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enot [183]
3 years ago
15

The psychologist who worked with Peter and helped him associate rabbits with positive things was:

Physics
2 answers:
vovangra [49]3 years ago
4 0

1. Mary Cover-Jones 

2.Knowing the formula for water and table salt is an example of episodic memory.    so d

3.Hippocampus


IRINA_888 [86]3 years ago
3 0

Answer:

1) Mary Cover-Jones

2) episodic

3) hippocampus

Hope this helped man!

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3. Compare the slope of the velocity-time graph to the average of all your acceleration values. Are they close? What does the sl
Marina86 [1]

The slope of a speed-time graph is the acceleration represented by the graph.

All other parts of this question refer to a lab experiment or exercise
where I was not present, but Zeesam16 was.  Therefore I have no data
with which to answer the rest of the question, and hope that Zeesam can
handle it.


6 0
3 years ago
Explain why is not advisable to use small values of I in performing an experiment on refraction through a glass prism?
MakcuM [25]
The angle of refraction would be further less 
3 0
3 years ago
A vehicle reaches a speed of 7.5 m/s over 15 seconds. What is its acceleration if it
denpristay [2]

acceleration = \frac{velocity}{time}  = \frac{7.5}{15}  = 0.5ms^-^2

So the answer is option b.

8 0
3 years ago
A 0.4000 kg sample of methanol at 16.0ºC is mixed with 0.4000 kg of water at 85.0ºC. Assuming no heat loss to the surroundings,
AVprozaik [17]

Answer:

T_finalmix = 59.5 [°C].

Explanation:

In order to solve this problem, a thermal balance must be performed, where the heat is transferred from water to methanol, at the end the temperature of the water and methanol must be equal once the thermal balance is achieved.

Q_{water}=Q_{methanol}

where:

Q_{water}=m_{water}*Cp_{water}*(T_{waterinitial}-T_{final})

mwater = mass of the water = 0.4 [kg]

Cp_water = specific heat of the water = 4180 [J/kg*°C]

T_waterinitial = initial temperature of the water = 85 [°C]

T_finalmix = final temperature of the mix [°C]

Q_{methanol}=m_{methanol}*Cp_{methanol}*(T_{final}-T_{initialmethanol})

Now replacing:

0.4*4180*(85-T_{final})=0.4*2450*(T_{final}-16)\\142120-1672*T_{final}=980*T_{final}-15680\\157800=2652*T_{final}\\T_{final}=59.5[C]

4 0
3 years ago
A boat race runs along a triangular course marked by buoys A, B, and C. The race starts with the boats headed west for 3700 mete
ale4655 [162]

Answer:

The  last two bearings are

49.50° and 104.02°

Explanation:

Applying the Law of cosine (refer to the figure attached):

we have

x² = y² + z² - 2yz × cosX

here,

x, y and z represents the lengths of sides opposite to the angels X,Y and Z.

Thus we have,

cos X=\frac{x^2-y^2-z^2}{-2yz}

or

cos X=\frac{y^2 + z^2-x^2}{2yz}

substituting the values in the equation we get,

cos X=\frac{2900^2 + 3700^2-1700^2}{2\times 2900\times 3700}

or

cos X=0.8951

or

X = 26.47°

similarly,

cos Y=\frac{1700^2 + 3700^2-2900^2}{2\times 1700\times 3700}

or

cos Y=0.649

or

Y = 49.50°

Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°

The bearing of 2 last legs of race are angels Y and Z.

7 0
3 years ago
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