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eimsori [14]
3 years ago
11

Many machines-including inclined planes such as ramps- increase the strength of the force put into the machine but __________ th

e distance over which the force is applied.
A.) Increase

B.) Decrease

C.) Does not change

D.) None of the above
Physics
1 answer:
Leviafan [203]3 years ago
5 0
<h3><u>Answer;</u></h3>

B.) Decrease

<h3><u>Explanation;</u></h3>
  • Many machines including inclined planes such as ramps; increase the strength of the force put into the machine but <u>decrease </u>the distance over which the force is applied.
  • Other machines increase the distance over which the force is applied but decrease the strength of the force. Also other machines change the direction of the force, with or without also increasing its strength or distance.
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A piece of copper wire is formed into a single circular loop of radius 9.1 cm. A magnetic field is oriented parallel to the norm
bazaltina [42]

Given Information:  

Radius of circular loop = r = 9.1 cm = 0.091 m  

Change in time = Δt = 0.66 seconds

Change in magnetic field = ΔB = 0.90 T

Resistance of wire per unit length = R = 2.9x10⁻²  Ω/m

Number of turns = N = 1

Required Information:  

Electrical energy dissipated = E = ?  

Answer:  

Electrical energy dissipated = 50.09x10⁻³ Joules  

Step-by-step explanation:  

We know that energy is given by

E = Pt

Where power is given by

P = ξ²/R

Where ξ is the induced EMF in the wire and is given by

ξ = -NΔΦ/Δt

Where ΔΦ is the change in flux and is given by

ΔΦ = ΔBAcosφ

Where φ is the angle between magnetic field and circular loop

A = πr² and R = 2.9x10⁻²*2πr

Substituting the above relations into the energy equation and simplifying yields,

E = [-Nπr²cosφ(ΔB/Δt)²]*t/R

E = [-1*π(0.091)²*cos(0)(0.90/0.66)²*0.66]/2.9x10⁻²*2π*(0.091)

E = 0.050094 Joules

E = 50.09x10⁻³ Joules

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6 0
3 years ago
A 1000-kg aircraft going 25 m/s collides with a 1500-kg aircraft that is parked. They stick
Sonbull [250]

Answer:

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Explanation:

Mass of moving aircraft = 1000 kg

Mass of stationary aircraft = 1500 kg

Velocity of moving aircraft = 25 m/s

Velocity by which they both move together = 10 m/s

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v=u+at\\\Rightarrow 0=10+a\times 9.7\\\Rightarrow \frac{-10}{9.7}=a\\\Rightarrow a=-1.03\ m/s^2

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-10^2}{2\times -1.03}\\\Rightarrow s=48.54\ m

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