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3 years ago

9

Which galaxy is the most stretched out?

2 answers:

azamat3 years ago

5 0

The answer is the oddball spiral galaxy. hope this help

RideAnS [48]3 years ago

4 0

Answer:

this is the anwser

Explanation:

The oddball spiral galaxy, called Messier 66, is one-thirdof the Leo Triplet, a group of three interacting galaxies about 35 millionlight-years from Earth (a light-year is the distance light can cover in ayear).

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A molecule of roughly spherical shape has a mass of 6.10 × 10-25 kg and a diameter of 0.70 nm. The uncertainty in the measured p

Ilia_Sergeevich [38]

Answer:

$\Delta v = 0.123 m/s$

Explanation:

As per the law of uncertainty we know that

$\Delta p \times \Delta x = \frac{h}{4\pi}$

now we know that

$\Delta x = 0.70 nm$

$m = 6.10 \times 10^{-25}$

also we have

$h = 6.626 \times 10^{-34} J.s$

now we will have

$m\Delta v \times \Delta x = \frac{h}{4\pi}$

$(6.10 \times 10^{-25})\Delta v \times (0.70 \times 10^{-9}) = \frac{6.626 \times 10^{-34}}{4\pi}$

$\Delta v= 0.123 m/s$

4 0

3 years ago

Which of the following is true about a family psychologist?

ikadub [295]

Are there answer choices?

3 0

3 years ago

Read 2 more answers

Atoms in a solid are not stationary, but vibrate about their equilibrium positions. Typically, the frequency of vibration is abo

hoa [83]

To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.

Our values are given as

$f = 4.11 *10^{12} Hz$

$A = 1.23 * 10^{-11}m$

The angular velocity of a body can be described as a function of frequency as

$\omega = 2\pi f$

$\omega = 2\pi 4.11 *10^{12}$

$\omega=2.582*10^{13} rad/s$

PART A) The expression for the maximum angular velocity is given by the amplitude so that

$V = A\omega$

$V =( 1.23 * 10^{-11})(2.582*10^{13})$

$V = = 317.586m/s$

PART B) The maximum acceleration on your part would be given by the expression

$a = A \omega^2$

$a =( 1.23 * 10^{-11})(2.582*10^{13})^2$

$a= 8.2*10^{15}m/s^2$

4 0

3 years ago

A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×

HACTEHA [7]

Answer:

$f_r = 150.47 N$

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

$\dfrac{mv^2}{r}= N sin \theta + f_r cos \theta$

∑ F_y = 0

$0 = N cos \theta - f_r sin \theta - mg$

$N = \dfrac{f_rsin \theta + mg}{cos \theta}$

$\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta$

= $f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta$

$f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}$

$f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}$

$f_r = 150.47 N$

8 0

3 years ago

Why aren't descriptive investigations repeatable ?

Ann [662]

Because the information cant be out of the investigation

4 0

3 years ago

Read 2 more answers