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koban [17]
2 years ago
9

Wood block 1 in (Figure 1), which has a mass of 1.0 kg, is at rest on a wood ramp. The angle of the ramp is 20º above horizontal

. What is the smallest mass of block 2 that will start block 1 sliding uphill? Do not neglect friction.
Please help I've been stuck on this for days...

Physics
1 answer:
Sergeeva-Olga [200]2 years ago
7 0

Answer:

Explanation:

For this problem we use the translational equilibrium condition. Our reference frame for block 1 is one axis parallel to the plane and the other perpendicular to the plane.

X axis

      -Aₓ - f_e +T = 0        (1)

Y axis

      N₁ - W_y = 0              ( 2)

let's use trigonometry for the weight components

      sin θ = Wₓ / W

      cos θ = W_y / W

      Wₓ = W sin θ

      W_y = W cos θ

We write the diagram for the second body.

Note that in the block the positive direction rd upwards, therefore for block 2 the positive direction must be downwards

      W₂ -T = 0                             (3)

we add the equations is 1 and 3

       - W₁ sin θ - μ N₁ + W₂ = 0

from equation 2

       N₁ = W₁ cos θ

       

we substitute

        -W₁ sin θ - μ (W₁ cos θ) + W₂ = 0

W₂ = m₁ g (without ea - very expensive)

This is the smallest value that supports the equilibrium system

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Velocity = frequency * wavelength

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Vinil7 [7]

Answer:

Power output: W=1426.9MW

Explanation:

The power output of the falls is given mainly by its change in potential energy:

Q=-P_{tot}=-(P_{2}-P_{1})

The potential energy for any point can be calculated as:

P=m*g*h

If we consider the base of the falls to be the reference height, at point 2 h=0, so P2=0, and height at point 1 equals 52m:

Q=P_{1}=m*g*h

If we replace m with the mass rate M we obtain the rate of change in potential energy over time, so the power generated:

W=M*g*h=2.8*10^{3}m^{3}/s*1*10^{3}kg/m^{3}*9.8m/s^{2}*52m =1426.9MW

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Which trailer has more downward pressure where it attaches to the car?
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The one that is loaded worst. The overall weight is not important; tongue weight is a matter of loading. Our 12,000 lb snow cat trailer, which has stops to position the cat properly, has under 100 lbs tongue weight. Excessive tongue weight is a Bad Thing because it reduces weight on the towing vehicle's front wheels, leading to instability.

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A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the
yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE          

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0     (initial KE = 0 because the ball starts from rest)  (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

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Consider an optical cavity of length 40 cm. Assume the refractive index is 1, and use the formula for Icavity vs wavelength to p
Bad White [126]

Answer:

Diode Lasers  

Consider a InGaAsP-InP laser diode which has an optical cavity of length 250  

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4. The optical gain bandwidth (as measured between half intensity points) will  

normally depend on the pumping current (diode current) but for this problem  

assume that it is 2 nm.  

(a) What is the mode integer m of the peak radiation?  

(b) What is the separation between the modes of the cavity? Please express your  

answer as Δλ.  

(c) How many modes are within the gain band of the laser?  

(d) What is the reflection coefficient and reflectance at the ends of the optical  

cavity (faces of the InGaAsP crystal)?  

(e) The beam divergence full angles are 20° in y-direction and 5° in x-direction  

respectively. Estimate the x and y dimensions of the laser cavity. (Assume the  

beam is a Gaussian beam with the waist located at the output. And the beam  

waist size is approximately the x-y dimensions of the cavity.)  

Solution:  

(a) The wavelength λ of a cavity mode and length L are related by  

n

mL

2

λ = , where m is the mode number, and n is the refractive index.  

So the mode integer of the peak radiation is  

1290

1055.1

10250422

6

6

= ×

××× == −

−

λ

nL

m .  

(b) The mode spacing is given by nL

c f 2

=Δ . As

λ

c f = , λ

λ

Δ−=Δ 2

c f .  

Therefore, we have nm

nL f

c

20.1

)10250(42

)1055.1(

2 || 6

2 2 26

= ×××

× ==Δ=Δ −

− λλ λ .  

(c) Since the optical gain bandwidth is 2nm and the mode spacing is 1.2nm, the  

bandwidth could fit in two possible modes.  

For mode integer of 1290, nm

m

nL 39.1550

1290

10250422 6

= ××× ==

−

λ

Take m = 1291, nm

m

nL 18.1549

1291

10250422 6

= ××× ==

−

λ

Or take m = 1289, nm

m

nL 59.1551

1289

10250422 6

= ××× ==

−

λ .

Explanation:

8 0
2 years ago
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