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3 years ago

If an object's kinetic energy is zero, what is its momentum?

1 answer:

6 0

If the object's kinetic energy is zero, then due to in multiplication factor, it's momentum will also be equal to zero 'cause the velocity of the object must be Nil

In short, Your Answer would be: "Zero"

Hope this helps!

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An output gear has 10 teeth and an input gear has 40 teeth. What is the mechanical advantage of gear combination? Is force or sp

yKpoI14uk [10]

Answer:

Mechanical advantage: 4. Speed is multiplied

Explanation:

The mechanical advantage of gear combination is

$MA = \frac{N_i}{N_o}$

where

$N_i$ is the number of teeth in the input gear

$N_o$ is the number of teeth in the output gear

Here we have

$N_i = 40\\N_o = 10$

So we find

$MA = \frac{40}{10}=4$

In this example, it is the speed to be multiplied: the speed of the output gear is 4 times larger than the speed of the input gear. In fact, the mechanical advantage can be rewritten as

$MA= \frac{\omega_o}{\omega_i}$

where

$\omega_i$ is the angular speed of the input gear

$\omega_o$ is angular speed of the output gear

So re-arranging the equation, we find

$\omega_o = MA \cdot \omega_i = 4 \omega_i$

so, the speed of the output gear is 4 times larger than the speed of the input gear.

7 0

3 years ago

Explain how levitt and dubner’s argument effectively uses logical

faltersainse [42]

Explanation:

Levitt and Dubner’s argument effectively uses logical, concrete evidence to arrive at conclusions about morality and cheating practices.the realities they present about sumo and the bagel business Their utilization of factual evidence in various illustrations shows morality and cheating practices are more common in high incentive situations where telling lies or an act of fooling people was rewarded awesomely. They use statistical evidence and different examples support the fact that Levitt and Dubner have arrived at a generalisation on moral grounds.

5 0

3 years ago

Read 2 more answers

A thin, uniform metal bar, 2.00 m long and weighing 90.0 N, is hanging vertically from the ceiling by a frictionless pivot. Sudd

matrenka [14]

Answer:

(a) The "angular speed" is 5.88 rad/s.

Explanation:

Given values,

The length of the bar is L = 2m

The weight of the bar is w = 90 N

The metal bar is hanging vertically from the ceiling by a frictionless pivot

The mass of the ball is m = 3kg

The distance between the ceiling and the ball is d = 1.5m

$\text { The "initial speed of the ball" is } V_{i}=10 \mathrm{m} / \mathrm{s}$

$\text { The "final speed of the ball" is } V_{f}=6 \mathrm{m} / \mathrm{s}$

(a) Calculating the angular speed:

$W_{\mathrm{f}}=\left(3 \mathrm{mg} \mathrm{d} \frac{V_{i}+V_{f}}{w l^{2}}\right)$

$W_{f}=\left(3 \times 3 \times 9.8 \times 1.5 \times \frac{10+6}{90 \times 2^{2}}\right)$

$W_{f}=\frac{2116.8}{360}$

$\mathrm{W}_{\mathrm{f}}=5.88 \mathrm{rad} / \mathrm{s}$

The angular speed is 5.88 rad/s.

(b) The "angular momentum" is conserved because the torque is not exerted by "the pivot" on the system about the "axis of rotation" but the "linear momentum" is not conserved because "the pivot" exerts a "vertical" and a "horizontal force" on the system during the collision.

8 0

4 years ago

A hydraulic jack has a small piston area A1 = 2 m^2 and the large piston area A2 = 20 m^2. A 1500 Kg car needs to be lifted. a)

Alex777 [14]

Answer:

1471.5 Newton

Explanation:

Small piston area = A₁ = 2 m²

Large piston area A₂ = 20 m

m = Mass of car = 1500 kg

g = Acceleration due to gravity = 9.81 m/s²

Force

F = mg = 1500×9.81 = 14715 N

Force applied by car is 14715 N

a) Pascal's law

$\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\Rightarrow F_1=F_2\frac{A_1}{A_2}\\\Rightarrow F_1=14715\frac{2}{20}=1471.5$

Force required is 1471.5 Newton

b) Mechanical advantage

$\frac{F_2}{F_1}=\frac{14715}{1471.5}=10$

Mechanical advantage is 10

5 0

3 years ago

A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/srev/s . The wheel can be consi

vagabundo [1.1K]

Answer:1.7 rev/s

Explanation:

Given

Frequency of wheel $N_1=2\ rev/s$

angular speed $\omega_1=2\pi N_1=4\pi\ rad/s$

mass of wheel $m_1=4.5\ kg$

diameter of wheel $d_1=0.30\ m=30\ cm$

radius of wheel $r_1=\frac{d_1}{2}=\frac{30}{2}=15\ cm$

mass of clay $m_2=2.8\ kg$

the radius of the chunk of clay $r_2=8\ cm$

Moment of inertia of Wheel

$I_1=\dfrac{m_1r_1^2}{2}=\dfrac{4.5\times 15^2}{2}\ kg-cm^2$

Combined moment of inertia of wheel and clay chunk

$I_2=\dfrac{m_1r_1^2}{2}+\dfrac{m_2r_2^2}{2}=\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2}\ kg-cm^2$

Conserving angular momentum

$\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \dfrac{4.5\times 15^2}{2}\cdot 4\pi=(\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2})\omega_2\\\\\Rightarrow \omega _2=\dfrac{4\pi }{1+\dfrac{2.8}{4.5}\times (\dfrac{8}{15})^2}=\dfrac{4\pi}{1+0.1769}=0.849\times 4\pi$

Common frequency of wheel and chunk of clay is

$\Rightarrow N_2=\dfrac{4\pi \times 0.849}{2\pi}=1.698\approx 1.7\ rev/s$

5 0

3 years ago