Use each of the following terms in a separate sentence: tissue organ,and function

1 answer:

3 0

"The organ proved to be a vital part of the body's metabolism" "The tissue was damaged from the scalpel but would heal" "The function of the heart is to pump blood"

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A car traveling 90km/hr is 100 m behind a truck traveling 50km/hr. How long will it take the car to reach the truck?

AlexFokin [52]

The faster car behind is catching up/closing the gap/gaining on
the slow truck in front at the rate of (90 - 50) = 40 km/hr.

At that rate, it takes (100 m) / (40,000 m/hr) = 1/400 of an hour
to reach the truck.

(1/400 hour) x (3,600 seconds/hour) = 3600/400 = <em>9 seconds</em>, exactly

4 0

3 years ago

Read 2 more answers

Calculate the average velocity of a motor cycle that travels 72km/hr in<br> 20 seconds

olchik [2.2K]

Velocity =displacement
Change in time
D=72km/hr
Time=20s
But the S.I unit of velocity is m/s so you woul have to change 72km/hr to m/s

Changing 72km to m

1 kilometer=1000meters
Then, 72 kilometers =?

72•1000/1
=72000m

Changing 72hours to seconds

If 1 hour = 3600 seconds
Then 72 hours=?

72•3600/1

=259200 seconds

Velocity =displacement
Change in time

V= 72,000
259,2005
=0.028m/s

3 0

3 years ago

Would the frequency of the angular simple harmonic motion (SHM) of the balance wheel increase or decrease if the dimensions of t

storchak [24]

Answer:

Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.

Explanation:

Considering the complete question attached in figure below.

Time period for balance wheel is:

$T=2\pi\sqrt{\frac{I}{K}}$

$I=mR^{2}$

m = mass of balance wheel

R = radius of balance wheel.

Angular frequency is related to Time period as:

$\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}$

As dimensions of new balance wheel are one-third of their original values

$R_{new}=\frac{R}{3}$

$\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega$