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Natali [406]
3 years ago
5

Rope BCA passes through a pulley at point C and supports a crate at point A. Rope segment CD supports the pulley and is attached

to an eye anchor embedded in a wall. Rope segment BC creates an angle of ϕ = 51.0 ∘ with the floor and rope segment CD creates an angle θ with the horizontal. If both ropes BCA and CD can support a maximum tensile force Tmax = 120 lb , what is the maximum weight Wmax of the crate that the system can support? What is the

Physics
1 answer:
RUDIKE [14]3 years ago
6 0

Answer:

Wmax = 63.65 ≈ 64 lb

Explanation:

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Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller
mrs_skeptik [129]

Answer:

surface charge density on each sphere is 440 \times 10^{-9} C

Explanation:

given data

radius of smaller sphere = 5 cm

radius of  larger sphere is 12 cm

electric field at surface of larger sphere = 660 kV/m = 660 × 1000 v/m

solution

we apply here electric field formula that is express as

E = (\frac{1}{4\pi\epsilon  })\times  (\frac{Q_{1} }{R^{2} } )    .................1

put here value

660000 = 9 \times 10^9 \times \frac{Q1}{0.12^2}  

Q1 = 1056 × 10^{-9}

and

here field inside a conductor is zero so that electric potential ( V ) is constant

\frac{Q{1} }{R} = \frac{Q{2} }{r}   ..................2

so Q2 will be

Q2 =  \frac{5}{12} \times 1056 \times 10^{-9}  

Q2 =  440 \times 10^{-9}  C

6 0
3 years ago
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Vlad1618 [11]
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7 0
3 years ago
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A 5.93 kg ball is attached to the top of a vertical pole with a 2.35 m length of massless string. The ball is struck, causing it
Delvig [45]

Answer

given,

mass of ball = 5.93 kg

length of the string = 2.35 m

revolve with velocity of 4.75 m/s

acceleration due to gravity = 9.81 m/s²

T cos θ = mg

T cos θ = 5.93\times 9.81

T cos θ = 58.17

T sin \theta =\dfrac{mv^2}{r}

T sin \theta =\dfrac{5.93\times 4.75^2}{2.35 sin \theta}

T sin^2 \theta =56.93

sin^2 \theta = 1 - cos^2 \theta

T (1 - cos^2 \theta) =56.93

T (1 - (\dfrac{58.17}{T})^2) =56.93

T² - 56.93T - 3383.75 = 0

T =  93.22 N

cos \theta = \dfrac{58.17}{93.22}

θ = 51.39°

6 0
3 years ago
The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr
MA_775_DIABLO [31]

Answer:

q=49Q/64

and

x =16R/15

Explanation:

See  attached figure.

E_{Q}= E due to sphere

E_{q}= E due to particule

E_{total}=E_{Q}-E_{q}=0  (1)

according to the law of gauss and superposition Law:

E_{Q}=E_{1}+E_{2}=E_{2} ; electric field due to the small sphere with r1=R/4

E_{Q}=kq_{2}/(r_{1}^{2})=

q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}

then: E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})  (2)

on the other hand, for the particule:

E_{q}=kq/(r_{p}^{2})

r_{p}=2R-R/4=7R/4   ⇒    E_{q}=16kq/(49R^{2})   (3)

We replace (2) y (3) in (1):

E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})

q=49Q/64

--------------------

if R<x<2R   AND E_{total}=E_{Q}-E_{q}=0

E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})

remember that  q=49Q/64

then:

Q(2R-x^{2})=49/64*x^{2}

solving:

x_{1} =16R/15

x_{2} =16R

but: R<x<2R  

so : x =16R/15

7 0
3 years ago
For a specific volume of 0.2 m3/kg, find the quality of steam if the absolute pressure is (a) 40 kPa and (b) 630 kPa. What is th
ICE Princess25 [194]

Answer:

x=0.0498

x'=0.659

Explanation:

Specific Volume V=0.2m_3/kg

Absolute Pressure (a) P_a= 40kpa

Giving

T_a=75.87

v_f=1.265*10^{-3}m^3/kg

v_g=3.993m^3/kg

                               (b) P_a= 630kpa

Giving

T_b=160.13C

v_f'=1.10282*10^{-3} m^3/kg

v_g'=0.30286 m^3/kg

(a)

Generally the equation for quality of Steam X  is mathematically given by

x=\frac{v-v_f}{v_g-v_f}

x=\frac{0.2-1.0265*10^{-3}}{3.993-1.0265*10^{-3}}

x=0.0498

(b)

Generally the equation for quality of Steam X  is mathematically given by

x'=\frac{v-v_f'}{v_g'-v_f'}

x'=\frac{0.2-1.10*10^{-3}}{3.30-1.1*10^{-3}}

x'=0.659

4 0
3 years ago
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