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Paul [167]
3 years ago
10

When Marcel finds the distance L from the previous part, it turns out to be greater than Lend, the distance from the pivot to th

e end of the seesaw. Hence, even with Jacques at the very end of the seesaw, the twins Gilles and Jean exert more torque than Jacques does. Marcel now elects to balance the seesaw by pushing sideways on an ornament (shown in red) that is at height h above the pivot. (Figure 3)With what force in the rightward direction, Fx, should Marcel push? If your expression would give a negative result (using actual values) that just means the force should be toward the left.Express your answer in terms of W, Lend, w, L2, L3, and h.
Physics
1 answer:
Llana [10]3 years ago
3 0

Answer:

Fx = - (1/h)( wL2 + wL3 - wLend )

Explanation:

Assuming The twins Gilles and Jean has a weight ( w ) each

The torque that would balance the equation would be = wL2 + wL3 -------- 1

THEREFORE the ccw torques are = wLend + Fh ----------- 2

hence equation 2 equals equation 1

= wLend + Fh = wL2 + wL3 --------- 3

equation 3 can as well be represented as

F = ( 1/h) ( wL2 + wL3 - wLend )---------- 4

From equation 4 it can be seen that F is on the left hand side therefore the value of Fx is negative

therefore equation 4 is represented as

 Fx = - (1/h)( wL2 + wL3 - wLend )

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The magnetic field at the center of a 1.50-cm-diameter loop is 2.70 mT . Part A. What is the current in the loop?
elixir [45]

Explanation:

It is given that,

Diameter of the circular loop, d = 1.5 cm

Radius of the circular loop, r = 0.0075 m

Magnetic field, B=2.7\ mT=2.7\times 10^{-3}\ T

(A) We need to find the current in the loop. The magnetic field in a circular loop is given by :

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I=\dfrac{2Br}{\mu_o}

I=\dfrac{2\times 2.7\times 10^{-3}\times 0.0075}{4\pi \times 10^{-7}}

I = 32.22 A

(b) The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

r=\dfrac{\mu_o I}{2\pi B}

r=\dfrac{4\pi \times 10^{-7}\times 32.22}{2\pi \times 2.7\times 10^{-3}}

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r=2.38\times 10^{-3}\ m

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8 0
3 years ago
Rosalind is an impulsive child who seldom waits for her turn when playing with other children. Rosalind's father is also an impu
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Answer:

Alburt Bandura

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A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it
Degger [83]

Answer:

a)\omega_1=8.168\,rad.s^{-1}

b)n_1=7.735 \,rev

c)\alpha_1 =0.6864\,rad.s^{-2}

d)\alpha_2=4.1454\,rad.s^{-2}

e)t_2=1.061\,s

Explanation:

Given that:

  • initial speed of turntable, N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}
  • full speed of rotation, N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}
  • time taken to reach full speed from rest, t_1=11.9\,s
  • final speed after the change,  N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}
  • no. of revolutions made to reach the new final speed,  n_2=11\,rev

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

\omega=\frac{2\pi.N}{60}\, rad.s^{-1}

where N = angular speed in rpm.

putting the respective values from case 1 we've

\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}

\omega_1=8.168\,rad.s^{-1}

(c)

using the equation of motion:

\omega_1=\omega_0+\alpha . t_1

here α is the angular acceleration

78=0+\alpha_1\times 11.9

\alpha_1 = \frac{8.168 }{11.9}

\alpha_1 =0.6864\,rad.s^{-2}

(b)

using the equation of motion:

\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1

8.168^2=0^2+2\times 0.6864\times n_1

n_1=48.6003\,rad

n_1=\frac{48.6003}{2\pi}

n_1=7.735\, rev

(d)

using equation of motion:

\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2

12.5664^2=8.168^2+2\alpha_2\times 11

\alpha_2=4.1454\,rad.s^{-2}

(e)

using the equation of motion:

\omega_2=\omega_1+\alpha_2 . t_2

12.5664=8.168+4.1454\times t_2

t_2=1.061\,s

4 0
3 years ago
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