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2 years ago

15

Which method would likely use less energy? (1) Pushing the heavy box at a slow, constant pace or (2) Pushing the heavy box in fa

st bursts with breaks in between?

1 answer:

MrMuchimi2 years ago

6 0

1 would use less energy. Please vote my answer brainliest! Thanks.

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Followed by the previous question: presume that the electron performs a uniform circular motion around the hydrogen nucleus. Wha

Ksivusya [100]

Answer:

$A_c=87.73*10^{21}m/s$

Explanation:

From the question we are told that

$r=5\times 10^{-11}$

$T=1.5 \times 10^{-16}$

Generally the equation for velocity is mathematically given as

$Velocity (v)=\frac{2 \pi r}{t}$

$V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}$

$V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}$

Generally the equation for Centripetal acceleration is mathematically given as

$A_c=\frac{V^2}{r}$

$A_c=(\frac{20.944*10^5)}{r5*10^{-11}}$

$A_c=87.73*10^{21}m/s$

8 0

2 years ago

The cycle that is the slowest because there is no gas phase is the _____ cycle.

topjm [15]

It would be B. phosphorus

4 0

3 years ago

Read 2 more answers

Which of the following correctly describe electric field lines?

Xelga [282]

-- Electric field lines DO never cross. <em>(A) </em>

-- Electric field lines that are close together DO indicate a stronger electric field. <em>(B) </em>

-- Electric field lines DO not affect the charge that created them. <em>(C)</em>

-- Electric field lines DON'T begin on north poles and end on south poles. North and South "poles" are the way we talk about magnets, not electric charges.

6 0

2 years ago

Read 2 more answers

Is glass weigh more than wood

inn [45]

It depends how large the object is. But, if it is 2 by 4, most likely the wood sieges more than the glass. Hope this helps.

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2 years ago

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A car is going south on I-69 at 33 m/s (74 mph). The car has good brakes so its maximum braking acceleration is – 8.5 m/s^2 . Tr

dmitriy555 [2]

Answer:

1.69515 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-33^2}{2\times -8.5}\\\Rightarrow s=64.06\ m$

The distance between the traffic and the car after braking is 120-64.06 = 55.94 m

Time = Distance / Speed

$\text{Time}=\frac{55.94}{33}\\\Rightarrow \text{Time}=1.69515\ seconds$

The reaction time cannot be more than 1.69515 seconds

4 0

2 years ago