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2 years ago

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If Bill threw a ball straight up on the Moon (g=1.6 m/s2) with a starting velocity of 22m/s from a cliff and it fell past him an

d reached a velocity of 56m/s as it impacted the Moon’s surface below the cliff, how long was the flight time?

1 answer:

kap26 [50]2 years ago

5 0

Answer:

Write two important of physical state

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Shadow is formed when an_____<br>object comes in the<br>way<br>of light.

Zanzabum

Answer:

shadow is formed when an opaque object comes on path of light shadows are formed because light travels in a straight line and it can't pass through an opaque object

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3 years ago

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Carbon dioxide being released from a fire extinguisher a good example of how volume __________.

kirill [66]

"<span>increases when pressure decreases". Pressure and volume of gasses are related from Boyle's law, which states that Pressure is proportional to 1/V, so as pressure decreases, volume increases. </span>

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3 years ago

There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co

mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

$q_1=q_2=2.06\mu C=2.06\times 10^{-6} C$

$q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C$

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force, $F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2$

$F_1=F_3=F$

$F_{net}=\sqrt{F^2+F^2+0}-F_2$

$F_{net}=\sqrt 2F-F_2$

$F=\frac{kq^2}{d^2}$

$F_2=\frac{Kq^2}{2d^2}$

$F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})$

Where $k=9\times 10^9$

$F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})$

$F_{net}=0.208 N$

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3 years ago

Can anyone tell me the ans of this question also please

Dmitry_Shevchenko [17]

Answer:

Hey there!

Stopwatch X recorded 40 seconds, and stopwatch Y recorded 50 seconds.

Stopwatch Y recorded 10 seconds longer than stopwatch X.

Hope this helps :)

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3 years ago

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A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di

Oksi-84 [34.3K]

<h2>82.353 km/hr</h2>

Explanation:

The driver travels 135 km towards East in 1.5 hr. He stops for 45 min. He again travels 215 km towards East in 2.0 hr.

The total displacement of the driver in the given time is ths sum of individual displacements, because all the displacements are in the same directon.

Total displacement = $135\text{ }km\text{ }East\text{ }+\text{ }0\text{ }km\text{ }+\text{ }215\text{ }km\text{ }East\text{ }=\text{ }350\text{ }km\text{ }East$

Total time travelled = $1.5\text{ }hr\text{ }+\text{ }45\text{ }min\text{ }+\text{ }2.0\text{ }hr\text{ }=\text{ }90\text{ }min\text{ }+\text{ }45\text{ }min\text{ }+\text{ }120\text{ }min\text{ }=\text{ }255\text{ }min\text{ }=\text{ }4\text{ }hr\text{ }15\text{ }min\text{ }=\text{ }4.25\text{ }hr$

$\text{Average velocity = }\dfrac{\text{Total displacement}}{\text{Total time taken}}=\dfrac{350\text{ }km}{4.25\text{ }hr}=82.353\text{ }\frac{km}{hr}$

∴ Driver's average velocity = $82.353\text{ }\frac{km}{hr}$

3 0

3 years ago