All categories

3 years ago

A 40-kg uniform semicircular sign 1.6 m in diameter is supported by two wires as shown. What

Physics

1 answer:

garri49 [273]3 years ago

4 0

Answer:

T1 = 131.4 [N]

T2 = 261 [N]

Explanation:

To solve this problem we must make a sketch of how will be the semicircle, for this reason we conducted an internet search, to find the scheme of the problem. This scheme is attached in the first image.

Then we make a free body diagram, with this free body diagram, we raise the forces that act on the body. Since it is a problem involving static equilibrium, the sum of forces in any direction and moments must be equal to zero.

By performing a sum of forces on the Y axis equal to zero we can find an equation that relates the forces of tension T1 & T2.

The second equation can be determined by summing moments equal to zero, around the point of application of the T1 force. In this way we find the T2 force.

The value of T2, is replaced in the first equation and we can find the value for T1.

Therefore

T1 = 131.4 [N]

T2 = 261 [N]

The free body diagram and the developed equations can be seen in the second attached image.

You might be interested in

Whiteout friction you could not, write, drive or fly and airplane. Why not?

aleksklad [387]

friction is the resistance that one surface or object encounters when moving over another. Due to gravity pulling everything down things need to friction in order to move

i hope this helps :/

7 0

3 years ago

Two particles are moving along the x axis. Particle 1 has a mass m₁ and a velocity v₁ = +4.7 m/s. Particle 2 has a mass m₂ and a

nirvana33 [79]

Answer:

m₁ / m₂ = 1.3

Explanation:

We can work this problem with the moment, the system is formed by the two particles

The moment is conserved, to simulate the system the particles initially move with a moment and suppose a shock where the particular that, without speed, this determines that if you center, you should be stationary, which creates a moment equal to zero

p₀o = m₁ v₁ + m₂ v₂

pf = 0

m₁ v₁ + m₂ v₂ = 0

m₁ / m₂ = -v₂ / v₁

m₁ / m₂= - (-6.2) / 4.7

m₁ / m₂ = 1.3

Another way to solve this exercise is to use the mass center relationship

Xcm = 1/M (m₁ x₁ + m₂ x₂)

We derive from time

Vcm = 1/M (m₁ v₁ + m₂v₂)

As they say the velocity of the center of zero masses

0 = 1/M (m₁ v₁ + m₂v₂)

m₁ v₁ + m₂v₂ = 0

m₁ / m₂ = -v₂ / v₁

m₁ / m₂ = 1.3

4 0

3 years ago

Match the major muscle group with its functional role.

Naily [24]

Answer:

Abdominal

Sitting up, postural alignment

Biceps

Lifting, pulling

Deltoids

Overhead lifting

Erector Spinae

Postural alignment

Gastronemius & Soleus

Push off for walking, standing on tiptoes

Gluteus

Climbing stairs, walking, standing up

Hamstrings

Walking

Latissimus Dorsi & Rhomboids

Postural alignment, pulling open a door

Obliques

Rotation and side flexion of body

Pectoralis

Push up, pull up, bench press

Quadriceps

Climbing stairs, walking, standing up

Trapezius

Moves head sideways

Triceps

Pushing

God bless you. Because my soul almost left my body when i had to do this.

7 0

3 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

3 years ago

What's earths most precious element?

atroni [7]

I believe the answer is californium because it is used in most metals and is very strong and expensive like gold and silver

3 0

3 years ago

Read 2 more answers