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konstantin123 [22]

3 years ago

5

john throws .4 kg ball with velocity of 18 m/s. Hits .2 kg bottle and bottle flies 25 m/s. How fast is ball traveling after hitt

ing the bottle?

1 answer:

UNO [17]3 years ago

6 0

Answer:

5.5 m/s

Explanation:

Assuming the bottle is initially stationary, we can write the law of conservation of momentum as follows:

$p_i = p_f\\m_b u_b = m_b v_b + m_B v_B$

where

$m_b = 0.4 kg$ is the mass of the ball

$u_b = 18 m/s$ is the initial velocity of the ball

$m_B = 0.2 kg$ is the mass of the bottle

$v_B = 25 m/s$ is the final velocity of the bottle

$v_b$ is the final velocity of the ball

Solving for $v_b$ ,

$v_b = \frac{m_b u_b - m_B v_B}{m_b}=\frac{(0.4 kg)(18 m/s)-(0.2 kg)(25 m/s)}{0.4 kg}=5.5 m/s$

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Write one advantage of MKS system over CGS system.

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More convenient
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3 years ago

9. Una jeringa contiene cloro gaseoso, que ocupa un volumen de 95 mL a una presión de 0,96 atm. ¿Qué presión debemos ejercer en

masha68 [24]

Answer:

2.61 atm

Ley de Boyle

Explanation:

$P_1$ = Presión inicial = 0.96 atm

$P_2$ = Presión final

$V_1$ = Volumen inicial = 95 mL

$V_2$ = Volumen final = 35 mL

En este problema usaremos la ley de Boyle.

$\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=\dfrac{0.96\times 95}{35}\\\Rightarrow P_1=2.61\ \text{atm}$

La presión ejercida sobre el émbolo para reducir su volumen es de 2.61 atm.

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3 years ago

A child and sled with a combined mass of 58.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed o

erma4kov [3.2K]

Answer:

The height is $h = 0.5224 \ m$

Explanation:

From the question we are told that

The combined mass of the child and the sled is $m = 58.0 \ kg$

The speed of the sled is $u = 3.20 \ m/s$

Generally applying SOHCAHTOA on the slope which the combined mass is down from

Here the length of the slope(L) where the combined mass slides through is the hypotenuses

while the height(h) of the height of the slope is the opposite

Hence from SOHCAHTOA

$sin (\theta) = \frac{h}{L}$

=> $Lsin(\theta) = h$

Generally from the kinematic equation we have that

$v^2 = u^2 + 2aL$

Here the u is the initial velocity of the combined mass which is zero since it started from rest

and a is the acceleration of the combined mass which is mathematically evaluated as

$a = g * sin (\theta )$

$v^2 = u^2 + 2 * g * sin (\theta )L$

=> $2Lsin(\theta ) = \frac{v^2 - u^2 }{g}$

=> $h = \frac{ v^2 - u^2}{2g}$

=> $h = \frac{ 3.20^2 - 0^2}{2 * 9.8 }$

=> $h = 0.5224 \ m$

6 0

3 years ago

2) A 20 kg mass moving at a speed of 3 m/s is stopped by a constant force of 15 N. How many seconds must the force act on the ma

hammer [34]

Take the object's starting direction of motion to be the positive direction, so that a stopping force acts in the opposite direction. By Newton's second law, the object undergoes an acceleration <em>a</em> such that

-15 N = (20 kg) <em>a</em>

Solve for <em>a</em> :

<em>a</em> = - (15 N) / (20 kg) = -0.75 m/s²

The object's velocity <em>v</em> at time <em>t</em> is then given by

<em>v</em> = 3 m/s + (-0.75 m/s²) <em>t</em>

so the time it takes for the object to slow to a rest is

0 = 3 m/s + (-0.75 m/s²) <em>t</em>

<em>t</em> = (3 m/s) / (0.75 m/s²) = 4.0 s

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VladimirAG [237]

Answer:

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Explanation:

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3 years ago

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