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3 years ago

Ejdjsjdjskskdkdkdkdkdkdkdkdkdkdkdkdjdjdjdjdjdjdjdjd

Physics

2 answers:

Inga [223]3 years ago

6 0

Answer:Ejdjsjdjskskdkdkdkdkdkdkdkdkdkdkdkdjdjdjdjdjdjdjdjd!

Explanation:

riadik2000 [5.3K]3 years ago

5 0

DOEOUEPUROYEOYEKYDKDLYDYLDULDULDULDLUDULDLYDLUDLUDLUDLUDLUDULSOYDYODYODOYXYL I LUV MY DADDY MY SUPER HIRRO

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Darwin is sitting on top of a 10 m tree and decided to drop a 3 kg baseball to the ground. What is the velocity when the PE turn

Nimfa-mama [501]

The AMOUNT of energy the ball has doesn't change. It's 294 joules in Darwin's hand, and it's still 294 joules when the ball hits the ground. It's all PE before he let's it go, and it steadily changes from PE to KE all the way down.

It BEGINS to turn into KE immediately, when Darwin lets go of the ball, and it starts to fall.

More and more PE turns into KE as the ball falls, all the way down.

When the ball hits the ground, it has no more PE left. All of its mechanical energy is then KE.

8 0

4 years ago

A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting th

-BARSIC- [3]

Answer:

Average acceleration on first part of the chunk is given as

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = -13.125 m/s^2$

Explanation:

By momentum conservation along x direction we will have

$mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2$

so we have

$v_1 + v_2 = 2v$

$v_1 + v_2 = 4.68$

also by energy conservation

$\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J$

$\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17$

$(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)$

$(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83$

$21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83$

$2v_2^2 - 9.36v_2 + 2.12 = 0$

by solving above equation we will have

$v_1 = 4.44 m/s$

$v_2 = 0.24 m/s$

Average acceleration on first part of the chunk is given as

$a_1 = \frac{4.44 - 2.34}{0.16}$

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = \frac{0.24 - 2.34}{0.16}$

$a_2 = -13.125 m/s^2$

5 0

4 years ago

It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, w

Charra [1.4K]

Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

F = Eq

And we also know that :

F = Bqv

F = $\frac{mv^{2} }{r}$

Bqv = $\frac{mv^{2} }{r}$

or Eq = $\frac{mv^{2} }{r}$

Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

6 0

3 years ago

Which of the following wavelengths will produce standing waves on a string that is 3.5 m long?

denpristay [2]

In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:

$\lambda=\frac{2}{n} L$