As long the sun will be in the red giant? Question about time period, not when

A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.

Elanso [62]

The tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

<h3>Tension in the cable</h3>

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ = ¹/₂WL

T sinθ = ¹/₂W

T = (W)/(2 sinθ)

T = (29 x 9.8)/(2 x sin57)

T = 169.43 N

<h3>Vertical component of the force</h3>

T + F = W

F = W - T

F = (9.8 x 29) - 169.43

F = 114.77 N

Thus, the tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

Learn more about tension here: brainly.com/question/24994188

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6 0

2 years ago

Three resistors, 6.0-W, 9.0-W, 15-W, are connected in parallel in a circuit. What is the equivalent resistance of this combinati

ycow [4]

Answer:

2.9Ω

Explanation:

Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;

1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn

Where;

Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.

Note that;

R1= 6.0Ω

R2 = 9.0Ω

R3= 15.0 Ω

Therefore;

1/Req = 1/6 + 1/9 + 1/15

1/Req= 0.167 + 0.11 + 0.067

1/Req= 0.344

Req= (0.344)^-1

Req= 2.9Ω

3 0

3 years ago

A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp

Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

$y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}$

1. Finding Ф from the initial conditions:

$-0.02 = 0.02cos(0 + \phi) => \phi = \pi$

2. Finding time t at position y = 1 cm:

$0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}$

3. Find velocity v at time t from equation 2:

$v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))$

5 0

3 years ago

Read 2 more answers

Five loops are formed of copper wire of the same gauge (cross-sectional area). Loops 1-4 are identical; Loop 5 has the same heig

nata0808 [166]

The answer to this question is A

3 0

4 years ago

A body on the surface of a planet with the same radius as Earths weighs 10 times more than it does on Earth. What is the mass of

nadya68 [22]

Answer:

Explanation:

Given

radius of Planet is equal to radius of Earth

$r_p=r_e$

Weight of body on Planet $F_p=mg_p$

where m=mass of body

$g_p=acceleration\ due\ to\ gravity\ on\ surface\ of\ Planet$

Weight of body on earth $F_e=mg_e$

$g_e=acceleration\ due\ to\ gravity\ on\ Earth$

acceleration due to gravity is given by

$g=\frac{GM}{r^2}$

where G=gravitational constant

M=mass of Planet

r=radius of planet

for earth $g_e=\frac{GM_e}{r_e^2}$

for planet $g_p=\frac{GM_p}{r_p^2}$

substituting these values in $F_e$ and $F_p$

$F_p=m\times \frac{GM_p}{r_p^2}---1$

$F_e=m\times \frac{GM_e}{r_e^2}---2$

divide 1 and 2

$\frac{F_p}{F_e}=\frac{m\times \frac{GM_p}{r_p^2}}{m\times \frac{GM_e}{r_e^2}}$

$10=\frac{M_p}{M_e}$

$M_p=10M_e$

6 0

3 years ago