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Doss [256]
4 years ago
10

BMW has introduced a program of covered scheduled maintenance for its cars. That is, when you bring in your vehicle for maintena

nce at a BMW dealership, BMW will cover the costs of everything (except for worn-out tires). This practice specifically targets which stage of the product life cycle?
Physics
1 answer:
Kipish [7]4 years ago
6 0

Answer:

Usage

Explanation:

A product life cycle refers to the duration from the introduction of the product into the market until it's taken off the shelves.

A program of covered scheduled maintenance for the cars refers to the practice in which the dealership will cover the costs of everything when you bring in your vehicle for maintenance.

In this question, the mentioned practice specifically targets the stage - Usage of the product life cycle

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It does not hit the students face because the speed of the balloon slows down as energy is lost through thermal.

Explanation:

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andrezito [222]

Answer:

the answer is false :)

Explanation:

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3 years ago
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To measure the acceleration due to gravity on a distant planet, an astronaut hangs a 0.070-kg ball from the end of a wire. The w
mote1985 [20]

Answer:

g = 1.64m/s²

Explanation:

1.5m in 0.078s

V = 15 / 0.078

= 19.23m/s

Tension = mg

μ = 3.10 × 10⁻⁴

T = V²μ

mg =  V²μ

g =  V²μ / m

g = ((19.23)²(3.10 × 10⁻⁴)) / (0.070)

g = 1.64m/s²

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3 years ago
What is true about valence electrons?
luda_lava [24]
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In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i
RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is 2.7 rad/s^2

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

a_c=\omega^2 r

where:

\omega is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

\omega=1.7 rad/s is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity (g=9.8 m/s^2), so:

a_c=3.2 g = 3.2(9.8)=31.4 m/s^2

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m

b)

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

a=4.4 g

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration (a_c) and the tangential acceleration (a_t):

a=\sqrt{a_c^2+a_t^2}

We know that:

a = 4.4g

a_c = 3.2 g

So, we can find the tangential acceleration:

a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2

The angular acceleration is related to the tangential acceleration by

\alpha = \frac{a_t}{r}

where r = 10.9 m is the length of the centrifuge. Substituting,

\alpha = \frac{29.6}{10.9}=2.7 rad/s^2

Learn more about centripetal and angular acceleration here:

brainly.com/question/2562955

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

8 0
3 years ago
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