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Svetllana [295]
3 years ago
11

A modified Atwood machine is at rest. The hanging block has a mass of 3kg. The black on wheels has unknown mass M1. When release

d the block m2 falls at 5.88m/s^2. If frictional forces are considered so small they are negligible, the block M1 must have a mass of _______ kg.

Physics
1 answer:
fgiga [73]3 years ago
7 0

Answer:

From the figure,

The free-body diagrams for  and  are shown in the figures below. The only forces on the blocks are the upward tension  and the downward gravitational forces  and  . Applying Newton’s second law, we obtain:   

which can be solved to yield 

Substituting the result back, we have 

(a) With  and , the acceleration becomes   

(b) Similarly, the tension in the cord is   

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How do objects and substances have kinetic energy??<br><br> I need help ASAP!!
9966 [12]

Objects and substances in motion have Kinetic Energy.

Explanation:

Kinetic Energy is the motion of objects, substances, molecules, atoms .

Whenever an object starts to move, it is in motion. Kinetic Energy is generated, when the object moves and its potential energy gets dropped with the movement. Kinetic Energy is found in every object that moves.

The Kinetic Energy of an object depends both on its mass and its speed. Kinetic Energy increases as mass and speed are increased.

Example:  By using the wind flow from wind turbines, the Kinetic energy in the wind can be transformed to generate electricity by using generators.

7 0
3 years ago
Apply the general results obtained in the full analysis of motion under the influence of a constant force in Section 2.5 to answ
zvonat [6]

Answer:

y(i) = h

v(y.i) = 0

Explanation:

See attachment for elaboration

3 0
3 years ago
The apparent height of a building 10.5 km away is 0.02 radians. What is the approximate height of the building to the nearest me
Ksenya-84 [330]

Answer:

Approximate height of the building is 23213 meters.

Explanation:

Let the height of the building be represented by h.

0.02 radians = 0.02 × \frac{180^{o} }{\pi }

                     = 0.02 x (180/\frac{22}{7})

0.02 radians  = 1.146°

10.5 km = 10500 m

Applying the trigonometric function, we have;

Tan θ = \frac{opposite}{adjacent}

So that,

Tan 1.146° = \frac{h}{10500}

⇒ h = Tan 1.146° x 10500

      = 2.21074 x 10500

      = 23212.77

h = 23213 m

The approximate height of the building is 23213 m.

8 0
2 years ago
The second minimum in the diffraction pattern of a 0.11-mm-wide slit occurs at 0.72°. What is the wavelength of the light?
Anit [1.1K]

Answer:

691.13 nm

Explanation:

d = width of the slit = 0.11 x 10⁻³ m

θ = angle of diffraction pattern = 0.72° degree

λ = wavelength of the light = ?

m = order = 2                              (since second minimum)

for the second minimum diffraction pattern we use the equation

d Sinθ = m λ

Inserting the values

(0.11 x 10⁻³) Sin0.72 = (2) λ

λ = 691.13 x 10⁻⁹ m

λ = 691.13 nm

6 0
3 years ago
A volumetric flask made of Pyrex is calibrated at 20.0°C. It is filled to the 285-mL mark with 40.5°C glycerin. After the flask
Charra [1.4K]

Answer:

V_f = 287.04 mL

Explanation:

We are given the initial/original volume of the glycerine as 285 mL.

Now, after it is finally cooled back to 20.0 °C , its volume is given by the formula;

V_f = V_i (1 + βΔT)

Where;

V_f is the final volume

V_i is the original volume = 285 mL

β is the coefficient of expansion of glycerine and from online tables, it has a value of 5.97 × 10^(-4) °C^(−1)

Δt is change in temperature = final temperature - initial temperature = 32 - 20 = 12 °C

Thus, plugging in relevant values;

V_f = 285(1 + (5.97 × 10^(-4) × 12))

V_f = 287.04 mL

7 0
3 years ago
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