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Nookie1986 [14]
3 years ago
13

Cognitive maps allow organisms to

Physics
1 answer:
RoseWind [281]3 years ago
5 0
B. Mentally represent their environment
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Rank the tensions in the ropes, t1, t2, and t3, from smallest to largest, when the boxes are in motion and there is no friction
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<span>AS T1,T2,T3 are the tensions in the ropes,assuming that there are Three blocks of mass 3m, 2m, and m.T3 is the string between 3m and 2m,T2 is the string between 2m and m ,T1 is the string attached to m thus T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass, so T3 < T2 < T1.</span>
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You could use an analytical or triple beam balance to determine a ___ called ____
GaryK [48]

Answer:

a and b are the correct answers

Explanation:

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3 years ago
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According to prevailing gender stereotypes describe the most important characteristics associated with men and woman
VLD [36.1K]

Answer

According to prevailing gender stereotypes most important characteristics associated with men and woman are as follows

1. Domestic behaviors is one of the basic stereotypes. Some people have thinking that women will  cook for family, she will take care of the children, she will do loundary and will clean the house. About man they thought he will take care of finances, will pay the bills and do the home repairs.

2. Some people thought that women are emotional and submissive, while men are thought to be self-confident and aggressive.

3.Some people thought that the most common occupation for women are teaching and nursing while the rest of the occupation are for men like pilots, doctors, and engineers are men.

4. Men are expected to be tall and strong while women are expected to be thin and graceful, while.

5. Dressing of men and women also thought to be different. Usually men expected to wear pants and short hairstyles whele  women are expected to wear goun, frouks  and make-up.

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3 years ago
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if the vessel in the sample problem accelerates fir 1.00 min, what will its speed be after that minute ?
LUCKY_DIMON [66]

Answers:

a) 154.08 m/s=554.68 km/h

b) 108 m/s=388.8 km/h

Explanation:

<u>The complete question is written below: </u>

<u></u>

<em>In 1977 off the coast of Australia, the fastest speed by a vessel on the water was achieved. If this vessel were to undergo an average acceleration of 1.80 m/s^{2}, it would go from rest to its top speed in 85.6 s.  </em>

<em>a) What was the speed of the vessel? </em>

<em> </em>

<em>b) If the vessel in the sample problem accelerates for 1.00 min, what will its speed be after that minute? </em>

<em></em>

<em>Calculate the answers in both meters per second and kilometers per hour</em>

<em></em>

a) The average acceleration a_{av} is expressed as:

a_{av}=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{\Delta t} (1)

Where:

a_{av}=1.80 m/s^{2}

\Delta V is the variation of velocity in a given time \Delta t, which is the difference between the final velocity V and the initial velocity V_{o}=0 (because it starts from rest).

\Delta t=85.6 s

Isolating V from (1):

V=a_{av}\Delta t + V_{o} (2)

V=(1.80 m/s^{2})(85.6 s) + 0 m/s (3)

V=154.08 \frac{m}{s} (4)

If 1 km=1000m and 1 h=3600 s then:

V=154.08 \frac{m}{s}=554.68 \frac{km}{h} (4)

b) Now we need to find the final velocity when \Delta t=1 min=60 s:

<em></em>

V=(1.80 m/s^{2})(60 s) + 0 m/s (5)

V=108 \frac{m}{s}=388.8 \frac{km}{h} (6)

5 0
3 years ago
A projectile is launched from ground level to the top of a cliff which is 195 m away and 155 m high. If the projectile lands on
muminat

Answer:

66.02m/s

Explanation:

the equation describing the distance covered in the horizontal direction is

x=ucos\alpha t-(1/2)gt^{2} but the acceleration in the horizontal path is zero, hence we have

x=ucos\alpha t

Since the horizontal distance covered is 155m at 7.6secs, we have ucos\alpha =\frac{155}{7.6} \\ucos\alpha =20.38.............equation 1

Also from the vertical path, the distance covered is expressed as

y=usin\alpha t-(1/2)gt^{2}

since the horizontal distance covered in 7.6secs is 195m, then we have

y=usin\alpha t-(1/2)gt^{2}\\y=7.6usin\alpha -4.9(7.6)^{2}\\478.02=7.6usin\alpha \\usin\alpha =62.9...........equation 2

Hence if we divide both equation 1 and 2 we arrive at

\frac{usin\alpha }{ucos\alpha } =\frac{62.9}{20.38} \\tan\alpha =3.08\\\alpha =tan^{-1}(3.08)\\\alpha =72.02^{0}\\

Hence if we substitute the angle into the equation 1 we have

ucos72.02=20.38\\u=66.02m/s

Hence the initial velocity is 66.02m/s

3 0
3 years ago
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