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klemol [59]
2 years ago
13

A small box of mass m1 is sitting on a board of mass m2 and length L (Figure 1) . The board rests on a frictionless horizontal s

urface. The coefficient of static friction between the board and the box is ?s. The coefficient of kinetic friction between the board and the box is, as usual, less than ?s.
Throughout the problem, use g for the magnitude of the acceleration due to gravity. In the hints, use Ff for the magnitude of the friction force between the board and the box.

Find Fmin, the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box (which will then fall off of the opposite end of the board).

Express your answer in terms of some or all of the variables ?s, m1, m2, g, and L. Do not include Ff in your answer.
Physics
1 answer:
chubhunter [2.5K]2 years ago
3 0

Explanation:

Whole system will accelerate under the action of applied force. The box will experience the force against the friction and when this force exceeds then the box will move. so

Ff = μs×m1×g

m1×a = μs×m1×g

a = μs×g

The applied force is given by

F = (m1 + m2)×a so

F = μs×g×(m1+m2)

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A very tall building has a height H0 on a cool spring day when the temperature is T0. You decide to use the building as a sort o
Vlada [557]

Answer:

The temperature is   T  =  \frac{h}{H_O \alpha_{steel} }  + T_O

Explanation:

From the question we are told that

      The height on a cool spring day is H_O

      The temperature on a cool spring day is  T_O

      The difference in height between a cool spring day and a summer day  is     h

     The coefficient of static friction is \alpha _{steel}

The mathematical relation for the linear expansion of the steel buiding is represented as

               h  =  H_o \alpha_{steel}  [T-T_O]

Where T is the temperature of the steel during summer

Now making T the subject we have

                T  =  \frac{h}{H_O \alpha_{steel} }  + T_O

5 0
2 years ago
a crowbar of 2 meter is used to lift an object of 800N if the effort arm is 160cm , calculste the effort applied
Vitek1552 [10]

Answer:

200 N

Explanation:

The crowbar is 2 meter, or 200 cm.  The effort arm is 160 cm, so the moment arm of the object is 40 cm.

(800 N) (40 cm) = F (160 cm)

F = 200 N

5 0
2 years ago
A proposed space elevator would consist of a cable stretching from the earth's surface to a satellite, orbiting far in space, th
NISA [10]

To solve this problem we will apply the concepts related to energy conservation. Here we will use the conservation between the potential gravitational energy and the kinetic energy to determine the velocity of this escape. The gravitational potential energy can be expressed as,

PE= \frac{GMm}{d}

The kinetic energy can be written as,

KE= \frac{1}{2} mv^2

Where,

G = 6.67*10^{-11}m^3/kg\cdot s^2Gravitational Universal Constant

m = 5.972*10^{24}kg Mass of Earth

h = 56*10^6m  Height

r = 6.378*10^6m Radius of Earth

From the conservation of energy:

\frac{1}{2} mv^2 = \frac{GMm}{d}

Rearranging to find the velocity,

v = \sqrt{\frac{2Gm}{d}} \rightarrow  Escape velocity at a certain height from the earth

If the height of the satellite from the earth is h, then the total distance would be the radius of the earth and the eight,

d = r+h

v = \sqrt{\frac{2Gm}{r+h}}

Replacing the values we have that

v = \frac{2(6.67*10^{-11})(5.972*10^{24})}{6.378*10^6+56*10^6}

v = 3.6km/s

Therefore the escape velocity is 3.6km/s

3 0
2 years ago
In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum
gulaghasi [49]

Answer:

Ratio of length will be \frac{L_2}{L_1}=1.108

Explanation:

We have given time period of the pendulum when length is L_1 is T_1=0.95sec

And when length is L_2 time period T_2=1sec

We know that time period is given by

T=2\pi \sqrt{\frac{L}{g}}

So 0.95=2\pi \sqrt{\frac{L_1}{g}}----eqn 1

And 1=2\pi \sqrt{\frac{L_2}{g}}-------eqn 2

Dividing eqn 2 by eqn 1

\frac{1}{0.95}=\sqrt{\frac{L_2}{L_1}}

Squaring both side

\frac{L_2}{L_1}=1.108

8 0
2 years ago
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jarptica [38.1K]
Calculate the weight of the table through the equation,

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where W is the weight, m is the mass, and g is the acceleration due to gravity. Substituting the known values,
   W = (0.44 kg)(9.8 m/s²) 
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The components of this weight can be calculated through the equation,

   Wx = W(sin θ) 

and Wy = W(cos θ)

x - component:
   Wx = W(sin θ)
Substituting,
  Wx = (4.312 N)(sin 150°) = <em>2.156 N</em>

  Wy = (4.312 N)(cos 150°) =<em> -3.734 N</em>
6 0
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