What refers to how large or strong is the force?

What is a measure of the amount of matter in an object? Question 2 options: Force Inertia Mass Acceleration

fredd [130]

The answer is C.) mass is the matter of an object

6 0

3 years ago

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What is a tsunami? How is it formed?

IrinaK [193]

A tsunami is a series of waves generated in an ocean or other body of water by a disturbance such as an earthquake, landslide, volcanic eruption, or meteorite impact. ... Undersea earthquakes, which typically occur at boundaries between Earth's tectonic plates, cause the water above to be moved up or down

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3 years ago

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A 1090 kg car has four 12.7 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to ro

max2010maxim [7]

Answer:

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

Explanation:

To develop this exercise we proceed to use the kinetic energy equations,

In the end we replace

$KE_{Total}=KE_{Translational}+KE_{Rotational}$

$KE_{Total}=\frac{1}{2}m_{car}+4*\frac{1}{2}*I*(\frac{v}{r})^2$

Here

$I=\frac{1}{2}m_{wheels}*r^2$ meaning the 4 wheels,

So replacing

$KE_{Rotational}=4\frac{1}{2}*(\frac{1}{2}m_{wheels}*r^2)*(\frac{v}{r})^2=m*v^2$

So,

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}*v^2}{\frac{1}{2}m_{car}*v^2+m_{wheels}*v^2}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}}{\frac{1}{2}m_{car}+m_{wheels}}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{10}{545+10}$

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

3 0

3 years ago

A tray of electronic components contains 15 components, 4 of which are defective. If 4 components are selected, what is the poss

dlinn [17]

Answer:

a) 0.0007326

b) 0.03223

c) 0.2418

d) 0.2418

Explanation:

To find different probabilities for the selection of components among eleven good and four defective components, we will use the Combination.

a) $C(4,4) = 1; C(15,4) = 1365$

$P = \frac{C(4,4)}{C(15,4)} = \frac{1}{1365} = 0.0007326$

b) $C(4,3) = 4; C(11,1) = 11$

$P = \frac{C(4,3)*C(11,1)}{C(15,4)} = \frac{4*11}{1365} = 0.03223$

c) $C(4,2) = 6; C(11,2) = 55$

$P = \frac{C(4,2)*C(11,2)}{C(15,4)} = \frac{6*55}{1365} = 0.2418$

d) $C(11,4) = 330$

$P = \frac{C(11,4)}{C(15,4)} = \frac{330}{1365} = 0.2418$

8 0

3 years ago

The energy in the typical thunderstorm is about 1.4×1011j. a typical lightning flash transfers 30 c across a potential differenc

iragen [17]

780 seconds, or 13 minutes.

In the future, please use proper capitalization. There's a significant difference in the meaning between mV and MV. One of them indicated millivolts while the other indicates megavolts. For this problem, I'll make the following assumptions about the values presented. They are:
Total energy = 1.4x10^11 Joules (J)
Current per flash = 30 Columbs (C)
Potential difference = 30 Mega Volts (MV)

First, let's determine the power discharged by each bolt. That would be the current multiplied by the voltage, so
30 C * 30x10^6 V = 9x10^8 CV = 9x10^8 J

Now that we know how many joules are dissipated per flash, let's determine how flashes are needed.
1.4x10^11 / 9x10^8 = 1.56E+02 = 156

Since each flash takes 5 seconds, that means that it will take about 5 * 156 = 780 seconds which is about 780/60 = 13 minutes.

3 0

3 years ago