Radiation with a wavelength of 238 nm shines on a metal surface and ejects electrons that have a maximum speedof 3.75 X 105 m/s.

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

grandymaker [24]

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.

5 0

3 years ago

Read 2 more answers

A student performs an experiment in measuring the period of a simple pendulum of known length 49.0 cm.He performed five trials a

Levart [38]

Correct Answer is Bb

4 0

3 years ago

Which of the following are homogeneous solutions?

Sindrei [870]

Salt water
Air]
Vinegar

3 0

4 years ago

As shown in the figure above, a 50 kg box is dragged across the floor with a pulling force (Fp) of 200 N which acts at an angle

Nat2105 [25]

Answer:

The acceleration of the box is 2.05 m/s²

Explanation:

The given parameters of the motion of the box are;

The mass of the box, m = 50 kg

The pulling force, $F_p$ acting on the box = 200 N

The angle at which the force acts, θ = 30° above the horizontal

The coefficient of kinetic friction, $\mu_k$ = 0.25

The normal reaction from the box resting on a flat surface, N = The weight of the box, W - The vertical component of the pulling force, $F_{py}$

N = W - $F_{py}$ = m·g - $F_p$ × sin(θ)

Where;

g = The acceleration due to gravity = 9.8 m/s²

∴ N = W - $F_{py}$ = m·g - $F_p$ × sin(θ) = 50 kg × 9.8 m/s² - 200 N × sin(30°)

∴ N = 490 N - 200 N × 0.5 = 390 N

The normal reaction, N = 390 N

The force of friction, $F_f$ = The coefficient of kinetic friction, $\mu_k$ × The normal reaction, N

∴ $F_f$ = $\mu_k$ × N = 0.25 × 390 N = 97.5 N

The net force, $F_{NET}$ , acting on the block = The pulling force, $F_p$ - The friction force, $F_f$

∴ $F_{NET}$ = $F_p$ - $F_f$ = 200 N - 97.5 N = 102.5 N

$F_{NET}$ = 102.5 N

According to Newton's second law of motion on the net force acting on an object, we have;

$F_{NET}$ = m × a

Where;

a = The acceleration of the box

∴ a = $F_{NET}$ /m = 102.5 N/(50 kg) = 2.05 m/s²

The acceleration of the box = a = 2.05 m/s².

8 0

3 years ago

A pen rolls off a 0.55–meter high table with an initial horizontal velocity of 1.2 meters/second. At what horizontal distance fr

levacccp [35]

To calculate the horizontal distance traveled by the ball, we first calculate the total time it takes to reach the ground as follows:

t = √[2y/g] = <span>√[2(0.55) / 9.81]
t = 0.33 s

The horizontal distance would be

</span><span>X = Vx*t = 1.2*√[2*.55/9.8] = 0.4 m
</span>
Hope this helps.

4 0

3 years ago

Read 2 more answers