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3 years ago

A temporary license permits the holder to drive for up to _ days while the application is reviewed

Physics

1 answer:

Ugo [173]3 years ago

7 0

60 days, tell me if I'm correct please.

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A heavy boy and a lightweight girl are balanced on a mass-less seesaw. If they both move forward so that they are one-half their

Rom4ik [11]

Answer:

b) Nothing will happen, the sea saw will still be balanced.

Explanation:

b) Nothing will happen, the sea saw will still be balanced.

Reason:-

When two kids are balanced, the sum of torques on the seesaw will be zero.

if each kid, reduces their distances by half, then the torque of each kid will be half and the sum of torque of each on the seesaw will be zero.

Therefore the seesaw is balanced

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3 years ago

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I need to show my work as well but on the computer so, please show work for how you got the answers. Thank you!

balandron [24]

Answer:

1) $1H_2 + 1Cl_2$ => 2HCl

2) $2Al + 6HCl$ => $2AlCl_3 + 3H_2$

3) $1Ca_2Br_2 + 2NaCO_3$ => $2CaCO_3 + 2NaBr$

4) $3NaOH + 1FeCl_3$ => $3NaCl + 1Fe(OH)_3$

Explanation:

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3 years ago

A circuit consists of a series combination of 6.50 −kΩ and 4.50 −kΩ resistors connected across a 50.0-V battery having negligibl

Vlada [557]

Answer:

Part A: 16.1 V

Part B: 20.5 V

Part C: 21.5%

Explanation:

The voltmeter is in parallel with the 4.5-kΩ resistor and the combination is in series with the 6.5-kΩ resistor. The equivalent resistance of the parallel combination is given as

$\dfrac{1}{R_E}=\dfrac{1}{4.50}+\dfrac{1}{10.0}$

$R_E=\dfrac{4.50\times10.0}{4.50+10.0} = 3.10$

Part A

The voltmeter reading is the potential difference across the parallel combination. This is found by using the voltage-divider rule.

$V_1 = \dfrac{3.10}{3.10+6.50}\times50.0 = \dfrac{3.10}{9.60}\times50.0 = 16.1 \text{ V}$

Part B

Without the voltmeter, the potential difference across the 4.5-kΩ resistor is found using the same rule as above:

$V_2 = \dfrac{4.50}{4.50+6.50}\times50.0 = \dfrac{4.50}{11.0}\times50.0 = 20.5 \text{ V}$

Part C

The error in % is given by

$\dfrac{20.5-16.1}{20.5}\times100\% = \dfrac{4.4}{20.5}\times100\% = 21.5\%$

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3 years ago

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Does exists the friction force in space?If yes,tell an full example

VashaNatasha [74]

Yes, friction does exist in space. Friction has nothing to do with the earth's atmosphere. It exists everywhere in the universe. <span />

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3 years ago

How to solve this step by step

victus00 [196]

Well, it goes 60 miles in one hour......so set up a ratio.....
60 miles/5 miles = 1 hr/x.....you'll get 60x = 5....then 5/60 would be 0.083

7 0

3 years ago