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3 years ago

12

in physics lab, a cube slides down a frictionless incline as shown in the figure below, and elastically strikes another cube at

the bottom that is only one-half its mass. if the incline is 30 cm high and the table is 90cm off the floor, where does each cube land [hint: both leave the incline moving horizontally]

1 answer:

Tema [17]3 years ago

3 0

<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:

</span>

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How are force and motion related

sp2606 [1]

Force is used to put things in motion you can’t have motion without force!
Even if it just pushing a piece of paper you still use force to put the paper in motion!

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3 years ago

. How much work in joules is done by a person who uses a force of 25 N to move a desk 3.0 m?

sergejj [24]

We Know, W = F * s
here, F = 25 N
s = 3 m

Substitute it into the expression,
W = 25 * 3
W = 75 Joule

So, your final answer is 75 J

Hope this helps!

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4 years ago

Which of the following is always true about the particles that make up matter? A. They can only be found in solid substances. B.

Neko [114]

B.They are too small to be seen with only our eyes

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3 years ago

Read 2 more answers

Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton

solmaris [256]

Answer:

a

$\lambda = 3.68 *10^{-36} \ m$

b

$\lambda_p = 1.28*10^{-14} \ m$

Explanation:

From the question we are told that

The mass of the person is $m = 180 \ kg$

The speed of the person is $v = 1 \ m/s$

The energy of the proton is $E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J$

Generally the de Broglie wavelength is mathematically represented as

$\lambda = \frac{h}{m * v }$

Here h is the Planck constant with the value

$h = 6.62607015 * 10^{-34} J \cdot s$

So

$\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }$

=> $\lambda = 3.68 *10^{-36} \ m$

Generally the energy of the proton is mathematically represented as

$E_p = \frac{1}{2} * m_p * v^2_p$

Here $m_p$ is the mass of proton with value $m_p = 1.67 *10^{-27} \ kg$

=> $8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2$

=> $v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }$

=> $v = 3.09529 *10^{7} \ m/s$

So

$\lambda_p = \frac{h}{m_p * v_p }$

so $\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }$

=> $\lambda_p = 1.28*10^{-14} \ m$

5 0

3 years ago

a train engineer drives a train 235 km north along a straight track . then he drives the train back south 126 km . finally , he

yuradex [85]

let here North direction is along +Y and South direction is along -Y

similarly East direction is towards +X and west direction is towards -X

now it is given that

first it moves 235 km north

$d_1 = 235 \hat j$

then it moves back 126 km south

$d_2 = - 126 \hat j$

then again he moves north

$d_3 = 45 \hat j$

now the total displacement is given as

$d = d_1 + d_2 + d_3$

$d = 235 - 126 + 45 = 154 km$

so final displacement of train is 154 km North

5 0

4 years ago