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NeTakaya
3 years ago
13

In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because _____.

Physics
2 answers:
Jlenok [28]3 years ago
8 0

Answer:

of the presence of friction.

Explanation:

grandymaker [24]3 years ago
5 0
In a real system of levers, wheel or pulleys, the AMA (actual mechanical advantage) is less than the IMA (ideal mechanical advantage) because of the presence of friction.

In fact, the IMA and the AMA of a machine are defined as the ratio between the output force (the load) and the input force (the effort):
IMA= \frac{F_{out}}{F_{in}}
however, the difference is that the IMA does not take into account the presence of frictions, while the AMA does. As a result, the output force in the AMA is less than the output force in the IMA (because some energy is dissipated due to friction), and the AMA is less than the IMA.
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Suppose that a simple pendulum consists of a small 60.0 g bob at the end of a cord of negligible mass. If the angle 0 between th
erik [133]

Based on the mass of the bob and the angle between the cord and the vertical, the pendulum length is 0.50m.

The maximum kinetic energy can be found to be 9.42 x 10⁻⁴J.

<h3>What is the pendulum length?</h3>

This can be found as:

= g-force / w²

Solving gives:

= 9.8 / 4.43²

= 0.4998 m

= 0.50 m

<h3>What is the maximum kinetic energy?</h3>

This can be found as:

= 0.5 × m × w² × A²

Maximum kinetic energy is:

= 0.5 × 60 × 10⁻³ × (4.43 × 0.4998 x 0.08 rad)²

= 9.42 x 10⁻⁴J

Find out more on maximum kinetic energy at brainly.com/question/24690095.

5 0
1 year ago
A 2.40 cm × 2.40 cm square loop of wire with resistance 1.20×10−2 Ω has one edge parallel to a long straight wire. The near edge
Norma-Jean [14]

Answer:

current in loops is 52.73 μA

Explanation:

given data

side of square a = b  = 2.40 cm = 0.024 m

resistance R = 1.20×10^−2 Ω

edge of the loop c  = 1.20 cm = 0.012 m

rate of current = 120 A/s

to find out

current in the loop

solution

we know current formula that is

current = voltage / resistance    .................a

so current = 1/R × d∅/dt

and we know here that

flux ∅ = ( μ×I×b / 2π ) × ln (a+c/c)    ...............b

so

d∅/dt = ( μ×b / 2π ) × ln (a+c/c) × dI/dt       ...........c

so from equation a we get here current

current = ( μ×b / 2πR ) × ln (a+c/c) × dI/dt

current = ( 4π×10^{-7}×0.024 / 2π(1.20×10^{-2}) × ln (0.024 + 0.012/0.012) × 120

solve it and we get current that is

current = 4 ×10^{-7}× 1.09861 × 120

current = 52.73 ×10^{-6}  A

so here current in loops is 52.73 μA

8 0
2 years ago
A second baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. The throw is m
Anit [1.1K]
 <span>(a) 

Taking the angle of the pitch, 37.5°, and the particle's initial velocity, 18.0 ms^-1, we get: 

18.0*cos37.5 = v_x = 14.28 ms^-1, the projectile's horizontal component. 

(b) 
To much the same end do we derive the vertical component: 

18.0*sin37.5 = v_y = 10.96 ms^-1 

Which we then divide by acceleration, a_y, to derive the time till maximal displacement, 

10.96/9.8 = 1.12 s 

Finally, doubling this value should yield the particle's total time with r_y > 0 

<span>2.24 s

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6 0
2 years ago
Body of a 218 kg starta from rest and attains the velocity of 63 m/s in 9 second
Vanyuwa [196]

Answer:

567m

Explanation:

By using formula of velocity.

7 0
3 years ago
A wire carrying a 29.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's
Dmitrij [34]

Answer:

2.59 T

Explanation:

Parameters given:

Current flowing through the wire, I = 29 A

Angle between the magnetic field and wire, θ = 90°

Magnetic force, F = 2.25 N

Length of wire, L = 3 cm = 0.03 m

The magnetic force, F, is related to the magnetic field, B, by the equation below:

F = I * L * B * sinθ

Inputting the given parameters:

2.25 = 29 * 0.03 * B * sin90

2.25 = 0.87 * B

=> B = 2.25/0.87

B = 2.59 T

The magnetic field strength between the poles is 2.59 T

4 0
3 years ago
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