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3 years ago

On my bike, I was able to travel 40 miles in one hour. From this information, we can determine the bike's

2 answers:

5 0

On my bike, I was able to travel 40 miles in one hour. From this information, we can determine the bike's
From This Info, We Can Determine The Bikes Speed and Velocity!

Vladimir [108]3 years ago

5 0

We can determine the Bike's SPEED

Hope this helps :))

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What distance does a car travel as its speed changes from 0 to 20 m/s in 17 s at constant acceleration?

love history [14]

Since acceleration is constant

So the distance moved in constant acceleration will be given as

$d = \frac{v_f + v_i}{2} t$

as we know that

$v_f = 20 m/s$

$v_i = 0$

t = 17 s

so distance moved is given by

$d = \frac{0 + 20}{2}(17)$

$d = 170 m$

so distance moved will be 170 m

4 0

3 years ago

How long would it take Jesse with an acceleration of -2.50 m/s^2 to bring his bicycle with an initial velocity of 13.5 m/s to a

zhannawk [14.2K]

Answer:

use first equation of motion v=u+at..v is 0..since we have to take him to rest and u is 13.5

7 0

3 years ago

Read 2 more answers

What is the law of conservation in ENERGY?

Alinara [238K]

Answer:

energy cannot be created nor destroyed

Explanation:

the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.

6 0

3 years ago

A mass m=4kg is attached to both a spring with spring constant k=577N/m and a dash-pot with damping constant c=4N⋅s/m . The mass

Ber [7]

Answer:

$x(t) = (5.034\,m)\cdot e^{-0.5\cdot t}\cdot \cos ((12\,\frac{rad}{s} )\cdot t + 0.116\,rad)$

Explanation:

The position and velocity functions have the following forms:

$x(t) = A \cdot e^{-\frac{c}{2\cdot m}\cdot t }\cdot \cos (\omega'\cdot t + \phi)$

$v(t) = -\omega' \cdot A \cdot e^{-\frac{c}{2\cdot m}\cdot t }\cdot \sin (\omega'\cdot t + \phi)$

Where:

$\omega' = \sqrt{\frac{k}{m} - \frac{c^{2}}{4\cdot m^{2}} }$

First, the angular frequency of oscillation is calculated:

$\omega' = \sqrt{\frac{577\,\frac{N}{m}}{4\,kg}-\frac{(4\,\frac{N\cdot s}{m} )^{2}}{4\cdot (4\,kg)^{2}} }$

$\omega'\approx 12\,\frac{rad}{s}$

Later, it is needed to determine if system is underdamped, critically damped or overdamped. Critic value is given by $2\cdot \sqrt{k\cdot m}$ . The value is: