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charle [14.2K]
3 years ago
15

A student is bicycling north along Main Street to school. Another student is timing the bicycling student in order to determine

the average velocity. The bicycling student is 450 m away when the timer starts, and is 200 m away when the timer stops. If the student traveled the distance in 60 seconds, what is the students average velocity?
Physics
1 answer:
MatroZZZ [7]3 years ago
8 0

The average velocity is -4.17 m/s

Explanation:

The average velocity of a body is given by:

v=\frac{d}{t}

where

d is the displacement of the body

t is the time elapsed

For the student in this problem, we have:

Initial position: x_i = 450 m

Final position: x_f = 200 m

So the displacement is

d=x_f -x_i = 200 - 450 = -250 m

The time elapsed is

t = 60 s

Therefore, the average velocity is

v=\frac{-250}{60}=-4.17 m/s

Where the negative sign means the student is moving towards the origin.

Learn more about average speed and velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

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A MEMS-based accelerometer has a mass of m = 2 grams, an equivalent spring constant of k = 5 N/m, and an equivalent damping coef
pychu [463]

Answer:

The natural frequency = 50 rad/s = 7.96 Hz

Damping ratio = 0.5

Explanation:

The natural frequency is calculated in this manner

w = √(k/m)

k = spring constant = 5 N/m

m = mass = 2 g = 0.002 kg

w = √(5/0.002) = 50 rad/s

w = 2πf

50 = 2πf

f = 50/(2π) = 7.96 Hz

Damping ratio = c/[2√(mk)] = 0.1/(2 × √(5 × 0.002)) = 0.5

5 0
3 years ago
A plane comes in for a landing at a velocity of 80 meters per second west. As it touches down, it decelerates at a constant rate
azamat

Answer:

Answer D : about 1067 meters

Explanation:

There are two steps to this problem:

1) First find the time it takes the plane to stop using the equation for the acceleration:

a=\frac{Vf-Vi}{t}

Where Vf is the final velocity of the plane (in our case: zero )

Vi is the initial velocity of the plane (in our case: 80 m/s)

a is the acceleration (in our case -3 m/s^2 - notice negative value because the velocity is decreasing)

a=\frac{Vf-Vi}{t}\\-3=\frac{0-80}{t}\\t=\frac{-80}{-3} = \frac{80}{3}

with units corresponding to seconds given the quantities involved in the calculation.

2) Second knowing the time it took the plane to stop, now use that time in the equation for the distance traveled under accelerated motion:

Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi= 80 (\frac{80}{3}) +\frac{1}{2} (-3) (\frac{80}{3}) ^{2}=1066.666666...

Where the answer results in units of meters given the quantities used in the calculation.

We round this to 1067 meters

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