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3 years ago

A ski jumper travels down a slope and

Physics

1 answer:

AleksandrR [38]3 years ago

5 0

Answer:

304.86 metres

Explanation:

The x and y cordinates are $dcos\theta$ and $dsin\theta$ respectively

The horizontal distance travelled, $x=v_{ox}t=dcos\theta$

Making t the subject, $t=\frac{dcos\theta}{v_{ox}}$

Since $y=0.5gt^2=dsin\theta$ , we substitute t with the above and obtain

$0.5g(\frac{dcos\theta}{v_{ox}})^2=dsin\theta$

Making d the subject we obtain

$d=\frac{2v_{ox}^2sin\theta}{gcos^2\theta}$

$d=\frac{2*30^2sin48}{9.8cos^248}$

d=304.8584

d=304.86m

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A car accelerates from 0 m/s to 20 m/s in 5 seconds,

____ [38]

Answer:

Explanation:

20-0=20 m/s

20/5=4 m/s2

8 0

3 years ago

awhite billiard ball with mass mw = 1.47 kg is moving directly to the right with a speed of v = 3.01 m/s and collides elasticall

pochemuha

Answer:

speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

initial kinetic energy = final kinetic energy

$KE = 6.66 J$

Explanation:

Since there is no external force on the system of two balls so here total momentum of two balls initially must be equal to the total momentum of two balls after collision

So we will have

momentum conservation along x direction

$m_1v_{1i} + m_2v_{2i} = m_1v_{1x} + m_2v_{2x}$

now plug in all values in it

$1.47 \times 3.01 + 0 = 1.47 v_1cos68 + 1.47 v_2cos22$

so we have

$3.01 = 0.375v_1 + 0.927v_2$

similarly in Y direction we have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1y} + m_2v_{2y}$

now plug in all values in it

$0 + 0 = 1.47 v_1sin68 - 1.47 v_2sin22$

so we have

$0 = 0.927v_1 - 0.375v_2$

$v_2 = 2.47 v_1$

now from 1st equation we have

$3.01 = 0.375 v_1 + 0.927(2.47 v_1)$

$v_1 = 1.13 m/s$

$v_2 = 2.78 m/s$

so speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

Also we know that since this is an elastic collision so here kinetic energy is always conserved to

initial kinetic energy = final kinetic energy

$KE = \frac{1}{2}(1.47)(3.01^2)$

$KE = 6.66 J$

5 0

4 years ago

If temperature and the number of particles remain constant, and increase in pressure will cause volume to

omeli [17]

The correct answer is remain the same because if the particles and temperature remain constant, when the pressure increases the temperature is not increasing so it would just stay the same

5 0

3 years ago

If Tango measures the speed of sound at its landing site as 240 m/s, while Foxtrot measures speed of sound as 258 m/s at it's la

elena-s [515]

Tango landed in a site of lower temperature while Foxtrot landed in a site of higher temperature.

<h3>Effect of temperature on speed of sound</h3>

Speed of sound increases with increase in temperature because air molecules acquires more kinetic energy which increases their rate of vibration.

From the information, the speed of sound picked by Foxtrot (258 m/s) is greater than the speed of sound picked by Tango (240 m/s).

We can conclude that, Tango landed in a site of lower temperature while Foxtrot landed in a site of higher temperature.

Learn more about temperature and speed of sound here: brainly.com/question/14400858

#SPJ1

7 0

2 years ago

It takes 3.0 eV of energy to excite an electron in a 1-dimensional infinite well from the ground state to the first excited stat

valentina_108 [34]

Answer:

0.614 nm

Explanation:

Energy of the nth state of one dimensional infinite wall is,

$E=\frac{n^{2} h^{2} }{8mL^{2} }$

Given the energy to excite an electron from ground state to first excited state is,

$\Delta E=3eV\\\Delta E=3(1.6\times10^{-19})J$

And the Plank's constant is, $h=6.626\times10^{-34}Js$

Mass of electron, $m=9.1\times10^{-31}kg$

Now the energy will of a 1 dimensional infinite wall which excite an electron from ground state to first excited state will be,

$\Delta E=\frac{2^{2} h^{2} }{8mL^{2} } -\frac{1^{2} h^{2} }{8mL^{2} }$

Put all the variables in above equation and rearrange it for L.

$L^{2} =\frac{3((6.626\times10^{-34}Js)^{2} )}{8\times9.1\times10^{-31}kg\times3(1.6\times10^{-19})J} \\L^{2} =0.376922012\times10^{-18} m^{2}\\ L=0.6139\times10^{-9}m\\ L=0.614nm$

Therefore the width of the box is 0.614 nm.

6 0

3 years ago