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3 years ago

11

If an object is an irregularly shaped solid and it is dropped into a graduated cylinder and it displaces 25 mL of water and has

a mass of 50 grams what is the density of this irregular shaped object.

1 answer:

zavuch27 [327]3 years ago

7 0

Answer:

1250

Explanation:

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Consider the circuit shown in the figure to find the power delivered to 6 Ohm resistance (in W). Given that Vs= 30

vekshin1

66.7 Watts

Explanation:

Let $R_{1}=1.0$ ohms, $R_{2}=3.0$ ohms and $R_{3}=6.0\:$ ohms. Since $R_{2}$ and $R_{3}$ are in parallel, their combined resistance $R_{23}$ is given by

$\dfrac{1}{R_{23}}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}$

or

$R_{23}=\dfrac{R_{2}R_{3}}{R_{2}+ R_{3}}=2.0\:ohms$

The total current flowing through the circuit <em>I</em> is given

$I=\dfrac{V_{s}}{R_{Total}}$

where

$R_{Total}=R_{1}+R_{23}= 3.0\:ohms$

Therefore, the total current through the circuit is

$I=\dfrac{30\:V}{3.0\:ohms}=10\:A$

In order to find the voltage drop across the 6-ohm resistor, we first need to find the voltage drop across the 1-ohm resistor $V_{1}$ :

$V_{1}=(10\:A)(1.0\:ohms)=10\:V$

This means that voltage drop across the 6-ohm resistor $V_{3}$ is 20 V. The power dissipated <em>P</em> by the 6-ohm resitor is given by

$P=I^{2}R_{3}= \dfrac{V^{2}}{R_{3}}=\dfrac{(20\:V)^{2}}{6\:ohms}= 66.7 W$

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3 years ago

A 81 kg man is riding on a 40 kg cart traveling at a speed of 2.3 m/s. He jumps off with zero horizontal speed relative to the g

Alexus [3.1K]

Answer:

$\Delta v= 4.66\frac{m}{s}$

Explanation:

In this case we have to use the Principle of conservation of Momentum:

<em>This principle says that in a system the total momentum is constant if no external forces act in the system. The formula is:</em>

$m_1v_1+m_2v_2=m_1u_1+m_2u_2$

<em>Where:</em>

$m_1:$ Mass of the first object.

$m_2:$ Mass of the second object.

$v_1:$ Initial velocity of the first object.

$v_2:$ Initial velocity of the second object.

$u_1:$ Final velocity of the first object.

$u_2:$ Final velocity of the second object.

In <u>this problem</u> we have:

$m_1=81kg\\m_2=40kg\\v_1_2=2.3\frac{m}{s}$

$u_1=0\frac{m}{s}$

Observation: $v_1_2:$ Is because the system has the same initial velocity.

First we have to find $u_2$ ,

$m_1v_1+m_2v_2=m_1u_1+m_2u_2$

We can rewrite it as:

$(m_1+m_2)v_1_2=m_1u_1+m_2u_2$

Replacing with the data:

$(m_1+m_2)v_1_2=m_1u_1+m_2u_2\\\\(81kg+40kg)2.3\frac{m}{s}=81kg(0\frac{m}{s})+40kg(u_2)\\\\(121kg)2.3\frac{m}{s}=40kg(u_2)\\\\\frac{(121kg)2.3\frac{m}{s}}{40kg}=u_2\\\\\frac{278.3}{40}\frac{m}{s}=u_2\\\\6.96\frac{m}{s}=u_2$

We found the final velocity of the cart, but the problem asks for the resulting change in the cart speed, this means:

$\Delta v=u_2-v_2\\\Delta v=6.96\frac{m}{s}-2.3\frac{m}{s}\\\Delta v= 4.66\frac{m}{s}$

Then, the resulting change in the cart speed is:

$\Delta v= 4.66\frac{m}{s}$

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Explanation:

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