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2 years ago

Find the instantaneous acceleration at t=ls for an object moving along a straight axis with velocity function:

1 answer:

8 0

The answer is A.
Explanation:
We know that the average acceleration a for an interval of time Δt is expressed as:

a = Δv
Δt
where Δv is the change in velocity that occurs during Δt.
e formula for the instantaneous acceleration a is almost the same, except that we need to indicate that we're interested in knowing what the ratio of Δv to Δt approaches as Δt approaches zero.

We can indicate that by using the limit notation.

So, the formula for the instantaneous acceleration is:

a = lim Δv
Δt→0 Δt

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Jeremiah is conducting an investigation about the water cycle. He is given the following

Vlada [557]

The water cycle outlines the continuous water movement in liquid, solid and gaseous state between locations on the Earth's surface.

The glass jar represents the lake while the atmosphere is represented by the space above the water, and the sky is represented by the (clear) plastic wrap

Arrangement description and Processes;

The processes of the water cycle includes;

Evaporation;

Condensation

Precipitation

Sublimation

Runoff

Infiltration

The arrangement of the materials is as follows;

Place the glass jar (the lake) containing water and the lamp (the Sun) side by side, such that the lamp light shines on the water surface

Cover the glass jar by wrapping the plastic wrap (the sky) around it to prevent the escape of water vapor when the water is hot.

Switch on the lamp so that it heats the water by radiation heat transfer

Observed processes;

The processes demonstrated by the above experiment includes;

1) Evaporation: As the water in the glass jar becomes warmer, the level of the water in the jar can be observed to decrease slightly due to evaporation

2) Condensation: Fog formation, Clouds

When hotter, the water surface as seen through the clear plastic wrap becomes less clearer due to evaporation, and condensation of the vapor while floating above the water surface, similar to the clouds seen in the sky.

3. Precipitation: Rain;

The clear plastic wrap covering the top of the glass jar, prevents the movement of the vapor further away, such that the tiny condensed vapor gather together, to form big droplets under the plastic wrap that falls back into the jar, which is similar to the process of rainfall

The above processes are repeated as more water evaporates from the jar condenses on the plastic wrap and falls back into the jar, showing the process by which water is recycled from the lake into the atmosphere and back to the lake.

Learn more here:

brainly.com/question/2430469

4 0

2 years ago

Which action would be most likely to reduce the noise level of people talking in a restaurant?

vazorg [7]

B.adding carpet and fabric wall covering to absorb sound

7 0

1 year ago

Read 2 more answers

(a) State Hook's law. [2]

murzikaleks [220]

Hookes law state that provided that the elastic limit is not exceeded, the extension is directly proportional to the force

3 0

3 years ago

A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angl

aliya0001 [1]

Answer:

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i: unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R= 131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

$D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2} }$

$D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2} }$

D₂= 167.21 cm

Direction of the second displacement

$\beta = tan^{-1} \frac{D_{y}}{D_{x} }$

$\beta = tan^{-1} \frac{62.18}{155.22 }$

β= 21.8°

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

6 0

3 years ago

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon

LiRa [457]

Answer:

241.8 N.

Explanation:

The force on branch provides a reaction to the ape's weight force plus the centripetal force needed to keep the gibbon in a circular motion of radius 0.60 m.

Centripetal force = mv^2/r

F = mg + mv²/r

F = m(g + v²/r)

where,

m = mass

= 9 kg

g = acceleration due to gravity

= 9.8 m/s²

v = 3.2 m/s

r = 0.60 m

F = 9 * (9.8 + 3.2²/0.60)

= 241.8 N.

3 0

3 years ago