A. because as the merry-go-round spins the child accelerates towards the center of the merry-go-round at a uniform rate.
The Ammeter is used to detect and measure current or amperage. Also a more common tool now used is a multimeter that detects and measures voltage, current, and resistance.
Any questions please just ask. Thank you.
To verify the identity, we can make use of the basic trigonometric identities:
cot θ = cos θ / sin θ
sec θ = 1 / cos <span>θ
csc </span>θ = 1 / sin θ<span>
Using these identities:
</span>cot θ ∙ sec θ = (cos θ / sin θ ) (<span> 1 / cos </span><span>θ)
</span>
We can cancel out cos <span>θ, leaving us with
</span>cot θ ∙ sec θ = 1 / sin θ
cot θ ∙ sec θ = = csc <span>θ</span>
Answer:
Explanation:
Theoretical efficiency = T₁ - T₂ / T₁ where T₁ and T₂ is absolute temperature of hot and cold end of the heat engine.
= 600 / (273 + 700 )
= 600 / 973
= .6166
operating efficiency = 40% of .6166
= .4 x .6166
= .2466 = 24.66 %
efficiency = work output / heat input
= 5000 / heat input = .2466
heat input = 5000 / .2466
= 20275.75 J .
HEAT EXTRACED = 20275.75 J.
Given that.
F=3•i+4•j
And it from point (0,0)m to (5,6)m
dx=final position - initial position
dx=(5,6)-(0,0)
dx=(5,6)m
dx=5•i +6•j
Work done by the force is give by
W = F•dx
W=F•dx
Note that i•i=j•j=1 and i•j=j•i=0
Then,
W=(3i+4j)•(5i+6j)
Therefore,
W=3i•(5i+6j)+4j•(5i+6j)
W=15i•i+18i•j+20j•i+24j•j
W=15+0+0+24
W=39J
Then the work done by the force is 39 Joules
