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8_murik_8 [283]

3 years ago

6

A man releases a stone ar the top edge of a tower during the last second of its travel the stone falls through a distance of (9/

25)H where H is the towes highr. Find H

2 answers:

Lapatulllka [165]3 years ago

7 0

equations of motion (EoM) use EoM <span><span>v2</span>=<span>u2</span>+2ax</span> to establish velocities at positions shown in blue in drawing from EoM <span>v=u+at</span> for final 1 second of flight time, we can say <span>v=u+g(1)</span> <span><span><span>2gH</span><span>−−−−</span>√</span>=<span><span>2g<span>1625</span>H</span><span>−−−−−−</span>√</span>+g</span><span> then, solve for H [in terms of g]</span>

Musya8 [376]3 years ago

4 0

Equations of motion (EoM) use EoM <span>v2=u2+2ax</span> to establish velocities at positions shown in blue in drawing from EoM v=u+at for final 1 second of flight time, we can say v=u+g(1) <span><span>2gH−−−−√</span>=<span><span>2g1625H</span>−−−−−−√</span>+g</span><span> then, solve for H [in terms of g]

</span>

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A submarine travels 25 km/h north for 3.2 hours. What is its displacement?

Vika [28.1K]

displ = velocity x time

25 x 3.2 = 75+5 km north.

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3 years ago

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Two forces act on an object. One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N direct

xeze [42]

Answer:

8 N North.

Explanation:

Given that,

One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N directed south.

We need to find the magnitude of net force acting on the object.

Let North is positive and South is negative.

Net force,

F = 10 N +(-2 N)

= 8 N

So, the magnitude of net force on the object is 8 N and it is in North direction (as it is positive). Hence, the correct option is (d) "8N north".

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3 years ago

A photon of red light (wavelength = 710 nm) and a ping-pong ball (mass = 2. 40 × 10^-3 kg) have the same momentum. At what speed

Paha777 [63]

The speed of the ball moving is

$v = 3.94 \times 10 {}^{ - 25}m/s$

what is momentum?

The momentum p of a classical object of mass m and velocity v is given by pclassical =mv.

For photons with wavelength λ,this equation does not hold.Instead, the momentum of the Photon is given by p Photon = h/λ

where,h is the planck's constant.

The momentum of the red Photon is

given:

$h = 6.626 \times 10 {}^{ - 34}kgm {}^{2}/s(plancks \: constant)$

$λ = 700 \times 10 {}^{ - 9} m(photons \: wavelength)$

$p \: photons = \frac{6.626 \times 10 {}^{ - 34}kgm {} ^{2}/s }{700 \times 10 {}^{ - 9}m }$

$= 9.47 \times 10 {}^{ - 28}kgm /s$

since,the Photon and the ping-pong ball have the same momentum,we have

$pball = pphotons = \frac{6.626 \times 10 {}^{ - 34}kgm/s }{700 \times 10 {}^{ - 9} }$

$pball = mv,m = 2.40 \times 10 {}^{ - 3}kg$

$v = 3.94 \times 10 {}^{ - 25} m/s$

Therefore, if the red photon and the ping-pong ball have the same momentum, the ping-pong ball must have a speed of approximately

$v = 3.94 \times 10 {}^{ - 25}m/s$

learn more about momentum of photon from here: brainly.com/question/28197406

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3 0

2 years ago

At the county fair, Chris throws a 0.12kg baseball at a 2.4kg wooden milk bottle, hoping to knock it off its stand and win a pri

viva [34]

Answer:

$v_{f2}$ =6.5% $v_{i1}$

Explanation:

Mass of the ball: $m_{1} =0.12kg$ ]

Initial velocity of the ball: $v_{i1}$

final velocity of the ball: $v_{f1}$ which is -30/100 of $v_{i1}$ = $-0.3v_{i1}$

Mass of the bottle: $m_{2} =2.4kg$

Initial velocity of the bottle: $v_{i2}=0m/s$

final velocity of the bottle: $v_{f2}$ is unknown (to find)

<em>by using conservation momentum, which stated that the initial momentum is equal to the final momentum.</em>

<em /> $m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}$ <em />

<em>so since the bottle is at rest firstly, therefore </em> $v_{i2} =0$ <em />

<em /> $m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}$ <em />

<em /> $m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}$ <em> </em><em>equation 1</em>

so now substitute $v_{f1}$ into equation 1

$m_{1} v_{i1} =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}$

<em /> $m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}$ <em />

<em>collect the like terms</em>

$m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}$

$1.3m_{1} v_{i1} =m_{2} v_{f2}$

divide both side by $m_{2}$

$v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }$

Now substitute

$v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2} =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}$

$v_{f2} =$ 6.5% $v_{i1}$

<em />

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3 years ago

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I could really use some help on this question guys! Will give brainliest!

aev [14]

Answer:

I think it’s the third one

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3 years ago

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