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MrRissso [65]
3 years ago
14

g A rod is 2.0 m long and lies along the x-axis, with one end at the origin. A force of 25 N is applied at the point x = 1.2 m,

and is directed 30° above the x-axis. What is the torque on the rod?
Physics
1 answer:
Nataly [62]3 years ago
6 0

Answer:

τ = 15N.m

Explanation:

Torque on the rod is given by:

\tau=F*sin(30\°)*d

\tau=25*0.5*1.2

\tau=15N.m

If you were told that gravity contributes as well, and you were given the mass of the rod, you would have to subtract the torque of the weight to this value.

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List the planet name and position from the Sun for each one (1, 2, 3)
IgorLugansk [536]

Answer:

The planets in order from the sun are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and finally the dwarf planet Pluto. Most people have at least heard about our solar system and the planets in it.

Explanation:

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8 0
2 years ago
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To stretch a certain nonlinear spring by an amount x requires a force F given by F = 40 x − 6 x 2 , where F is in Newtons and x
strojnjashka [21]

Answer:

64 J

Explanation:

The potential energy change of the spring ∆U = -W where W = work done by force, F.

Now W = ∫F.dx

So, ∆U = - ∫F.dx = - ∫Fdxcos180 (since the spring force and extension are in opposite directions)

∆U = - ∫-Fdx

=  ∫F.dx

Since F = 40x - 6x² and x moves from x = 0 to x = 2 m, we integrate thus, ∆U =  ∫₀²F.dx

=  ∫₀²(40x - 6x²).dx

=  ∫₀²(40xdx - 6x²dx)

=  ∫₀²(40x²/2 - 6x³/3)

=  ∫₀²(20x² - 2x³)

= [20x² - 2x³]₀²

= [(20(2)² - 2(2)³) - (20(0)² - 2(0)³)

= [(20(4) - 2(8)) - (0 - 0))

= [80 - 16 - 0]

= 64 J

5 0
3 years ago
If a bag holds 70.874 grams ,how many itemsare in the bag?​
Alina [70]

Answer:

The answer is not able to be solved, because we dont know what objects are in it, and how heavy they are. More information please!

Explanation:

7 0
3 years ago
How much thermal energy is needed to melt 1.25 kg of water at its melting point? Use Q = masslaten heat of fusion.
Amanda [17]

Answer:

Latent heatnof fusion = 417.5 J

Explanation:

Specific latent heat of fusion of water is 334kJ.kg-1.

The heat required to melt water when it's ice I called latent heat because there is no temperature change, the only change observed is change in physical structure.

The amount of heat required to change 1 kg of solid to its liquid state (at its melting point) at atmospheric pressure is called Latent heat of Fusion.

Latent heat = ML

Latent heat= 1.25 kg * 334kJ.kg-1

Latent heat = 1.25*334 *(J/kg)*kg

Latent heat = 417.5 J

8 0
3 years ago
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