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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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A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18
Annette [7]

Answer:

Part a)

Induced EMF when length vector is along Z direction is 0.72 V

Part b)

Induced EMF when length vector is along Y direction is ZERO

Explanation:

As we know that the motional EMF induced in the wire is given as

E = (v\times B). L

1)

As we know that

v = 18\hat i + 24\hat j + 72\hat k

B = 0.080 \hat j

L = 0.50 \hat k

now we have

\vec v \times \vec B = 1.44\hat k - 5.76 \hat i

so we have

E = 1.44 (0.50) = 0.72 V

2)

If the length vector is along Y direction then we have

L = 0.50 \hat j

so again we have

\vec v \times \vec B = 1.44\hat k - 5.76 \hat i

so we have

EMF = 0

6 0
2 years ago
The human eye can respond to as little as 10^−18 J of light energy. For a wavelength at the peak of visual sensitivity, 550 nm,
Bumek [7]
<span>The key equation is going to come from Mr Planck: E=h \nu

Where h is Plancks constant; and ν is the frequency. This equation gives you the energy per photon at a given frequency. Alas, you're given wavelength, but that's easy enough to convert to frequency given the following equation:

c= lambda / nu

where c is the speed of light; λ (lambda) is the wavelength; and ν is again frequency. As soon as you know the energy of a photon with a wavelength of 550nm, you should know how many photons you would require to accumulate 10^-18J. Be careful with your units.</span>
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2 years ago
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The maximum tension that a 0.80 m string can tolerate is 15 N. A 0.35-kg ball attached to this string is being whirled in a vert
zimovet [89]

Answer:

v=5.86 m/s

Explanation:

Given that,

Length of the string, l = 0.8 m

Maximum tension tolerated by the string, F = 15 N

Mass of the ball, m = 0.35 kg

We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

F=\dfrac{mv^2}{r}

v is the maximum speed

v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{15\times 0.8}{0.35}} \\\\v=5.86\ m/s

Hence, the maximum speed of the ball is 5.86 m/s.

3 0
2 years ago
Physics in motion unit 6a the nature of waves
mylen [45]
What’s the question?
7 0
2 years ago
A steel tank of weight 600 lb is to be accelerated straight upward at a rate of 1.5 ft/sec2. Knowing the magnitude of the force
VikaD [51]

Answer:

a) the values of the angle α is 45.5°

b) the required magnitude of the vertical force, F is 41 lb

Explanation:

Applying the free equilibrium equation along x-direction

from the diagram

we say

∑Fₓ = 0

Pcosα - 425cos30° = 0

525cosα - 368.06 = 0

cosα = 368.06/525

cosα = 0.701

α = cos⁻¹ (0.701)

α = 45.5°

Also Applying the force equation of motion along y-direction

∑Fₓ = ma

Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)

525sin45.5° + F + 212.5 - 600 = 27.95

374.46 + F + 212.5 - 600 = 27.95

F - 13.04 = 27.95

F = 27.95 + 13.04

F = 40.99 ≈ 41 lb

8 0
2 years ago
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