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Lapatulllka [165]
3 years ago
9

Bacteria move back and forth by using their flagella (structures that look like little tails). Speeds of up to 50 μm/s ⎛ ⎝50×10−

6 m/s⎞ ⎠ have been observed. The total distance traveled by a bacterium is large for its size, while its displacement is small. Why is this?
Physics
1 answer:
Rzqust [24]3 years ago
4 0

Answer:

Explained

Explanation:

Displacement is the change in the position of an object with reference to a starting point. for displacement to occur the position of the object must change.

Here the bacteria although moving but it is moving back and forth, meaning its initial and final positions are the same and hence no displacement. Whereas distance is the total distance traveled no matter in what direction. Hence, The total distance traveled by a bacterium is large for its size, while its displacement is small.

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5 0
3 years ago
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How is humidity related to air pressure?
Vlad [161]
Humid air has higher pressure because of the heaviness of the water
7 0
3 years ago
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At t=0 a ball is projected vertically upward from a roof of a building of height 68.1 meters with a velocity of 42.4 m/s. The ba
Finger [1]

Answer:

7.45 s.

Explanation:

Given:

h = 68.1 m

vi = 0 m/s

vf = 42.4 m/s

g = 9.81 m/s^2

Using,

h = vi*t +1/2*(a*t^2)

68.1 = 1/2 * (9.81*t^2)

t = sqrt((68.1*2)/9.81)

= 3.726 s.

Total time of flight = 2*t

= 2 * 3.726

= 7.45 s.

4 0
3 years ago
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A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =
natali 33 [55]

With acceleration

\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j

and initial velocity

\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i

the velocity at time <em>t</em> (b) is given by

\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j

We can get the position at time <em>t</em> (a) by integrating the velocity:

\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du

The particle starts at the origin, so \mathbf x(0)=\mathbf0.

\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du

\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}

\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j

Get the coordinates at <em>t</em> = 8.00 s by evaluating \mathbf x(t) at this time:

\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j

\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j

so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).

Get the speed at <em>t</em> = 8.00 s by evaluating \mathbf v(t) at the same time:

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j

This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}

5 0
3 years ago
A 1200 kg car is brought from 25 m/s to 10m/s over a time period of 5 seconds. Determine the force experienced by the car
aleksklad [387]
It's simple.
We know force is the rate of change in momentum.
So F=(mv-mu)/t or F=m(v-u)/t
=1200*(25-10)/5=3600N
7 0
3 years ago
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