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egoroff_w [7]
3 years ago
6

A 28 kg mass suspends from a light rope 18 m long & is held to one side by the horizontal force, F, as shown below.

Physics
1 answer:
frutty [35]3 years ago
8 0

Answer: 215.15 N

Explanation:

If we draw a free body diagram of the mass we will have the following:

\sum{F_{x}}=-Tcos\theta + F=0 (1)

\sum{F_{y}}=Tsin\theta - mg=0 (2)

Where T is the tension force of the rope, m=28 kg the mass, g=9.8 m/s^{2} the acceleration due gravity and mg is the weight.

On the other hand, we can calculate \theta as follows:

cos\theta=\frac{s}{l}

\theta=cos^{-1}(\frac{s}{l})

Where s=11.1 m and l=18 m

\theta=cos^{-1}(\frac{11.1 m}{18 m})

\theta=51.9\° (3)

Now, we firstly need to find T from (2):

T=\frac{mg}{sin\theta} (4)

T=\frac{(28 kg)(9.8 m/s^{2})}{sin(51.9\°)}

T=348.69 N (5)

Substituting (5) in (1):

F=Tcos\theta (6)

F=348.69 N cos(51.9\°)

Finally:

F=215.15 N

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A positive charge of 18nC is evenly distributed along a straight rod of length 4.0 m that is bent into a circular arc with a rad
sergey [27]

Explanation:

Formula for angle subtended at the center of the circular arc is as follows.

           \theta = \frac{S}{r}

where,   S = length of the rod

              r = radius

Putting the given values into the above formula as follows.      

                \theta = \frac{S}{r}

                             = \frac{4}{2}

                             = 2 radians (\frac{180^{o}}{\pi})

                             = 114.64^{o}

Now, we will calculate the charge density as follows.

                 \lambda = \frac{Q}{L}

                            = \frac{18 \times 10^{-9} C}{4 m}

                            = 4.5 \times 10^{-9} C/m

Now, at the center of arc we will calculate the electric field as follows.

                 E = \frac{2k \lambda Sin (\frac{\theta}{2})}{r}

                     = \frac{2(9 \times 10^{9} Nm^{2}/C^{2})(4.5 \times 10^{-9}) Sin (\frac{114.64^{o}}{2})}{2 m}

                      = 34.08 N/C

Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.

5 0
3 years ago
If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her
Marysya12 [62]

Answer:

 μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

          fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ  = 0

          fr sin θ  - cos θ  (W / 2 + 0,3 W_painter) = 0

          fr = cotan θ  (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

     

X axis

        F1 - fr = 0

        F1 = fr

the friction force has the expression

       fr = μ N

Y Axis

       N - W - W_painter = 0

       N = W + W_painter

       

we substitute

      fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

     μ (W + W_painter) = cotan θ  (W / 2 + 0,3 W_painter)

      μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values ​​they give

      μ = cotan θ  (12/2 + 0.3 55) / (12 + 55)

      μ = cotan θ  (22.5 / 67)

      μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

       cotan 45 = 1 / tan 45 = 1

the result is

    μ = 0.336

5 0
3 years ago
How do noise and vibration affect you when operating a boat?
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If a roller coaster cart, with a mass of 100 kg, traveled this coaster, how much kinetic energy would it have at point 'E'?
zzz [600]

Answer:

Explanation:

Assuming no friction between the roller coaster car and the hill, and neglecting air resistance, the kinetic energy the roller coaster car would have at the bottom of the hill would be equal to its gravitational potential energy at the top of the hill, by conservation of energy.

8 0
3 years ago
A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r
Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

(c) at = 4 m/s²  ,  v = 42 m/s

Explanation:

The uniformly accelerated circular movement,  is a circular path movement in which the angular acceleration is constant.

There is tangential acceleration (at ) and is constant.

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t  Formula (1)

θ=  ω₀*t + (1/2)*α*t²  Formula (2)

at = α*R  Formula (3)

v= ω*R  Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s  : wheel’s initial angular velocity

R = 1.0 m  : wheel’s radium

(a)  Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ=  ω₀*t + (1/2)*α*t²

θ=  (2)*(10) + (1/2)*(4)*(10)²

θ=  220 rad  

θ=  220 rad  

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

6 0
3 years ago
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