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3 years ago

(Please Help! Match the sentence to the number down below! Please I'll give brainlist and points!)

Physics

1 answer:

dmitriy555 [2]3 years ago

5 0

A.10

B.3

C.5

D.7

E.8

F.9

G.2

H.1

I.4

J.6

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Suzanne makes a tower out of six identical plastic box. Then she uses a pan balance to measure the mass of the tower. She finds

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Xplain the difference between aerobic and anaerobic exercise and give 1 example of each?

Lilit [14]

Answer:

The term "aerobic" refers to the body's ability to provide energy through the use of oxygen. The term anaerobic refers to the body's ability to provide energy without the use of oxygen.

An example of an aerobic exercise is swimming. An example of an anaerobic exercise is sprinting.

Explanation:

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On February 15, 2013, a superbolide meteor (brighter than the Sun) entered Earth's atmosphere over Chelyabinsk, Russia, and expl

fredd [130]

Answer:

156.67 m/s

0.45676 times the speed of sound

Explanation:

Distance from the ground = 23.5 km = 23500 m

Time taken by the blast waves to reach the ground = $2\ minutes\ 30\ seconds=2\times 60+30=150\ s$

Spedd of the wave would be

$Speed=\dfrac{Distance}{Time}\\\Rightarrow v_b=\dfrac{23500}{150}\\\Rightarrow v-b=156.67\ m/s$

The velocity of the blast wave is 156.67 m/s

v = Velocity of sound = 343 m/s

$\dfrac{v_b}{v}=\dfrac{156.67}{343}\\\Rightarrow v_b=v\dfrac{156.67}{343}\\\Rightarrow v_b=0.45676v$

The blast wave is 0.45676 times the speed of sound

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3 years ago

What would be most useful to help make a simple compass a nonmetal bar,a round metal can,a small iron nail,or a Quartz needle

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Iron nail. the rest of those are not iron or some form of magnetic material.

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A JFET has a drain current of 5mA. If IDSS = 10mA and VGS ( off )= -6 v. find The Value Of

levacccp [35]

$\underline {\huge \boxed{ \sf \color{skyblue}Answer : }}$

Given :

$\tt \large {\color{purple} ↬ } \: \: \: \: \: I_{D} = 5mA$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: I_{DSS} = 10mA$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: V_{GS(off)} = -6V$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: V_{GS} = {?}$

$\: \: \:$

Let's Slove :

$\tt \large I_{D} = I_{(DSS)} (1 - \frac {V_{GS}}{V_{GS(off)}} )^{2}$

$\: \: \:$

$\tt \large \: V_{GS} = (1 - \frac{ \sqrt{I_D} }{ \sqrt{I_{DSS}} } ) \times V_{GS(off)}$

$\: \: \:$

$\tt \large \: V_{GS} = (1 - \frac{ \sqrt{5m} }{ \sqrt{10m} } ) \times { - 6}$

$\: \:$

$\underline \color{red} {\tt \large \boxed {\tt V_{GS} = 1.75 ✓}}$

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2 years ago