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2 years ago

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Which of the following describes the treatment in an investigation?

1 answer:

insens350 [35]2 years ago

5 0

Answer:

Live training (whether through videoconference or a live course on a learning management system) or an eLearning course with knowledge checks is suggested so that learners can receive immediate feedback. Inclusion of a post-test – as well as an electronic guide describing jurisdiction-specific protocols – is strongly recommended.

Explanation:

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Describe how tree rings indicate time. Do the same for ice cores and varves?

tatyana61 [14]

Answer:

counting tree rings can indicate age

Explanation:

hope i helped at least a little bit :)

7 0

2 years ago

At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.5 m/s2. At the same insta

Licemer1 [7]

Answer:

6.96 s

Explanation:

<u>Given:</u>

u = initial speed of the automobile = 0 m/s

a = constant acceleration of the automobile = $2.5\ m/s^2$
v = constant speed of the truck = 8.7 m/s

<u>Assume:</u>

t = time instant at which the automobile overtakes the truck.

At the moment the automobile and the truck both meat each other the distance travel by both vehicles must be the same.

$\therefore \textrm{Distance traveled by the automobile }=\textrm{Distance traveled by the truck}\\\Rightarrow ut+\dfrac{1}{2}at^2=vt\\\Rightarrow (0)t+\dfrac{1}{2}\times 2.5\times t^2=8.7t\\\Rightarrow 1.25t^2=8.7t\\\Rightarrow 1.25t^2-8.7t=0\\\Rightarrow t(1.25t-8.7)=0\\\Rightarrow t = 0\,\,\,or\,\,\, t = \dfrac{8.7}{1.25}\\\Rightarrow t = 0\,\,\,or\,\,\, t = 6.96\\$

Since t = 0 s is the initial condition. So, they both meet again at t = 6.96 s such that the automobile overtakes the truck.

6 0

2 years ago

The machine in the figure is ideal and an effort force of

DENIUS [597]

Answer:

1.5 m

Explanation:

Let the distance from the box to the pivot be c.

Let the distance from the pivot to the effort be y.

From the question given above, the following data were obtained:

Effort force (Fₑ) = 7 N

Force of resistance (Fᵣ) = 14 N

Distance from the box to the pivot (c) = 0.75 m

Distance from the pivot to the effort (y) =?

Clockwise moment = Fₑ × y

Anticlock wise moment = Fᵣ × c

Clockwise moment = Anticlock wise moment

Fₑ × y = Fᵣ × c

7 × y = 14 × 0.75

7 × y = 10.5

Divide both side by 7

y = 10.5 / 7

y = 1.5 m

Therefore, the distance from the pivot to the effort is 1.5 m

5 0

2 years ago

A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, $\alpha=3.3 rad/s^2$

Diameter of the wheel, d=21 cm

Radius of wheel, $r=\frac{d}{2}=\frac{21}{2}$ cm

Radius of wheel, $r=\frac{21\times 10^{-2}}{2} m$

1m=100 cm

Magnitude of total linear acceleration, a= $1.7 m/s^2$

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration, $a_t=\alpha r$

$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

$a_t=34.65\times 10^{-2}m/s^2$

Radial acceleration, $a_r=\frac{v^2}{r}$

We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0

2 years ago

Which of the following Illustrates 2 resistors in a series circult? A, B, C, D.

Kaylis [27]

Answer:

It's either B or D, I'm not positive which it is

Explanation:

6 0

1 year ago