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3 years ago

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Rafael swings a baseball bat with a force of 25 N. The baseball bat exerts a force of 5 N on the baseball. What is the mechanica

l advantage of the baseball bat? The mechanical advantage of the baseball bat is

2 answers:

goldfiish [28.3K]3 years ago

7 0

Answer:

0.2

Explanation:

Mechanical Advantage is defined as the ratio of force output with respect to the force input. Thus,

MA = output / input

MA = 5 / 25

MA = 1/5

MA = 0.2

Since this value is less then one. Even though it's tagged a mechanical advantage, in reality it would be mechanical disadvantage.

Therefore, we can say that the mechanical advantage of the baseball bat is 0.2

Firdavs [7]3 years ago

5 0

Answer:

0.2

Explanation:

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A 5.30 g bullet moving at 963 m/s strikes a 610 g wooden block at rest on a frictionless surface. The bullet emerges, traveling

Tems11 [23]

Answer:

a) $V_{wf}$ = 4.67m/s

b) V = 8.29 m/s

Explanation:

Givens:

The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

a) From conservation of linear momentum

Pi = Pf

$m_{b}V{b_{i} } + V_{wi} = m_{w} V_{wf} + m_{b}V_{bf}$

where $m_{b},V{b_{i}$ are the mass and the initial velocity of the bullet, $m_{w}$ and $V_{wi}$ are the mass and the initial velocity of the wooden block, and $V_{wf}$ and $V_{bf}$ are the final velocities of the wooden block and the bullet

The wooden block is initial at rest $(V_{wi} = 0)$ this yields

$m_{b}V{b_{i} } = m_{w} V_{wf} + m_{b}V_{bf}$

By solving for $V_{wf}$ adn substitute the givens

$V_{wf}$ = $\frac{m_{b}V_{bi} - m_{b} V_{bf} }{m_{W} }$

= $\frac{5.3(g)(963(m/s)-426(m/s) }{610(g)}$

$V_{wf}$ = 4.67m/s

b) The center of mass speed is defined as

$V = \frac{m_{b} }{m_{b}+m_{w} } V_{bi}$

substituting:

$V = \frac{5.3(g)}{5.3(g)+610(g)} X 963(m/s)$

V = 8.29 m/s

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3 years ago

Camera flashes charge a capacitor to high voltage by switching the current through an inductor on and off rapidly. In what time

Ber [7]

Answer:

The value is $\Delta t = 4.0 *10^{-7} \ s$

Explanation:

From the question we are told that

The current is $\Delta I = 0.100 \ A$

The inductor is $L = 2.0mH = 2.0*10^{-3} \ H$

The voltage induced is $\epsilon = 500 V$

Generally the emf induced is mathematically represented as

$\epsilon = L * \frac{\Delta I }{\Delta t }$

Here $\Delta t$ is the time taken

=> $\Delta t = \frac{L * \Delta I }{\epsilon }$

=> $\Delta t = \frac{2*10^{-3} * 0.100 }{500 }$

=> $\Delta t = 4.0 *10^{-7} \ s$

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Answer:

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