to 12:00 p.m. Which choice corre

Vanyuwa [196]

The wind speed increased but in the same direction.
It is A/1 because the wind speed did increase and it is still going the same way clockwise because it also did not move that much.

3 0

3 years ago

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How is the behavior of waves affected by a medium?

Vlad1618 [11]

They are slowed down.

even air is a medium and will slow down waves (slightly) compared to a vacuum/ space

if the wave hits the medium at an angle it can be 'bent' (refracted) like light through a glass prism

6 0

3 years ago

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The force P is applied to the 45-kg block when it is at rest. Determine the magnitude and direction of the friction force exerte

cluponka [151]

Answer:

Check attachment for free body diagram of the question.

I used the free body diagram and the angles given in the diagram missing in the question above, but I used the data given in the above question.

Explanation:

Let frictional force be Fr acting down the plane

Let analyze the structure before inserting values

Using Newton's second law along the y-axis

ΣFy = Fnet = m•ay

Since the body is not moving in the y-direction, then ay = 0

N+PSinβ — WCosθ = 0

N+PSin20—441.45Cos15 = 0

N+PSin20—426.41 = 0

N = 426.41 — PSin20 , equation 1

The maximum Frictional force to be overcome is given as

Fr(max) = μsN

Fr(max) = 0.25(426.41 — PSin20)

Fr(max)= 106.6 —0.25•PSin20

Fr(max) = 106.6 — 0.08551P, equation 2

This is the maximum force that must be overcome before the body starts to move

Using Newton's law of motion in the x direction

Note, we took the upward direction up the plane as the direction of motion since the force want to move the block upward

Fnetx = ΣFx

Fnetx = P•Cosβ —W•Sinθ — Fr

Fnetx = P•Cos20—441.45•Sin15—Fr

Fnetx = 0.9397P — 114.256 — Fr

Equation 3

When Fnetx is positive, then, the body is moving up the plane, if Fnetx is negative, then, the maximum frictional force has not yet being overcome and the object is still i.e. not moving

a. When P = 0

From equation 2

Fr(max) = 106.6 — 0.08551P

Fr(max) = 106.6 — 0.08551(0)

Fr(max)= 106.6 N

So, 106.6N is the maximum force to be overcome

So, here the only force acting on the body is the weight and it acting down the plane, trying to move the body downward.

Wx = WSinθ

Wx = 441.45× Sin15

Wx = 114.256 N.

Since the force trying to move the body downward is greater than the maximum static frictional force, then the body is not in equilibrium, it is moving downward.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — PSin20 , equation 1

N = 426.41 N, since P=0

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 426.41

Fr = 93.81 N.

b. Now, when P = 190N

From equation 2

Fr(max) = 106.6 — 0.08551(190)

Fr(max) = 106.6 —16.2469

Fr(max)= 90.353 N

So, 90.353 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 190Cos20 — 441.45Sin15

Fnetx = 64.29N

So the force moving the body up the incline plane is 64.29N

Fnetx < Fr(max)

Then, the frictional force has not being overcome yet.

Then, the body is in equilibrium.

Then, applying equation 3.

Fnetx = 0.9397P — 114.256 — Fr

Fnetx = 0, since the body is not moving

0 = 0.9397(190) —114.246 — Fr

Fr = 64.297 N

Fr ≈ 64.3N

c. When, P = 268N

From equation 2

Fr(max) = 106.6 — 0.08551(268)

Fr(max) = 106.6 —16.2469

Fr(max)= 83.68 N

So, 83.68 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 268Cos20 — 441.45Sin15

Fnetx = 137.58 N

So the force moving the body up the incline plane is 137.58 N

Fnetx > Fr(max)

Then, the frictional force has being overcome.

Then, the body is not equilibrium.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — 268Sin20 , equation 1

N = 334.75 N, since P=268N

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 334.75

Fr = 73.64 N

d. The required force to initiate motion is the force when the block want to overcome maximum frictional force.

So, Fnetx = Fr(max)

Px — Wx = Fr(max)

From equation 1

Fr(max) = 106.6 — 0.08551P,

P•Cosβ-W•Sinθ = 106.6 — 0.08551P

P•Cos20 — 441.45•Sin15 = 106.6 — 0.08551P

P•Cos20—114.256=106.6 - 0.08551P

PCos20+0.08551P =106.6 + 114.256

1.025P=220.856

P = 220.856/1.025

P = 215.43 N

3 0

3 years ago

Calculate the binding energy e of the boron nucleus 11 5b (1ev=1.602×10−19j). express your answer in millions of electron volts

nignag [31]

<span>Depends on the precision you're working to. proton mass ~ 1.00728 amu neutron mass ~ 1.00866 amu electron mass ~ electron mass = 0.000549 amu Binding mass is: mass of constituents - mass of atom Eg for nitrogen: (7*1.00728)-(7*1.00866)-(7*0.000549) -14.003074 = 0.11235amu Binding energy is: E=mc^2 where c is the speed of light. Nuclear physics is usually done in MeV[1] where 1 amu is about 931.5MeV/c^2. So: 0.11235 * 931.5 = 104.6MeV Binding energy per nucleon is total energy divided by number of nucleons. 104.6/14 = 7.47MeV This is probably about right; it sounds like the right size! Do the same thing for D/E/F and recheck using your numbers & you shouldn't go far wrong :) 1 - have you done this? MeV is Mega electron Volts, where one electronVolt (or eV) is the change in potential energy by moving one electron up a 1 volt potential. ie energy = charge * potential, so 1eV is about 1.6x10^-19J (the same number as the charge of an electron but in Joules). It's a measure of energy, but by E=mc^2 you can swap between energy and mass using the c^2 factor. Most nuclear physicists report mass in units of MeV/c^2 - so you know that its rest mass energy is that number in MeV.</span>

7 0

3 years ago

Two people push on a boulder to try to move it. The mass of the boulder is 825 kg. One person pushes north with a force of 64 N.

stira [4]

Answer:

0.09 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s².

From the question, expression for acceleration is given as

F' = ma

Using Pythagoras Theory,

√(F₁²+F₂²) = ma................... Equation 1

Where F₁ = Force of the First person on the boulder, F₂ = Force of the Second person on the boulder, F' = resultant force acting on the boulder, m = mass of the boulder, a = acceleration of the boulder.

make a the subject of the equation

a = √(F₁²+F₂²) /m................ Equation 2

Given: m = 825 kg, F₁ = 64 N, F₂ = 38 N,

Substitute into equation 2

a = [√(64²+38²)]/825

a = {√(5540)}/825

a = 74.43/825

a = 0.09 m/s²

4 0

3 years ago