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IceJOKER [234]
2 years ago
3

A sprinter is running a 100m sprint race. The race begins, and stopwatch is started. The sprinter passes the 12m [N] mark at 1.9

2 s and passes the 59m [N] mark at 7.98s, What was the sprinter's average velocity between the two time marks? Show all your work.
Physics
1 answer:
Sergio [31]2 years ago
3 0

Answer:

velocity = 7.7558 m/s

Explanation:

s = u × t

s - distance u - velocity t - time

59 - 12 = (7.98 - 1.92) *u

47 = ( 6.06) u

u = 7.7558 m/s

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Explain how cognitive psychologists combine traditional conditioning models with cognitive processes.
user100 [1]
Behaviorists generally claimed that conditioning occurred without thinking or reasoning ans was simply a result of consequences or reinforcement. Cognitive psychologists demonstrated that thinking and reasoning (cognition) influences the conditioning processes and that many behaviors that are conditioned depend on the type of cognitive reasoning that occurs during conditioning. Therefore, as one is being conditioned to respond to environmental stimuli or is responding to a consequence, they are also pondering and thinking about the process occuring. Cognition is often the reason individuals are not all conditioned in the same manner.
4 0
2 years ago
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two negative charge of -2.0 c and a positive charge of 3.0 c are separated by 80 m what is the force between two charges
kozerog [31]
The Coulomb force is equal to the constant k times the product of charge one and charge two over radius.

F=k((q1q2)/r)
6 0
2 years ago
A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between
Oduvanchick [21]

Answer:

Net electric field, E_{net}=91406.24\ N/C

Explanation:

Given that,

Charge 1, q_1=7.5\ nC=7.5\times 10^{-9}\ C

Charge 2, q_2=-2.9\ nC=-2.9\times 10^{-9}\ C

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

E_1=\dfrac{kq_1}{r^2}

E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}

E_1=65917.96\ N/C

Electric field due to charge 2 is given by :

E_2=\dfrac{kq_2}{r^2}

E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}

E_2=25488.28\ N/C

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

E_{net}=E_1+E_2

E_{net}=65917.96+25488.28

E_{net}=91406.24\ N/C

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0
2 years ago
Stu wanted to calculate the resistance of a light bulb connected to a 4.0-V battery, with a resulting current of 0.5 A. He used
Margarita [4]

Answer:

The answer is not correct.

Explanation:

Stu's answer is not correct, the equation to use is known as the law of ohm. In which the voltage is defined as the product of the current by the resistance, then we will see this equation.

V = I*R\\where:\\I = current [amp]\\R = resistance [ohm]\\V = voltage [volts]\\

In order to find resistance, this term is found multiplying the current on the right side of the equation, therefore the current will be divided on the left side of the equation.

R=\frac{V}{I} \\replacing:\\R=\frac{4}{0.5} \\R=8[ohms]

That is the reason that the result found by Stu is not correct.

6 0
2 years ago
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The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If th
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<h2><u>Answer:</u></h2>

The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If the trials were run on a real track with real gliders, using stopwatches and meter sticks for measurement, the data compared by the following statements:

1. (There would be variables that would be hard to control, leading to less reliable data.)

3. (Meter sticks may lack precision or may be read incorrectly.)

4. (Real glider data may vary since real collisions may involve loss of energy.)

5. (Human error in recording or plotting the data could be a factor.)


6 0
2 years ago
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