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3 years ago

What are the requirements for an inertial frame of reference?

1 answer:

3 0

As simply put as I believe is possible, it's the immediate space-time co-oridinate in which the observer is not in motion. All measurements by the observer are in relation to that 'place'

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When do you turn your wheels sharply left before backing slowly in a parallel parking maneuver?

Alex_Xolod [135]

Answer:

What is C - when your front bumper is even with the front vehicles back bumper.

Explanation:

Good Luck

8 0

2 years ago

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

5 0

3 years ago

A location on land has the conditions stated below.

ycow [4]

I believe the answer is D. Thunderstorm
Unstable air indicates that <span>the lowest layers of an </span>air<span> mass is low in temperature. Thunderstorms is more likely to happen if there is unstable air mass because it provide a situation where there is more energy for the Thunderstorms to feed off of.</span>

5 0

3 years ago

Read 2 more answers

Determine the binding energy of an F-19 nucleus. The F-19 nucleus has a mass of 18.99840325 amu. A proton has a mass of 1.00728

Anvisha [2.4K]

Answer:

Energy = 1.38*10^13 J/mol

Explanation:

Total number of proton in F-19 = 9

Total number of neutron in F-19 = 10

Expected Mass of F-19

= 9*1.007 + 10*1.008 = 19.152 u

Actual mass of F-19 = 18.998 u

Energy of one particle of F-19 = 931.5*Δm = 931.5*(19.152-18.998)

= 143.234 MeV

Energy of one mole of F-19 = 143.234*10^6*1.6*10^-19*6.022*10^23

= 1.38*10^13 J/mol

8 0

3 years ago

A dog is walking back home and its motion is modeled by the graph above. What position was the dog at 3 seconds after it started

hammer [34]

6 meters because 6/3 = 10/5

4 0

3 years ago