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3 years ago

5

ERCRSOVOS - Table tennis footwork that is useful in this moment to cover a large distance quickly (SCRAMBLED)PE)

1 answer:

nata0808 [166]3 years ago

4 0

Answer:

it's CROSSOVER I knew it because I'm a table tennis player

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Radar, the first link in the cosmic distance chain, is used to establish the baseline distance necessary for the second link, pa

Elena L [17]

Answer:

the Earth-Sun distance.

Explanation:

To measure parallax the base line distance that we need to know is the Earth-Sun distance. This distance is called 1 astronomical unit. This distance is 1.496×10^8 Km. So, this base line distance is necessary for the second link parallax.

7 0

3 years ago

A car is traveling at vx = 30 m/s . The driver applies the brakes and the car decelerates at ax = -5.0 m/s2 . What is the stoppi

____ [38]

Answer:

90 m

Explanation:

We use an ecuation of uniformly accelerated motion, which allows us to find the distance traveled by the car from the moment that driver applies the brakes until it stops completely, that is, when its final speed is zero.

$v_{f}^2=v_{o}^2+2ad$

We isolate the variable d, knowing that the final speed $v_{f}$ is zero

$d=\frac{-v_{o}^2}{2a}\\d=\frac{-(30m/s)^2}{2(-5m/s^2)}=90 m$

6 0

3 years ago

Really confused on this. Any help will be great

pav-90 [236]

C is what i would go with

8 0

3 years ago

M84, M87, and NGC 4258 all have accretion disks around their central black holes for which the rotational velocities have been m

givi [52]

Answer:

<u>For M84:</u>

M = 590.7 * 10³⁶ kg

<u>For M87:</u>

M = 2307.46 * 10³⁶ kg

Explanation:

1 parsec, pc = 3.08 * 10¹⁶ m

The equation of the orbit speed can be used to calculate the doppler velocity:

$v = \sqrt{\frac{GM}{r} }$

making m the subject of the formula in the equation above to calculate the mass of the black hole:

$M = \frac{v^{2} r}{G}$ .............(1)

<u>For M84:</u>

r = 8 pc = 8 * 3.08 * 10¹⁶

r = 24.64 * 10¹⁶ m

v = 400 km/s = 4 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

$M = \frac{( 4*10^{5}) ^{2} *24.64* 10^{16} }{6.674 * 10^{-11} }$

M = 590.7 * 10³⁶ kg

<u>For M87:</u>

r = 20 pc = 20 * 3.08 * 10¹⁶

r = 61.6* 10¹⁶ m

v = 500 km/s = 5 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

$M = \frac{( 5*10^{5}) ^{2} *61.6* 10^{16} }{6.674 * 10^{-11} }$

M = 2307.46 * 10³⁶ kg

The mass of the black hole in the galaxies is measured using the doppler shift.

The assumption made is that the intrinsic velocity dispersion is needed to match the line widths that are observed.

3 0

3 years ago

a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by thet

ANTONII [103]

Answer:

Explanation:

Given

radius of circular path $r=4\ in.$

Position is given by

$\theta =\cos 2t---1$

Differentiate 1 to angular velocity we get

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2$

Differentiate 2 to get angular acceleration

$\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3$

Net acceleration is the vector summation of tangential and centripetal force

$a_t=\alpha \times r$

$a_t=-4\cos 2t\times 4=-16\cos 2t$

$a_r=\omega ^2\cdot r$

$a_r=(-2\sin 2t)^2\cdot 4$

$a_r=16\sin^2(2t)$

$a_{net}=\sqrt{a_r^2+a_t^2}$

$a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}$

$a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}$

6 0

3 years ago