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2 years ago

Which can lead to an inaccurate tympanic temperature reading?

1 answer:

6 0

Significant ear drainage or a scarred tympanic membrane can lead to inaccurate results on a tympanic temperature reading.

<h3>Why inaccurate tympanic temperature reading?</h3>

Significant ear drainage or a scarred tympanic membrane can lead to inaccurate results on a tympanic temperature reading.

Although an ear infection or the presence of an ear infection will not significantly affect a tympanic thermometer reading.

If the client has been sleeping on one side, take the temperature on the other side as heat may be increased on the side against the pillow.

Recent consumption of a cold beverage will not affect tympanic temperature.

To learn more about temperature readings, refer https://brainly.ph/question/20039492

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Space satellites in the inner solar system (such as those in orbit around Earth) are often powered by solar panels. Satellites t

erastova [34]

Answer:

Explanation:

See attached file

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4 years ago

A student that jumps a vertical height of 50 cm during the hang time activity.

muminat

The hang time of the student is 0.64 seconds, and he must leave the ground with a speed of 3.13 m/s

Why?

To solve the problem, we must consider the vertical height reached by the student as max height.

We can use the following equations to solve the problem:

<u>Initial speed calculations:</u>

$v_{f}^{2}=v_{o}^{2}+2*a*d$

At max height, the speed tends to zero.

So, calculating, we have:

<u> $v_{f}^{2}=v_{o}^{2}+2*a*d\\\\0=v_{o}^{2}+2*(-9.81\frac{m}{s^{2}})*0.5m\\\\v_{o}^{2}=9.81\frac{m^{2} }{s^{2}}\\\\v_{o}=\sqrt{9.81\frac{m^{2} }{s^{2}}}=3.13\frac{m}{s}$ </u>

<u>Hang time calculations:</u>

We must remember that the total hang time is equal to the time going up plus the time going down, and both of them are equal,so, calculating the time going down, we have have:

$y-yo=vo.t+\frac{1}{2}*9.81\frac{m}{s^{2}}*t^{2} \\\\0.5m-0=0*t+4.95*t^{2}\\\\t^{2}=\frac{0.5}{4.95}\\\\t=\sqrt{0.101}=0.318s$

Then, for the total hang time, we have:

$TotalHangTime=2*0.318seconds=0.64seconds$

Have a nice day!

3 0

4 years ago

What happens to light when it strikes the inside surface of a smooth, curved mirror?

zubka84 [21]

When light strikes the inside surface of a smooth, curved mirror the light will bounce back toward a single spot. The reason it does this is because mirrors reflect light and other objects.

Hope this helps! Please make my answer the brainliest!

5 0

4 years ago

Read 2 more answers

An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the a

Liono4ka [1.6K]

Answer:

$16.25^{\circ}$

Explanation:

R = Horizontal range of projectile = 75 m

v = Velocity of projectile = 37 m/s

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Horizontal range is given by

$R=\dfrac{v^2\sin2\theta}{g}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{Rg}{v^2}}{2}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{75\times 9.81}{37^2}}{2}\\\Rightarrow \theta=16.25^{\circ}$

The angle at which the arrow is to be released is $16.25^{\circ}$ .

4 0

3 years ago

A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log float

storchak [24]

Answer:

t ’= $\frac{1450}{0.6499 + 2 v_r}$ , v_r = 1 m/s t ’= 547.19 s

Explanation:

This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.

By the time the boat goes up the river

v_b - v_r = d / t

By the time the boat goes down the river

v_b + v_r = d '/ t'

let's subtract the equations

2 v_r = d ’/ t’ - d / t

d ’/ t’ = 2v_r + d / t

$t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }$

In the exercise they tell us

d = 1.22 +1.45 = 2.67 km= 2.67 10³ m

d ’= 1.45 km= 1.45 1.³ m

at time t = 69.1 min (60 s / 1min) = 4146 s

the speed of river is v_r

t ’= $\frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \ + 2 \ v_r}$

t ’= $\frac{1450}{0.6499 + 2 v_r}$

In order to complete the calculation, we must assume a river speed

v_r = 1 m / s

let's calculate

t ’= $\frac{ 1450}{ 0.6499 + 2 \ 1}$

t ’= 547.19 s

8 0

3 years ago