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kvv77 [185]
3 years ago
15

When the starter motor on a car is engaged, there is a 320 A current in the wires between the battery and the motor. Suppose the

wires are made of copper and have a total length of 1.5 m . For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. What minimum diameter can the wires have if the voltage drop along the wires is to be less than 0.50 V?
Physics
1 answer:
Salsk061 [2.6K]3 years ago
8 0

Answer:

0.5

Explanation:

because it is V uwbsusvegwjosnfvehdbuxbdusndgdghqbwbwbbeuehdbdhdhdhdudhdhdudhdnskowoqllqlqlqnebe

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_____ is when a sound wave bounces off a surface back into the medium from which it came.
denpristay [2]
Correct answer is, Reflection

8 0
3 years ago
Read 2 more answers
A 4.4 kg object is being pushed along a surface, causing it to accelerate at a rate of 1.5 m/s2 . Th e coeffi cient of kinetic f
emmasim [6.3K]
Friction = 0.25 * m * g = 10.8
Push = m * a + Friction = 4.4 * 1.5 + 10.8

8 0
3 years ago
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
What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (such deceleration caused on
Wewaii [24]

The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is  252.52\ m/s^2.

The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.

The given data:

time, t = 1.1 s

initial speed, u = 1000 km/h = \frac{2500}{9}\ m/s

final speed, v = 0 m/s

So we will be using the equation of motion, that is,

v = u + at

\therefore 0=\frac{2500}{9} + a(1.1)

\Rightarrow a=-\frac{2500}{9(1.1)}

\therefore a = - 252.52 \ m/s^2

Hence , the deceleration of the rocket is  252.52\ m/s^2.

To learn more about Attention here:

brainly.com/question/28500124

#SPJ4

6 0
2 years ago
6. DRAW 3 3/8° LINE<br> 7. DRAW A 5* LINE
Fiesta28 [93]

Answer:

Each part so obtained will represent the fraction 1/8 and the number line obtained will be of the form: To mark 3/8; move three parts on the right-side of zero. To mark 5/8; move five parts on the right-side of zero. To mark -1 3/8 i.e. -11/8; move eleven parts on the left-side of zero.

Explanation:

5 0
4 years ago
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