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AlexFokin [52]
3 years ago
11

Which Of The Following Research Methods Are Widely Used By Psychologists?

Physics
1 answer:
goldenfox [79]3 years ago
4 0

Answer: As you said D.

Explanation:

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If I keep F constant in F=ma, what is the relationship between m and a?
Jobisdone [24]

Answer:

If F is a constant, we can take f = 1

f = m×a

ma = 1

therefore we can say that force is hence proportinal to the product of mass and acceleration.

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If I wave my arm does it cause a convection wave
lana [24]

Answer:

no mam

Explanation:

the answer to the question is

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Two pulses are moving along a string. One pulse is
raketka [301]

Answer:

The answer is the 3rd option!

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3 years ago
A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p
anzhelika [568]

Answer:

The final temperature of the air is T_2= 605 K

Explanation:

We can start by doing an energy balance for the closed system

\Delta KE+\Delta PE+ \Delta U = Q - W

where

\Delta KE = the change in kinetic energy.

\Delta PE = the change in potential energy.

\Delta U = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so \Delta KE = 0 and \Delta PE=0

and our energy balance equation is \Delta U = Q - W

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change \Delta U=W+Q using the energy balance equation.

We convert the interval of time to seconds t = 5 \:min = 300\:s

\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t

\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ

We can use the change in specific internal energy \Delta U = m(u_2-u_1) to find the final temperature of the air.

We are given that T_1=300 \:K and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy u_1

u_1=214.07\:\frac{kJ}{kg}

Next, we will calculate the final specific internal energy u_2

\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1

\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}

u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}

With the value u_2=439.07 \:\frac{kJ}{kg} and the ideal gas tables for air we make a regression between the values u = 434.78 \:\frac{kJ}{kg},T=600 \:K and u = 442.42 \:\frac{kJ}{kg}, T=610 \:K and we find that the final temperature T_2 is 605 K.

3 0
3 years ago
Action-reaction forces are always the same ____ but are in opposite ____​
aivan3 [116]

Answer:

Magnitude, direction

Explanation:

Magnitude, direction

5 0
3 years ago
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