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Akimi4 [234]
2 years ago
14

Derive an expression for the workdone by a torque?​

Physics
2 answers:
vaieri [72.5K]2 years ago
7 0

Explanation:

<em>Work done by torque = magnitude of force * perpendicular distance between the point of action and point(axis) of rotation.</em>

Nikolay [14]2 years ago
3 0

Answer:

Work done by torque = magnitude of force * perpendicular distance between the point of action and point(axis) of rotation.

Explanation:

We can also write this as :-

W = F x d

where,

W->Work Done

F->Force

d->displacement

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An 8.5 kg crate is pulled 5.1 m up a 30 degree incline by a rope angled 17 degrees above the incline. The tension in the rope is
Ulleksa [173]

Answer:

1. a W_t=746.63 J

  b E_p=212.415 J

  c W_n=183.96J

2. T_e=99.71J

Explanation:

a). The work done by the tension is:

W_t=T*dt

dt=\frac{5.1}{cos(17)}

W_t=140N*\frac{5.1}{cos(17)}

W_t=746.63 J

b). The work done potential of gravity

E_p=m*g*h

h=5.1*sin(17)

E_p=8.5kh*9.8*5.1*sin(30)

E_p=212.415 J

c). The work done by the normal force

W_n=N*d_n

d_n=5.1*sin(30)=2.55

W_n=8.5kg*9.8*cos(30)*2.55

W_n=183.96J

2. The increase in thermal energy is:

T_e=F*d

F_k=u_k*m*g=0.271*8.5kg*9.8*cos(30)

F_k=19.5N

T_e=19.55*5.1m

T_e=99.71J

4 0
3 years ago
Need some help with this!
zvonat [6]

Answer: what’s it asking

Explanation:

8 0
3 years ago
Read 2 more answers
Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ
Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of   opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

Fn_{3} = \sqrt{405^{2}+701.5^{2}  }

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0
3 years ago
The driving force for an electric current is called
qaws [65]
Its called electro motive force . let me know if its right

6 0
3 years ago
Read 2 more answers
Discuss the differences between Seismic Reflection and Seismic Refraction
Maru [420]

Explanation:

Seismic refraction is the bending of the seismic waves as they pass geological layers of the earth due to different densities. This is especially true for Primary waves because they can pass through all the layers of the earth both liquid and solid.

Seismic reflectivity is the bouncing back of seismic waves at a boundary of geological layers due to different densities or subsurface formation. This is especially true for secondary waves that are unable to pass through liquid layers of the earth like the outer core and mantle. When they reach the boundary of these layers they bounce back towards the earth's surface.

These properties of P and S waves are used to ‘auscultate’ the epicenter of an earthquake by triangulation.

Learn More:

For more on Seismic Reflection and Seismic Refraction check out;

brainly.com/question/13502364

#LearnWithBrainly

6 0
3 years ago
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