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Murljashka [212]
3 years ago
14

Which one of the two potential differences emf or terminal voltage will be greater in magnitude?Why?​

Physics
1 answer:
tresset_1 [31]3 years ago
8 0

Answer:

True, yes terminal voltage of a battery can never be greater than the emf of the battery. Because due to internal resistance of the current terminal

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Identify the prefix that would be used to express<br> 2,000,000,000 bytes of computer memory?
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It would be 2 GB

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A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal
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Answer

given,

mass of the ball = 3 kg

swing in vertical circle with radius = 2 m

   work done by the gravity = ?          

   work done by the tension = ?            

Work done by the gravity = - m g Δh            

 Δ h = 2 + 2 = 4 m                                                                

Work done by the gravity =- 3 \times 9.8 \times 4

                                           = -117.6 J                  

work done by gravity is equal to -117.6 J            

Work done by tension will be equal to zero.        

Zero because tension is always perpendicular to velocity

work done by tension is equal to 0 J                          

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A place where things are baked <br>​
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Determine the magnitude of the resultant force acting on a 1.5 −kg particle at the instant t=2 s, if the particle is moving alon
Phoenix [80]

Answer:

F = 63N

Explanation:

M= 1.5kg , t= 2s, r = (2t + 10)m and

Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

7 0
3 years ago
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