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Mrrafil [7]
2 years ago
13

Ions, giving canoe 1 a speed of 0.58 m>s and canoe 2 a speed of 0.42 m>s. If the

Physics
1 answer:
WINSTONCH [101]2 years ago
6 0

Answer:

Draw a table showing the different forms of the first law for the four processes described.

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Every time there is a full moon, Mrs. Cook insists that students in her classes display strange behavior. What would be the best
lidiya [134]
Here is the answer to the question. Every time when there is a full moon, Mrs. Cook insists that students in her classes display strange behavior and the best way for her to prove the validity of her theory is to r<span>esearch data to find concrete evidence of a link between the moon cycle and human behavior. Hope this helps.</span>
4 0
3 years ago
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PLZ HELP ON #22-26!!!! <br><br>Please explain why and how you got your answer.
AleksAgata [21]
22. a - (vf^2 - vi^2)/(2d) 
a = (0 - 23^2)/(170) 
a = -3.1 m/s^2

23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3 
33 = 3t 
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long. 

24. The sprinter starts from rest. The average acceleration is found from: 
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s

25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m 

26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s 
acceleration, a = -7.00m/s^2
displacement, s - 92m 
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0^2 - u^2 + 2 (-7.00) (92) 
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This is the speed pf the car just bore braking. 

I hope this helps!! 

5 0
3 years ago
A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a
Rasek [7]

Answer:

W=173.48J

Explanation:

information we know:

Total force: F=45N

Weight: w=100N

distance: 4m

vertical component of the force: F_{y}=12N

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: F_{x}=Fcos\theta

vertical component: F_{y}=Fsen\theta

but from the given information we know that F_{y}=12N

so, equation these two F_{y}=Fsen\theta and F_{y}=12N

Fsen\theta =12N

and we know the force F=45N, thus:

45sen\theta=12

now we clear for \theta

sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

F_{x}=Fcos\theta

F_{x}=45cos(15.466)\\F_{x}=43.37N

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J

the work done is W=173.48J

7 0
3 years ago
What is the formula for lithium and chlorine?
siniylev [52]
LiCl, because lithium (Li) has one positively charged ion (1+), and
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cancel each other out.
7 0
3 years ago
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A railroad track and a road cross at right angles. An observer stands on the road and watches an eastbound train traveling at 60
mamaluj [8]

Answer:

After 4 s of passing through the intersection, the train travels with 57.6 m/s

Solution:

As per the question:

Suppose the distance to the south of the crossing watching the east bound train be x = 70 m

Also, the east bound travels as a function of time and can be given as:

y(t) = 60t

Now,

To calculate the speed, z(t) of the train as it passes through the intersection:

Since, the road cross at right angles, thus by Pythagoras theorem:

z(t) = \sqrt{x^{2} + y(t)^{2}}

z(t) = \sqrt{70^{2} + 60t^{2}}

Now, differentiate the above eqn w.r.t 't':

\frac{dz(t)}{dt} = \frac{1}{2}.\frac{1}{sqrt{3600t^{2} + 4900}}\times 2t\times 3600

\frac{dz(t)}{dt} = \frac{1}{sqrt{3600t^{2} + 4900}}\times 3600t

For t = 4 s:

\frac{dz(4)}{dt} = \frac{1}{sqrt{3600\times 4^{2} + 4900}}\times 3600\times 4 = 57.6\ m/s

4 0
3 years ago
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