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3 years ago

9

Two postal delivery workers have different routes. They both travel from the post office, to neighborhoods to deliver mail, and

then back to the post office. Which statement must be true about the two postal delivery workers?

1 answer:

SOVA2 [1]3 years ago

7 0

they travel the same distance

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Consider a 2250-lb automobile clocked by law-enforcement radar at a speed of 85.5 mph (miles/hour). if the position of the car i

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As per law of Heisenberg uncertainty law

product of uncertainty in position and uncertainty in momentum will be constant

$\Delta x . \Delta P = \frac{h}{4\pi}$

$\Delta x . m \Delta v = \frac{h}{4\pi}$

now plug in all data

$(5\times 0.3048). (2250 \times 0.454) \Delta v = \frac{6.6 \times 10^{-34}}{4\pi}$

$\Delta v = 3.37 \times 10^{-38} m/s$

So above is the uncertainty in velocity of the object

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Based on this information, a psychologist would most likely conclude that Jonah's stress led him toward positive personal changes.

<h3>What is stress?</h3>

When a person is afraid or tensed about something, he overthinks about him and become stressed.

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2 years ago

You're at rest in a hammock when a hungry mosquito sees an opportunity for lunch. A mild 2 m/s breeze is blowing. If the mosquit

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Answer:

A) against the breeze at 2 m/s

Explanation:

Given that air is blowing at 2 m/s.An a mosquito eat lunch.If mosquito want to eat lunch he have to move opposite to the direction of air because air is blowing at 2 m/s.So we can say that mosquito have to move against the air.

But in we have to protect our self from mosquito because if any mosquito bite us then it may lead to fever .So the protection is very important from mosquito .

Therefore the answer is --

A

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3 years ago

An 85.0 kg fisherman jumps from a dock at a speed of 4.30 m/s onto their 135.0 kg boat. If the boat was at rest to begin but mov

jeka94

Answer:

Final speed of boat + man is 1.66 m/s

Explanation:

As we know that there is no friction on the system or there is no external force on this system

So here we can use momentum conservation here

$mv = (m + M)v_f$

so we have

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now we have

$85 \times 4.30 = (85 + 135) v$

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3 years ago

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

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Let, the maximum height covered by projectile be $\sf{H_m}$

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$\qquad$ ______________________________

Then –

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$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

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2 years ago

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