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frozen [14]
3 years ago
6

Fluid of intelligent​

Physics
1 answer:
GarryVolchara [31]3 years ago
8 0

"Fluid intelligence involves being able to think and reason abstractly and solve problems. This ability is considered independent of learning, experience, and education. Examples of the use of fluid intelligence include solving puzzles and coming up with problem-solving strategies."  

- Verywell Mind

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An incline plane has Ф = 40.0° and μ k = 0.15. Starting from rest, how long will it take a 4.0 kg block to reach a speed of 12 m
lesya [120]
     The block on the action of two forces, the Force of Friction and the Tangential Weight. Using the Newton's Secound Law, we have:

P_{t}-Fat=ma \\ mgsen\O-mgucos\O=ma \\ a=g(sen\O-ucos\O)
 
     Using the Velocity Hourly Equation, we get:

V=V_{o}+a\Delta t \\ \Delta t= \frac{V}{a}
 
     Uniting the equations:

\Delta t=  \frac{V}{g(sen\O-ucos\O)}
 
     Entering the unknowns:

\Delta t= \frac{V}{g(sen\O-ucos\O)} \\ \Delta t= \frac{12}{10(sen40^o-0.15cos^o)}  \\ \Delta t= \frac{12}{10(0,64-0.15x0.77)}  \\ \Delta t= \frac{12}{5.27}  \\ \boxed {\Delta t=2.28s}

Obs: Approximate results

If you notice any mistake in my english, please let me know, because i am not native.

5 0
3 years ago
A 12 volt battery in a motor vehicle is capable of supplying the starter motor with 150 A. It is noticed that the terminal volta
qwelly [4]

Answer:0.0133 \Omega

Explanation:

Given

Voltage=12 V

Current(I)=150 A

V_{terminal}=10 V

r_{internal}=\frac{\Delta V}{I}

r_{internal}=\frac{12-10}{150}

r_{internal}=\frac{2}{150}=0.0133 \Omega

4 0
3 years ago
What is the approximate weight of a 20-kg cannonball on Earth? 2 N 20 N 196 N 1,960 N
Margaret [11]
Since weight is the force an object is exerting on another object, and the formula to calculate force is Force = Mass * Acceleration, the answer to your question is 196 N, since the mass of the cannonball times Earth's gravitational pull equals 196 N.
5 0
4 years ago
Read 2 more answers
The magnitude J of the current density in a certain wire with a circular cross section of radius R = 2.11 mm is given by J = (3.
oksano4ka [1.4K]

Answer:

i = 2.84 \times 10^{-3} A

Explanation:

As we know that current density is ratio of current and area of the crossection

now we have

J = \frac{di}{dA}

so the current through the wire is given as

i = \int J dA

now we have

i = \int_{0.921R}^R J dA

here we have

J = (3.25 \times 10^8)r^2

now plug in the values in above equation

i = \int_{0.921R}^R (3.25 \times 10^8)r^2 2\pi r dr

now we have

i = \int_{0.921R}^R 2\pi (3.25 \times 10^8)r^3 dr

i = (2.04 \times 10^9) \frac{r^4}{4}

now plug in both limits as mentioned

i = (2.04 \times 10^9)(\frac{R^4}{4} - \frac{(0.921R)^4}{4})

i = (2.04\times 10^9)(0.07 R^4)

here R = 2.11 mm

i = (2.04 \times 10^9)(0.07 (2.11 \times 10^{-3})^4)

i = 2.84 \times 10^{-3} A

8 0
3 years ago
A boat starts moving across a river at velocity v perpendicular to the river bank.
Rina8888 [55]
D because you are dump to ask that questions that is so East
5 0
3 years ago
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