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2 years ago

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Of the two forms of friction you studied,_____ friction is the one that must be overcome in order to start something moving.

1 answer:

bekas [8.4K]2 years ago

5 0

Answer:

static friction will be your answer

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2) The horizontal and vertical components of the initial velocity of a football are 16 m/s and 20 m/s respectively. How long doe

ValentinkaMS [17]

Answer: 2.04 s

Explanation:

Let the initial velocity be v, Angle of projectile be

Then the horizontal component = v cos θ = 16 m/s

Vertical component of velocity = v sin θ = 20 m/s

Time taken to reach the highest point is half the time taken for total flight.

Time for total flight,

$t = \frac{2vsin \theta}{g}$

$t'=\frac{vsin \theta}{g} = \frac {20 m/s}{9.8 m/s^2} = 2.04 s$

Thus, the football takes 2.04 s to rise to the highest point of its trajectory.

5 0

3 years ago

Which could describe the motion of an object?

Volgvan

mhm mhmhmhmhmmhmhmmh

3 0

2 years ago

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What kind of lens is there in our eyes? where does it form the image of an object?

Goshia [24]

Answer:

convex lens

Explanation:

An image is form in retina with light rays converging most at cornea and upon entering and existing the lens.Rays from top and bottom of the object are traced and produce an inverted image on the retina

8 0

2 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

<h3>a.</h3>

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

<h3>b.</h3>

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

<h2>b</h2>

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

3 0

3 years ago

A rock is thrown straight up with an initial velocity of 19.6 m/s. What time interval elapses between the rock’s being thrown an

inna [77]

Answer:

It will take 4 sec rock to comes its original point

Explanation:

It is given that the rock comes to its original point

So displacement S = 0 m

Initial velocity u = 19.6 m/sec

Acceleration due to gravity $g=9.8m/sec^2$

According to second equation of motion $h=ut+\frac{1}{2}gt^2$

$0=19.6\times t+\frac{1}{2}\times 9.8t^2$

$19.6=4.9t$

t = 4 sec

3 0

3 years ago

Read 2 more answers