3 years ago

The long handle on a rake is a _____.

Physics

1 answer:

artcher [175]3 years ago

6 0

From what i know it is c. it is a lever

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a charge of 30. coulombs passes through a 24-ohm resistor in 6.0 seconds. what is the current through the resistor? (1) 1.3 a (3

DochEvi [55]

As I = Q/t (charge/time) the answer is 5a.

30/6 = 5

7 0

3 years ago

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Would water molecules in Venus’ atmosphere, whose temperature is 740 K, escape into outer space? A water molecule has a mass tha

Akimi4 [234]

Answer:

The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus.

Explanation:

The average speed of gas molecules is given by:

$v_{rms}=\sqrt{\frac{3RT}{M}}$

R is the gas constant, T is the temperature and M the molar mass of the gas.

We know that a water molecule has a mass that is 18 times that of a hydrogen atom:

$M_H=1.01*10^{-3}\frac{kg}{mol}\\M_{H2O}=18M_H=0.02\frac{kg}{mol}$

So, we have:

$v_{rms}=\sqrt{\frac{3(8.314\frac{J}{mol \cdot K})740K}{0.02\frac{kg}{mol}}}\\v_{rms}=960.65\frac{m}{s}*\frac{1km}{1000m}=0.96\frac{km}{s}$

The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus:

$10\frac{km}{s}*\frac{1}{6}=1.6\frac{km}{s}\\0.96\frac{km}{s}$

4 0

3 years ago

A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

Nady [450]

The first collision because a greater amount of momentum must be taken and used in order to push the cart back, giving it a greater mass and impulse

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3 years ago

The North Star is located in the Big Dipper. True or False

harina [27]

Answer:

False

Explanation:

It is located in the little dipper whose stars are more faint.

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3 years ago

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What is the pendulum length whose period is 2.0s ?

Mashutka [201]

$Formula\ for\ period:\\\ T=2 \pi \sqrt{\frac{L}{g}}\\\ g-gravity=9,8 \frac{m}{s^2} ,\ L-pendulum \ length \\\\ \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\\\ \frac{T^2}{2 \pi } = \frac{L}{g} \\\\\ \frac{T^2}{2 \pi }*g=L\\\\ L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24m$ $T=2 \pi \sqrt{\frac{L}{g}} \\
 \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\
 \frac{T^2}{2 \pi } = \frac{L}{g} \\
 \frac{T^2}{2 \pi }*g=L\\
L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24m
$

7 0

3 years ago