Because: Some of the work done by the machine is used to overcome the friction created by the use of the machine. ... Work output can never be greater than work input. Machines allow force to be applied over a greater distance, which means that less force will be needed for the same amount of work.
Answer:
Yes, the relationships you observe in childhood affects the quality of your current relationships.
Explanation:
The saying,<em> "Children see, children do,"</em> is evidently true. A child's lifestyle and future relationships are greatly affected by the kind of relationships you allow him to see. Thus, it is very important that parents become role models for their kids.
For example, when your parents handle stress by shouting at each other, then most likely you'd also handle stress the same way in the future. So,<em> it is essential that the role models will show the child how to handle stress positively.</em> The experiences that children have at a young age affects their adult life.<u> It influences them and shapes them into who they will become.</u>
If people at home show positive relationships with each other and also supports a child regarding his education, then the child will most likely set positive relationships with his peers and achieve higher degree standards.
Thus, this explains the answer.
1. The iris regulates the amount of light entering the eye
2. The retina receives and organises visual information
3. The lens refracts light rays in a camera
Answer:
1.24 C
Explanation:
We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.
The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m
So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.
Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.
So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²
So, 4iρD/d² = 0.750πD²/4Δt.
iΔt = 0.750πD²/4 ÷ 4iρD/d²
iΔt = 0.750πD²d²/16ρ.
So the charge Q = iΔt
= 0.750π(Dd)²/16ρ
= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)
= 123.76 × 10⁻² C
= 1.2376 C
≅ 1.24 C
Answer:
false im just trying to get it if you'd like to give it to me
