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BARSIC [14]
3 years ago
5

A frictionless pulley of negligible mass is hung from the ceiling using a rope, also of negligible mass. Two masses, m1 and m2 (

m1 < m2), are connected to the rope over the pulley. The masses are free to drop. The magnitude of the tension Ttop is ____ the sum of the tensions T1 and T2.
Physics
1 answer:
kirza4 [7]3 years ago
4 0

Answer:

Ttop = T1 + T2

Explanation:

The three tension forces act on the pulley. With the assumption that the pulley has a negligible mass, the force of gravity on it is zero. So only forces Ttop, T1 and T2 act on the pulley.

The tension force Ttop acts upwards on the pulley while the tension forces T1 and T2 act on the pulley downwards. The pulley is in equilibrium so the sum of upward forces acting on the pulley and the sum of downward forces acting on the pulley must be equal.

So

Ttop = T1 + T2.

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Block on inclined plane experience a force due to gravity of 300N straight down. If the slope is inclined at 67.8°to the horizon
Tems11 [23]

Answer:

The component of the force due to gravity perpendicular and parallel to the slope is  113.4 N and 277.8 N respectively.

Explanation:

Force is any cause capable of modifying the state of motion or rest of a body or of producing a deformation in it. Any force can be decomposed into two vectors, so that the sum of both vectors matches the vector before decomposing. The decomposition of a force into its components can be done in any direction.

Taking into account the simple trigonometric relations, such as sine, cosine and tangent, the value of their components and the value of the angle of application, then the parallel and perpendicular components will be:

  • Fparallel = F*sinα =300 N*sin 67.8° =300 N*0.926⇒ Fparallel =277.8 N
  • Fperpendicular = F*cosα =  300 N*cos 67.8° = 300 N*0.378 ⇒ Fperpendicular= 113.4 N

<u><em>The component of the force due to gravity perpendicular and parallel to the slope is  113.4 N and 277.8 N respectively.</em></u>

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3 years ago
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A piano string having a mass per unit length equal to 4.70 10-3 kg/m is under a tension of 1 400 N. Find the speed with which a
sweet [91]

Answer:

The speed of the sound wave on the string is 545.78 m/s.

Explanation:

Given;

mass per unit length of the string, μ = 4.7 x 10⁻³ kg/m

tension of the string, T = 1400 N

The speed of the sound wave on the string is given by;

v = \sqrt{\frac{T}{\mu} }

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v is the speed of the sound wave on the string

Substitute the given values and solve for speed,v,

v = \sqrt{\frac{T}{\mu} }\\\\v = \sqrt{\frac{1400}{4.7*10^{-3}} }\\\\v = \sqrt{297872.34}\\\\v = 545.78 \ m/s

Therefore, the speed of the sound wave on the string is 545.78 m/s.

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The ground of an antenna needs to be attached to the car in such a way that the voltage of the antenna can connect to the earth. This connection allows the earth to receive and store excess voltage rather than the extra voltage bouncing back through your antenna and radio.

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