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snow_tiger [21]
3 years ago
6

At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is

300 th all How much earker does the thrown rock strike the ground?
Physics
1 answer:
Helen [10]3 years ago
6 0

Answer:

The thrown rock will strike the ground 2.42s earlier than the dropped rock.

Explanation:

<u>Known Data</u>

  • y_{i}=300m
  • y_{f}=0m
  • v_{iD}=0m/s
  • v_{iT}=-29m/s, it is negative as is directed downward

<u>Time of the dropped Rock</u>

We can use y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}, to find the total time of fall, so 0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}, then clearing for t_{D}.

t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s

<u>Time of the Thrown Rock</u>

We can use y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}, to find the total time of fall, so 0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}, then, 0=-4.9t_{T}^{2}-29t_{T}+300, as it is a second-grade polynomial, we find that its positive root is t_{T}=5.4s

Finally, we can find how much earlier does the thrown rock strike the ground, so t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s

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You are a member of a geological team in Central Africa. Your team comes upon a wide river that is flowing east. You must determ
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Answer:

(a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

Explanation:

Given that,

Constant speed = 6.00 m/s

Time = 20.1 sec

Speed = 9.00 m/s

Time = 11.2 sec

We need to write a equation for to travel due north across the river,

Using equation for north

v^2-c^2=\dfrac{w^2}{t^2}

Put the value in the equation

6.00^2-c^2=\dfrac{w^2}{(20.1)^2}

36-c^2=\dfrac{w^2}{404.01}....(I)

We need to write a equation for to travel due south across the river,

Using equation for south

v^2-c^2=\dfrac{w^2}{t^2}

Put the value in the equation

9.00^2-c^2=\dfrac{w^2}{(11.2)^2}

81-c^2=\dfrac{w^2}{125.44}....(II)

(a). We need to calculate the wide of the river

Using equation (I) and (II)

45=\dfrac{w^2}{125.44}-\dfrac{w^2}{404.01}

45=w^2(0.00549)

w^2=\dfrac{45}{0.00549}

w=\sqrt{\dfrac{45}{0.00549}}

w=90.5

We need to calculate the current speed

Using equation (I)

36-c^2=\dfrac{(90.5)^2}{(20.1)^2}

36-c^2=20.27

c^2=20.27-36

c=\sqrt{15.73}

c=3.96\ m/s

(b). We need to calculate the shortest time

Using formula of time

t=\dfrac{d}{v}

t=\dfrac{90.5}{6}

t=15.0\ sec

We need to calculate the distance

Using formula of distance

d=vt

d=3.96\times15.0

d=59.4\ m

Hence, (a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

3 0
3 years ago
The rate (in liters per minute) at which water drains from a tank is recorded at half-minute intervals. Use the average of the l
lana [24]

Answer:

see explanation

Explanation:

You are missing the chart with the rates and time to do this, however, I wll do it with a similar exercise here, and you only need to replace the procedure with your data:

See the attached table.

From the left we have:

r = 1/2 (50 + 48 + 46 + 44 + 42 + 40) = 135 L/min

From the right we have:

r = 1/2 (48 + 46 +44 + 42 + 40 + 38) = 129 L/min.

And this should be the correct answer. Watch your chart and replace if it's neccesary.

3 0
3 years ago
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3 years ago
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8 0
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2. A 55 kg woman has a momentum of 200 kg m/s. What is her velocity?
NeTakaya

Answer:

\boxed {\tt 3.63636364 \ m/s}

Explanation:

Velocity can be found using the following formula:

v=\frac{p}{m}

where p is the momentum and m is the mass.

The woman has a mass of 55 kilograms and a momentum of 200 kilogram meters per second.

p= 200 \ kgm/s\\m=55 \ kg

Substitute the values into the formula.

v=\frac{200 \ kg m/s}{55 \ kg}

Divide. Note that the kilograms, or kg, will cancel each other out.

v=\frac{200 \ m/s}{55}

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3 years ago
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