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The acceleration of automobiles is often given in terms of the time it takes to go from 0 mi/h to 60 mi/h. One of the fastest st

Rudiy27

Answer:

9.934 m/s²

Explanation:

Given:

Initial speed of the Bugatti Veyron Super Sport = 0 mi/h

Final speed of the Bugatti Veyron Super Sport = 60 mi/h

Now,

1 mi/h = 0.44704 m / s

thus,

60 mi/h = 0.44704 × 60 = 26.8224 m/s

Time = 2.70 m/s

Now,

The acceleration (a) is given as:

$a=\frac{\textup{Change in speed}}{\textup{Time}}$

thus,

$a=\frac{26.8224 - 0 }{2.70}$

or

a = 9.934 m/s²

6 0

2 years ago

A driver in a 2000 kg Porsche wishes to pass to pass a slow-moving school bus on a four-lane road. What is the average power in

Kryger [21]

The average power is $3.0\cdot 10^6 W$

Explanation:

First of all, we calculate the work done to accelerate the car; according to the work-energy theorem, the work done is equal to the change in kinetic energy of the car:

$W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where :

$K_f = \frac{1}{2}mv^2$ is the final kinetic energy of the car, with

m = 2000 kg is the mass of the car

v = 60 m/s is the final speed of the car

$K_i = \frac{1}{2}mu^2$ is the initial kinetic energy of the car, with

u = 30 m/s is initial speed of the car

Soolving:

$W=\frac{1}{2}(2000)(60)^2 - \frac{1}{2}(2000)(30)^2=2.7\cdot 10^6 J$

Now we can find the power required for the acceleration, which is given by

$P=\frac{W}{t}$

where

t = 9 s is the time elapsed

Solving:

$P=\frac{2.7\cdot 10^6}{9}=3.0\cdot 10^6 W$

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0

3 years ago

In the diagram, q1 = +6.60*10^-9 C and q2 = +3.10*10^-9 C. Find the magnitude of the total electric field at point P. pls help?

kirza4 [7]

Answer:

|E(t)| = 1258,46 [N/C]

α = 67,5⁰ (angle with respect x-axis)

Explanation:

E(t) Electric Field is a vector, so we need to determine module and direction

E(t) = E(q₁) + E (q₂) Where E(q₁) and E (q₂) are electric fields due to electric charge q₁ and q₂ respectively.

E(q₁) = K * q₁/ (d₁)² K = 9 *10⁹ [N*m²/C²] d₁ = 0,350 m

E(q₁) = 9 *10⁹ * 6,6*10⁻⁹ / 0,1225 [N*m²/C²] *C/m²

E(q₁) = 484,9 [N/C]

E(q₂) = 9 *10⁹ * 3,1*10⁻⁹ / 0,024025

E(q₂) = 1161,29

Then

|E(t)| = √ |Eq₁|² + |Eq₂|²

|E(t)| = √ ( 484,9)² +( 1161,29)²

|E(t)| = √ 235128 + 1348594,46

|E(t)| = 1258,46 [N/C]

And tanα = 1161,29/484,9 tanα = 2,395 α = 67,5⁰

The angle of the vector electric field with the x-axis

3 0

3 years ago

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Which is the main difference between cardio and strength training activities

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