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3 years ago

6

Which phenomenon would cease to exist in the absence of this axial tilt?

1 answer:

konstantin123 [22]3 years ago

8 0

If the Earth didn't tilt then we wouldn't have seasons.

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Two motorcycles are traveling in opposite directions at the same speed when one of the cyclists blasts her horn, which has frequ

Galina-37 [17]

Answer:

6ms^-1

Explanation:

Given that the frequency difference is

( 563- 544) = 19

So alsoThe wavelength of each wave is = v/f = 344 /544

and there are 19 of this waves

So it is assumed that each motorcycle has moved 0.5 of this distance

in one second thus the speed of the motorcycles will be

=> 19/2 x 344/544 = 6.0 m/s

5 0

2 years ago

Two forces of magnitude 8N and 4N act at right angle

serg [7]

Answer: 26.6

Explanation:

5 0

2 years ago

Using the definition of moment of inertia, calculate icm, the moment of inertia about the center of mass, for this object.

Step2247 [10]

<span> The masses have no inertia about their own CM, and "the object" is the two masses. </span>
<span>1. Icm (at point A) = 2mr^2
hope this helps</span>

3 0

3 years ago

An atom in the ground state has a collision with an electron, then emits a photon with a wavelength of 1240 nm. What conclusion

anyanavicka [17]

Answer:

attached below is the free body diagram of the missing illustration

Initial kinetic energy of the electron = 3 eV

Explanation:

The conclusion that can be drawn about the kinetic energy of the electron is

$E_{e} = E_{3} - E_{1}$

E $_{e}$ = initial kinetic energy of the electron

E $_{1}$ = -4 eV

E $_{3}$ = -1 eV

insert the values into the equation above

$E_{e}$ = -1 -(-4) eV

= -1 + 4 = 3 eV

6 0

3 years ago

A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl

zysi [14]

Answer: $\alpha =9.302\ rad/s^2$

Explanation:

Given

mass of sphere $m=250\ gm$

diameter of sphere $d=4.30\ cm$

radius $r=\frac{4.30}{2}\ cm$

$f=0.0200\ N$

friction will provide resisting torque so

$f\times r=I\times \alpha$

where $I=\text{moment of Inertia}$

$f=\text{friction force}$

$\alpha =\text{angular acceleration}$

$I=\frac{2}{5}mr^2$

$0.02\times r=\frac{2}{5}mr^2\times \alpha$

$\alpha =\frac{5}{2r}\times f$

$\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02$

$\alpha =9.302\ rad/s^2$

(b)time taken to decrease its rotational speed by $21\ rad/s$

$t=\dfrac{\Delta \omega }{\alpha }$

$t=\dfrac{21}{9.302}$

$t=2.25\ s$

6 0

3 years ago